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	<title>Mathematics in HKCEE, HKAL, and Beyond</title>
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		<title>Mathematics in HKCEE, HKAL, and Beyond</title>
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		<title>A nice coordinate geometry problem (suitable for Form 7 pure)</title>
		<link>http://koopakoo.wordpress.com/2009/10/28/a-nice-coordinate-geometry-problem-suitable-for-form-7-pure/</link>
		<comments>http://koopakoo.wordpress.com/2009/10/28/a-nice-coordinate-geometry-problem-suitable-for-form-7-pure/#comments</comments>
		<pubDate>Wed, 28 Oct 2009 17:35:18 +0000</pubDate>
		<dc:creator>koopakoo</dc:creator>
				<category><![CDATA[HKAL]]></category>
		<category><![CDATA[Pure Mathematics]]></category>
		<category><![CDATA[Teaching]]></category>

		<guid isPermaLink="false">http://koopakoo.wordpress.com/?p=701</guid>
		<description><![CDATA[Let  be a point on the curve   Let  be the tangent line at  Given  intersects the curve  again at  Prove that 
       <img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=koopakoo.wordpress.com&blog=4676682&post=701&subd=koopakoo&ref=&feed=1" />]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p>Let <img src='http://s3.wordpress.com/latex.php?latex=A%28a%2C+a%5E3+%2B+pa+%2B+q%29&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='A(a, a^3 + pa + q)' title='A(a, a^3 + pa + q)' class='latex' /> be a point on the curve <img src='http://s1.wordpress.com/latex.php?latex=C%3A++y+%3D+x%5E3+%2B+px+%2B+q.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='C:  y = x^3 + px + q.' title='C:  y = x^3 + px + q.' class='latex' /> <span id="more-701"></span> Let <img src='http://s2.wordpress.com/latex.php?latex=L&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='L' title='L' class='latex' /> be the tangent line at <img src='http://s3.wordpress.com/latex.php?latex=A.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='A.' title='A.' class='latex' /> Given <img src='http://s1.wordpress.com/latex.php?latex=L&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='L' title='L' class='latex' /> intersects the curve <img src='http://s2.wordpress.com/latex.php?latex=C&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='C' title='C' class='latex' /> again at <img src='http://s3.wordpress.com/latex.php?latex=B%28b%2C+b%5E3+%2B+pb+%2B+q%29.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='B(b, b^3 + pb + q).' title='B(b, b^3 + pb + q).' class='latex' /> Prove that <img src='http://s1.wordpress.com/latex.php?latex=b+%3D+-2a.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='b = -2a.' title='b = -2a.' class='latex' /></p>
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		<slash:comments>1</slash:comments>
	
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			<media:title type="html">Koopa</media:title>
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		<title>For my IMO Students (Basic and Advanced)</title>
		<link>http://koopakoo.wordpress.com/2009/10/28/for-my-imo-students-basic-and-advanced/</link>
		<comments>http://koopakoo.wordpress.com/2009/10/28/for-my-imo-students-basic-and-advanced/#comments</comments>
		<pubDate>Wed, 28 Oct 2009 16:51:07 +0000</pubDate>
		<dc:creator>koopakoo</dc:creator>
				<category><![CDATA[IMO]]></category>
		<category><![CDATA[Pure Mathematics]]></category>
		<category><![CDATA[Teaching]]></category>

		<guid isPermaLink="false">http://koopakoo.wordpress.com/?p=699</guid>
		<description><![CDATA[Here is good problem for my IMO students. 
Given a real number . Find the greatest area of triangle ABC with circum-radius 
       <img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=koopakoo.wordpress.com&blog=4676682&post=699&subd=koopakoo&ref=&feed=1" />]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p>Here is good problem for my IMO students. </p>
<p>Given a real number <img src='http://s1.wordpress.com/latex.php?latex=R+%3E+0&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='R &gt; 0' title='R &gt; 0' class='latex' />. Find the greatest area of triangle ABC with circum-radius <img src='http://s2.wordpress.com/latex.php?latex=R.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='R.' title='R.' class='latex' /></p>
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		<slash:comments>0</slash:comments>
	
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			<media:title type="html">Koopa</media:title>
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		<title>For my Form 7 Pure students: A proof of the Rational Root Theorem</title>
		<link>http://koopakoo.wordpress.com/2009/10/19/for-my-form-7-pure-students-a-proof-of-the-rational-root-theorem/</link>
		<comments>http://koopakoo.wordpress.com/2009/10/19/for-my-form-7-pure-students-a-proof-of-the-rational-root-theorem/#comments</comments>
		<pubDate>Mon, 19 Oct 2009 17:54:32 +0000</pubDate>
		<dc:creator>koopakoo</dc:creator>
				<category><![CDATA[HKAL]]></category>
		<category><![CDATA[Pure Mathematics]]></category>
		<category><![CDATA[Teaching]]></category>

		<guid isPermaLink="false">http://koopakoo.wordpress.com/?p=695</guid>
		<description><![CDATA[Let  Suppose , where  is a rational root. 

Then we have . 
Multiplying both sides by  yields: 
 Since , we must have  divides  
Now, from  multiplying both sides by  and argue similarly yields  divides  Hence the theorem. 
       <img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=koopakoo.wordpress.com&blog=4676682&post=695&subd=koopakoo&ref=&feed=1" />]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p>Let <img src='http://s1.wordpress.com/latex.php?latex=f%28x%29+%3D+a_nx%5En+%2B+a_%7Bn-1%7Dx%5E%7Bn-1%7D+%2B+%5Cdots+%2B+a_1x+%2B+a_0+%5Cin+%5Cmathbb%7BZ%7D%5Bx%5D.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='f(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0 \in \mathbb{Z}[x].' title='f(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0 \in \mathbb{Z}[x].' class='latex' /> Suppose <img src='http://s2.wordpress.com/latex.php?latex=p%2Fq&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='p/q' title='p/q' class='latex' />, where <img src='http://s3.wordpress.com/latex.php?latex=%5Cgcd%28p%2C+q%29+%3D+1&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\gcd(p, q) = 1' title='\gcd(p, q) = 1' class='latex' /> is a rational root. </p>
<p><span id="more-695"></span></p>
<p>Then we have <img src='http://s1.wordpress.com/latex.php?latex=f%28p%2Fq%29+%3D+0&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='f(p/q) = 0' title='f(p/q) = 0' class='latex' />. </p>
<p>Multiplying both sides by <img src='http://s2.wordpress.com/latex.php?latex=q%5E%7Bn-1%7D&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='q^{n-1}' title='q^{n-1}' class='latex' /> yields: </p>
<p><img src='http://s3.wordpress.com/latex.php?latex=q%5E%7Bn-1%7Df%28p%2Fq%29+%3D+%5Cfrac%7Ba_np%5En%7D%7Bq%7D+%2B+%28sum+of+integers%29+%3D+0.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='q^{n-1}f(p/q) = \frac{a_np^n}{q} + (sum of integers) = 0.' title='q^{n-1}f(p/q) = \frac{a_np^n}{q} + (sum of integers) = 0.' class='latex' /> Since <img src='http://s1.wordpress.com/latex.php?latex=%5Cgcd%28p%2C+q%29+%3D+1&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\gcd(p, q) = 1' title='\gcd(p, q) = 1' class='latex' />, we must have <img src='http://s2.wordpress.com/latex.php?latex=q&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='q' title='q' class='latex' /> divides <img src='http://s3.wordpress.com/latex.php?latex=a_n.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='a_n.' title='a_n.' class='latex' /> </p>
<p>Now, from <img src='http://s1.wordpress.com/latex.php?latex=f%28p%2Fq%29+%3D+0%2C&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='f(p/q) = 0,' title='f(p/q) = 0,' class='latex' /> multiplying both sides by <img src='http://s2.wordpress.com/latex.php?latex=p%5E%7Bn-1%7D&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='p^{n-1}' title='p^{n-1}' class='latex' /> and argue similarly yields <img src='http://s3.wordpress.com/latex.php?latex=p&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='p' title='p' class='latex' /> divides <img src='http://s1.wordpress.com/latex.php?latex=a_0.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='a_0.' title='a_0.' class='latex' /> Hence the theorem. </p>
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			<media:title type="html">Koopa</media:title>
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		<title>To my Form 7 students: I am impressed!</title>
		<link>http://koopakoo.wordpress.com/2009/10/15/to-my-form-7-students-i-am-impressed/</link>
		<comments>http://koopakoo.wordpress.com/2009/10/15/to-my-form-7-students-i-am-impressed/#comments</comments>
		<pubDate>Thu, 15 Oct 2009 17:58:18 +0000</pubDate>
		<dc:creator>koopakoo</dc:creator>
				<category><![CDATA[HKAL]]></category>
		<category><![CDATA[Pure Mathematics]]></category>
		<category><![CDATA[Teaching]]></category>

		<guid isPermaLink="false">http://koopakoo.wordpress.com/?p=688</guid>
		<description><![CDATA[Just finished my CB class tonight and I am impressed by two students. They both expressed interests in the following problem: 
Suppose  is a polynomial of degree . Suppose further that 
(i)  for all  and 
(ii)  
Find the leading coefficient of  and the degree of 

I won&#8217;t discuss the solution [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=koopakoo.wordpress.com&blog=4676682&post=688&subd=koopakoo&ref=&feed=1" />]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p>Just finished my CB class tonight and I am impressed by two students. They both expressed interests in the following problem: </p>
<p>Suppose <img src='http://s2.wordpress.com/latex.php?latex=p%28x%29&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='p(x)' title='p(x)' class='latex' /> is a polynomial of degree <img src='http://s3.wordpress.com/latex.php?latex=n&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='n' title='n' class='latex' />. Suppose further that </p>
<p>(i) <img src='http://s1.wordpress.com/latex.php?latex=p%28x%29+-+p%28x+-+1%29+%3D+x%5E%7B100%7D&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='p(x) - p(x - 1) = x^{100}' title='p(x) - p(x - 1) = x^{100}' class='latex' /> for all <img src='http://s2.wordpress.com/latex.php?latex=x%2C&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='x,' title='x,' class='latex' /> and </p>
<p>(ii) <img src='http://s3.wordpress.com/latex.php?latex=p%281%29+%3D+1.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='p(1) = 1.' title='p(1) = 1.' class='latex' /> </p>
<p>Find the leading coefficient of <img src='http://s1.wordpress.com/latex.php?latex=p%28x%29&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='p(x)' title='p(x)' class='latex' /> and the degree of <img src='http://s2.wordpress.com/latex.php?latex=p%28x%29.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='p(x).' title='p(x).' class='latex' /></p>
<p><span id="more-688"></span></p>
<p>I won&#8217;t discuss the solution I presented in class here. Here are two innovative ways suggested by the two students.  </p>
<p>1) Using Taylor series. One of the students based his solution using the Taylor&#8217;s formula. The ingenious thought is that he centered the expansion at <img src='http://s3.wordpress.com/latex.php?latex=x.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='x.' title='x.' class='latex' /> Namely, he writes the expansion as: </p>
<p><img src='http://s1.wordpress.com/latex.php?latex=p%28x+-+1%29+%3D+p%28x%29+%2B+p%27%28x%29%28-1%29+%2B+p%27%27%28x%29%28-1%29%5E2%2F2%21+%2B+%5Cdots+%2B+f%5E%7B%28k%29%7D%28x%29%28-1%29%5Ek%2Fk%21&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='p(x - 1) = p(x) + p&#039;(x)(-1) + p&#039;&#039;(x)(-1)^2/2! + \dots + f^{(k)}(x)(-1)^k/k!' title='p(x - 1) = p(x) + p&#039;(x)(-1) + p&#039;&#039;(x)(-1)^2/2! + \dots + f^{(k)}(x)(-1)^k/k!' class='latex' /></p>
<p>This method yields the correct answer and I am very impressed. </p>
<p>2) Using the Mean Value Theorem. Another student suggested using the Mean-Value theorem, but he could not finish the proof. I gave this idea some thought and eventually made it to work. Here is my solution whose idea is inspired by my student. </p>
<p>From <img src='http://s2.wordpress.com/latex.php?latex=p%28x%29+-+p%28x+-+1%29+%3D+x%5E%7B100%7D.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='p(x) - p(x - 1) = x^{100}.' title='p(x) - p(x - 1) = x^{100}.' class='latex' /> Differentiating both sides <img src='http://s3.wordpress.com/latex.php?latex=100&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='100' title='100' class='latex' /> times gives: </p>
<p><img src='http://s1.wordpress.com/latex.php?latex=p%5E%7B%28100%29%7D%28x%29+-+p%5E%7B%28100%29%7D%28x+-+1%29+%3D+100%21&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='p^{(100)}(x) - p^{(100)}(x - 1) = 100!' title='p^{(100)}(x) - p^{(100)}(x - 1) = 100!' class='latex' /> for all <img src='http://s2.wordpress.com/latex.php?latex=x.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='x.' title='x.' class='latex' /></p>
<p>Now applying the mean value theorem to <img src='http://s3.wordpress.com/latex.php?latex=p%5E%7B%28100%29%7D%28t%29&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='p^{(100)}(t)' title='p^{(100)}(t)' class='latex' /> yields:</p>
<p><img src='http://s1.wordpress.com/latex.php?latex=%5Cfrac%7Bp%5E%7B%28100%29%7D%28x%29+-+p%5E%7B%28100%29%7D%28x+-+1%29%7D%7Bx+-+%28x+-+1%29%7D+%3D+p%5E%7B%28101%29%7D%28c_x%29&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\frac{p^{(100)}(x) - p^{(100)}(x - 1)}{x - (x - 1)} = p^{(101)}(c_x)' title='\frac{p^{(100)}(x) - p^{(100)}(x - 1)}{x - (x - 1)} = p^{(101)}(c_x)' class='latex' />, where <img src='http://s2.wordpress.com/latex.php?latex=c_x+%5Cin+%28x-+1%2C+x%29.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='c_x \in (x- 1, x).' title='c_x \in (x- 1, x).' class='latex' /> </p>
<p>This gives: <img src='http://s3.wordpress.com/latex.php?latex=p%5E%7B%28101%29%7D%28c_x%29+%3D+100%21.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='p^{(101)}(c_x) = 100!.' title='p^{(101)}(c_x) = 100!.' class='latex' /></p>
<p>Now note that <img src='http://s1.wordpress.com/latex.php?latex=p%5E%7B%28101%29%7D%28t%29+-+100%21&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='p^{(101)}(t) - 100!' title='p^{(101)}(t) - 100!' class='latex' /> is a polynomial with infinitely many distinct roots, namely, <img src='http://s2.wordpress.com/latex.php?latex=c_1%2C+c_2%2C+%5Cdots&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='c_1, c_2, \dots' title='c_1, c_2, \dots' class='latex' />. [Note that they are indeed distinct because the intervals <img src='http://s3.wordpress.com/latex.php?latex=%280%2C+1%29%2C+%281%2C+2%29%2C+%5Cdots&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='(0, 1), (1, 2), \dots' title='(0, 1), (1, 2), \dots' class='latex' /> are disjoint. Therefore, <img src='http://s1.wordpress.com/latex.php?latex=p%5E%7B%28101%29%7D%28t%29+-+100%21+%5Cequiv+0&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='p^{(101)}(t) - 100! \equiv 0' title='p^{(101)}(t) - 100! \equiv 0' class='latex' /> by the identity theorem. </p>
<p>Hence, we have <img src='http://s2.wordpress.com/latex.php?latex=p%5E%7B%28101%29%7D%28x%29+%3D+100%21&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='p^{(101)}(x) = 100!' title='p^{(101)}(x) = 100!' class='latex' /> for all <img src='http://s3.wordpress.com/latex.php?latex=x.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='x.' title='x.' class='latex' /> This immediately yields the degree of <img src='http://s1.wordpress.com/latex.php?latex=p%28x%29&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='p(x)' title='p(x)' class='latex' /> is <img src='http://s2.wordpress.com/latex.php?latex=101&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='101' title='101' class='latex' />, and that the leading coefficient can be found by integrating <img src='http://s3.wordpress.com/latex.php?latex=101&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='101' title='101' class='latex' /> times, which yields the answer to be <img src='http://s1.wordpress.com/latex.php?latex=%5Cfrac%7B1%7D%7B101%7D.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\frac{1}{101}.' title='\frac{1}{101}.' class='latex' /></p>
<p>For example, integrating once yields: <img src='http://s2.wordpress.com/latex.php?latex=p%5E%7B%28100%29%7D%28x%29+%3D+100%21x+%2B+c_1&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='p^{(100)}(x) = 100!x + c_1' title='p^{(100)}(x) = 100!x + c_1' class='latex' />, integrating twice yields <img src='http://s3.wordpress.com/latex.php?latex=p%5E%7B%2899%29%7D%28x%29+%3D+100%21x%5E2%2F2%21+%2B+c_1x+%2B+x_2.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='p^{(99)}(x) = 100!x^2/2! + c_1x + x_2.' title='p^{(99)}(x) = 100!x^2/2! + c_1x + x_2.' class='latex' /> </p>
<p>Likewise, integrating 101 times yields: </p>
<p><img src='http://s1.wordpress.com/latex.php?latex=p%28x%29+%3D+100%21x%5E%7B101%7D%2F%28101%29%21+%2B+c_1x%5E%7B100%7D%2F100%21+%2B+c_2x%5E%7B99%7D%2F99%21+%2B+...+%2B+c_%7B101%7D.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='p(x) = 100!x^{101}/(101)! + c_1x^{100}/100! + c_2x^{99}/99! + ... + c_{101}.' title='p(x) = 100!x^{101}/(101)! + c_1x^{100}/100! + c_2x^{99}/99! + ... + c_{101}.' class='latex' /></p>
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		<title>Selected solutions for my F7 Pure (Polynomials)</title>
		<link>http://koopakoo.wordpress.com/2009/10/03/selected-solutions-for-my-f7-pure-polynomials/</link>
		<comments>http://koopakoo.wordpress.com/2009/10/03/selected-solutions-for-my-f7-pure-polynomials/#comments</comments>
		<pubDate>Sat, 03 Oct 2009 04:20:30 +0000</pubDate>
		<dc:creator>koopakoo</dc:creator>
				<category><![CDATA[HKAL]]></category>
		<category><![CDATA[Pure Mathematics]]></category>
		<category><![CDATA[Teaching]]></category>

		<guid isPermaLink="false">http://koopakoo.wordpress.com/?p=680</guid>
		<description><![CDATA[Book 2, pg 14. (00IQ12)(c)
(i) Let y = x^2, then we have ay^2 &#8211; by + a.
Considering the discriminant yields: b^2 &#8211; 4a^2 &#62; 0. Therefore, it has two distinct roots:  If they are both positive, then we are done. Otherwise, they must be both negative since  (Note  since .)
Suppose  Then [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=koopakoo.wordpress.com&blog=4676682&post=680&subd=koopakoo&ref=&feed=1" />]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p>Book 2, pg 14. (00IQ12)(c)<br />
(i) Let y = x^2, then we have ay^2 &#8211; by + a.<br />
Considering the discriminant yields: b^2 &#8211; 4a^2 &gt; 0. Therefore, it has two distinct roots: <img src='http://s3.wordpress.com/latex.php?latex=%5Calpha%2C+%5Cbeta.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\alpha, \beta.' title='\alpha, \beta.' class='latex' /> If they are both positive, then we are done. Otherwise, they must be both negative since <img src='http://s1.wordpress.com/latex.php?latex=%5Calpha+%5Cbeta+%3D+a%2Fa+%3D+1.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\alpha \beta = a/a = 1.' title='\alpha \beta = a/a = 1.' class='latex' /> (Note <img src='http://s2.wordpress.com/latex.php?latex=a+%5Cnot+%3D+0&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='a \not = 0' title='a \not = 0' class='latex' /> since <img src='http://s3.wordpress.com/latex.php?latex=ab+%3E+0&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='ab &gt; 0' title='ab &gt; 0' class='latex' />.)<br />
Suppose <img src='http://s1.wordpress.com/latex.php?latex=%5Calpha%2C+%5Cbeta+%3C+0.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\alpha, \beta &lt; 0.' title='\alpha, \beta &lt; 0.' class='latex' /> Then we have <img src='http://s2.wordpress.com/latex.php?latex=x+%3D+%5Cpm+i+%5Csqrt%7B%7C%5Calpha%7C%7D%2C+%5Cpm+i+%5Csqrt%7B%7C%5Cbeta%7C%7D.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='x = \pm i \sqrt{|\alpha|}, \pm i \sqrt{|\beta|}.' title='x = \pm i \sqrt{|\alpha|}, \pm i \sqrt{|\beta|}.' class='latex' /></p>
<p><span id="more-680"></span></p>
<p>By Viete&#39;s theorem, we have (after simplification) <img src='http://s3.wordpress.com/latex.php?latex=%7C%5Calpha%7C+%2B+%7C%5Cbeta%7C+%3D+-b%2Fa+%3D+-ab%2Fa%5E2++0&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='|\alpha| + |\beta| = -b/a = -ab/a^2  0' title='|\alpha| + |\beta| = -b/a = -ab/a^2  0' class='latex' />), and <img src='http://s1.wordpress.com/latex.php?latex=f%281%29+%3D+a+-+b+%2B+a+%3D+2a+-+b+%5Cnot%3D+0&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='f(1) = a - b + a = 2a - b \not= 0' title='f(1) = a - b + a = 2a - b \not= 0' class='latex' /> since <img src='http://s2.wordpress.com/latex.php?latex=b%5E2+-+4a+%3D+%28b+-+2a%29%28b+%2B+2a%29+%3E+0.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='b^2 - 4a = (b - 2a)(b + 2a) &gt; 0.' title='b^2 - 4a = (b - 2a)(b + 2a) &gt; 0.' class='latex' /></p>
<p>(ii) This part is actually easy, just applying the previous parts. However, one must pay attention to note (and you have to explicitly state this fact) that <img src='http://s3.wordpress.com/latex.php?latex=%5Cbeta_i&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\beta_i' title='\beta_i' class='latex' /> are not equal to <img src='http://s1.wordpress.com/latex.php?latex=0&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='0' title='0' class='latex' /> or <img src='http://s2.wordpress.com/latex.php?latex=1.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='1.' title='1.' class='latex' /> Since dividing by something potentially zero is a fatal error in HKAL.  </p>
<p>Book 4, 02IQ11.<br />
(a) (ii), (Method 1) Here is a better way to present the solution. (to avoid adding the extra line explaining why <img src='http://s3.wordpress.com/latex.php?latex=f%27+%5Cge+0&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='f&#039; \ge 0' title='f&#039; \ge 0' class='latex' /> might not be good enough.)<br />
<img src='http://s1.wordpress.com/latex.php?latex=f%27%28x%29+%3D+3x%5E2+-+p+%3E+0&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='f&#039;(x) = 3x^2 - p &gt; 0' title='f&#039;(x) = 3x^2 - p &gt; 0' class='latex' /> for all <img src='http://s2.wordpress.com/latex.php?latex=x+%5Cnot+%3D+0.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='x \not = 0.' title='x \not = 0.' class='latex' /> This would imply <img src='http://s3.wordpress.com/latex.php?latex=f&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='f' title='f' class='latex' /> is strictly increasing for all <img src='http://s1.wordpress.com/latex.php?latex=x&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='x' title='x' class='latex' /> since f has no local max or local min.  </p>
<p>(method 2): Suppose it has more than one real root, then all roots are real since complex roots come in pairs. Let the roots be <img src='http://s2.wordpress.com/latex.php?latex=a%2C+b%2C+c.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='a, b, c.' title='a, b, c.' class='latex' /> Then by Viete&#8217;s theorem, we have <img src='http://s3.wordpress.com/latex.php?latex=a+%2B+b+%2B+c+%3D+0&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='a + b + c = 0' title='a + b + c = 0' class='latex' />, and <img src='http://s1.wordpress.com/latex.php?latex=ab+%2B+bc+%2B+ca+%3D+-3p.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='ab + bc + ca = -3p.' title='ab + bc + ca = -3p.' class='latex' /> Now, <img src='http://s2.wordpress.com/latex.php?latex=a%5E2+%2B+b%5E2+%2B+c%5E2+%3D+%28a+%2B+b+%2B+c%29%5E2+-+2%28ab+%2B+bc+%2B+ca%29+%3D+6p+%5Cle+0.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca) = 6p \le 0.' title='a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca) = 6p \le 0.' class='latex' /> This means <img src='http://s3.wordpress.com/latex.php?latex=a+%3D+b+%3D+c+%3D+0&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='a = b = c = 0' title='a = b = c = 0' class='latex' />, which is absurd. </p>
<p>The rest are as discussed in class. </p>
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		<title>Selected solutions for my A.maths students (theory of quadratics)</title>
		<link>http://koopakoo.wordpress.com/2009/10/02/selected-solutions-for-my-a-maths-students-theory-of-quadratics/</link>
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		<pubDate>Fri, 02 Oct 2009 16:57:43 +0000</pubDate>
		<dc:creator>koopakoo</dc:creator>
				<category><![CDATA[Teaching]]></category>

		<guid isPermaLink="false">http://koopakoo.wordpress.com/?p=678</guid>
		<description><![CDATA[Notes L1b, Page 14.
Given  are roots of  Suppose , determine the values of 

Solution: We consider two cases.
Case 1,  Then the roots are  By Viete&#8217;s theorem, we have , which means  Hence  which gives 
Case 2,  Then we have  This gives  and therefore,  which gives [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=koopakoo.wordpress.com&blog=4676682&post=678&subd=koopakoo&ref=&feed=1" />]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p>Notes L1b, Page 14.</p>
<p>Given <img src='http://s1.wordpress.com/latex.php?latex=%5Calpha%2C+%5Cbeta&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\alpha, \beta' title='\alpha, \beta' class='latex' /> are roots of <img src='http://s2.wordpress.com/latex.php?latex=x%5E2+%2B+%281+-+x%29%5E2+-+m.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='x^2 + (1 - x)^2 - m.' title='x^2 + (1 - x)^2 - m.' class='latex' /> Suppose <img src='http://s3.wordpress.com/latex.php?latex=%7C%5Calpha%7C+%3D+%7C2%5Cbeta%7C&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='|\alpha| = |2\beta|' title='|\alpha| = |2\beta|' class='latex' />, determine the values of <img src='http://s1.wordpress.com/latex.php?latex=m.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='m.' title='m.' class='latex' /></p>
<p><span id="more-678"></span></p>
<p>Solution: We consider two cases.<br />
Case 1, <img src='http://s2.wordpress.com/latex.php?latex=%5Calpha+%3D+2%5Cbeta.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\alpha = 2\beta.' title='\alpha = 2\beta.' class='latex' /> Then the roots are <img src='http://s3.wordpress.com/latex.php?latex=%5Cbeta%2C+2%5Cbeta.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\beta, 2\beta.' title='\beta, 2\beta.' class='latex' /> By Viete&#8217;s theorem, we have <img src='http://s1.wordpress.com/latex.php?latex=%5Cbeta+%2B+2%5Cbeta+%3D+1&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\beta + 2\beta = 1' title='\beta + 2\beta = 1' class='latex' />, which means <img src='http://s2.wordpress.com/latex.php?latex=%5Cbeta+%3D+1%2F3.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\beta = 1/3.' title='\beta = 1/3.' class='latex' /> Hence <img src='http://s3.wordpress.com/latex.php?latex=%281+-+m%29%2F2+%3D+%281%2F3%29%282%2F3%29+%3D+2%2F9%2C&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='(1 - m)/2 = (1/3)(2/3) = 2/9,' title='(1 - m)/2 = (1/3)(2/3) = 2/9,' class='latex' /> which gives <img src='http://s1.wordpress.com/latex.php?latex=m+%3D+5%2F9.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='m = 5/9.' title='m = 5/9.' class='latex' /></p>
<p>Case 2, <img src='http://s2.wordpress.com/latex.php?latex=%5Calpha+%3D+-2%5Cbeta.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\alpha = -2\beta.' title='\alpha = -2\beta.' class='latex' /> Then we have <img src='http://s3.wordpress.com/latex.php?latex=%5Calpha+%2B+%5Cbeta+%3D+-%5Cbeta+%3D+1.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\alpha + \beta = -\beta = 1.' title='\alpha + \beta = -\beta = 1.' class='latex' /> This gives <img src='http://s1.wordpress.com/latex.php?latex=%5Cbeta+%3D+-1&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\beta = -1' title='\beta = -1' class='latex' /> and therefore, <img src='http://s2.wordpress.com/latex.php?latex=%281+-+m%29%2F2+%3D+-2&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='(1 - m)/2 = -2' title='(1 - m)/2 = -2' class='latex' /> which gives <img src='http://s3.wordpress.com/latex.php?latex=m+%3D+5.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='m = 5.' title='m = 5.' class='latex' /> Done.</p>
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		<title>IMO phase II training Homework Solution</title>
		<link>http://koopakoo.wordpress.com/2009/09/27/imo-phase-ii-training-homework/</link>
		<comments>http://koopakoo.wordpress.com/2009/09/27/imo-phase-ii-training-homework/#comments</comments>
		<pubDate>Sun, 27 Sep 2009 14:51:22 +0000</pubDate>
		<dc:creator>koopakoo</dc:creator>
				<category><![CDATA[IMO]]></category>
		<category><![CDATA[Number Theory]]></category>
		<category><![CDATA[Teaching]]></category>

		<guid isPermaLink="false">http://koopakoo.wordpress.com/?p=669</guid>
		<description><![CDATA[Here is my solution to the following problem given in IMO phase II training for the basic group: 
Problem(1): Find all positive integer solutions to: 

Proof: As I have said in l class, it is clear that . Let , where 
Then we have . 
Rearranging gives 
Since , we have  and 
Therefore,  [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=koopakoo.wordpress.com&blog=4676682&post=669&subd=koopakoo&ref=&feed=1" />]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p>Here is my solution to the following problem given in IMO phase II training for the basic group: </p>
<p>Problem(1): Find all positive integer solutions to: <img src='http://s1.wordpress.com/latex.php?latex=x%5E3+-+y%5E3+%3D+xy+%2B+61.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='x^3 - y^3 = xy + 61.' title='x^3 - y^3 = xy + 61.' class='latex' /></p>
<p><span id="more-669"></span></p>
<p>Proof: As I have said in l class, it is clear that <img src='http://s2.wordpress.com/latex.php?latex=x+%3E+y&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='x &gt; y' title='x &gt; y' class='latex' />. Let <img src='http://s3.wordpress.com/latex.php?latex=x+%3D+y+%2B+d&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='x = y + d' title='x = y + d' class='latex' />, where <img src='http://s1.wordpress.com/latex.php?latex=d+%5Cge+1.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='d \ge 1.' title='d \ge 1.' class='latex' /><br />
Then we have <img src='http://s2.wordpress.com/latex.php?latex=3y%5E2d+%2B+3yd%5E2+%2B+d%5E3+-+y%5E2+-+dy+%3D+61&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='3y^2d + 3yd^2 + d^3 - y^2 - dy = 61' title='3y^2d + 3yd^2 + d^3 - y^2 - dy = 61' class='latex' />. </p>
<p>Rearranging gives <img src='http://s3.wordpress.com/latex.php?latex=+%283d+-+1%29y%5E2+%2B+%283d%5E2+-+d%29y+%2B+d%5E3+%3D+61.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt=' (3d - 1)y^2 + (3d^2 - d)y + d^3 = 61.' title=' (3d - 1)y^2 + (3d^2 - d)y + d^3 = 61.' class='latex' /><br />
Since <img src='http://s1.wordpress.com/latex.php?latex=d+%3E%3D+1&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='d &gt;= 1' title='d &gt;= 1' class='latex' />, we have <img src='http://s2.wordpress.com/latex.php?latex=3d+-+1&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='3d - 1' title='3d - 1' class='latex' /> and <img src='http://s3.wordpress.com/latex.php?latex=3d%5E2+-+d+%5Cge+0.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='3d^2 - d \ge 0.' title='3d^2 - d \ge 0.' class='latex' /><br />
Therefore, <img src='http://s1.wordpress.com/latex.php?latex=61+%3D+%283d+-+1%29y%5E2+%2B+%283d%5E2+-+d%29y+%2B+d%5E3+%5Cge+d%5E3.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='61 = (3d - 1)y^2 + (3d^2 - d)y + d^3 \ge d^3.' title='61 = (3d - 1)y^2 + (3d^2 - d)y + d^3 \ge d^3.' class='latex' /> Hence <img src='http://s2.wordpress.com/latex.php?latex=d+%3D+1%2C+2&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='d = 1, 2' title='d = 1, 2' class='latex' /> or <img src='http://s3.wordpress.com/latex.php?latex=3.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='3.' title='3.' class='latex' /></p>
<p>Finally, solving the three quadratics from <img src='http://s1.wordpress.com/latex.php?latex=d+%3D+1%2C+2&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='d = 1, 2' title='d = 1, 2' class='latex' /> and <img src='http://s2.wordpress.com/latex.php?latex=3&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='3' title='3' class='latex' /> respectively yields <img src='http://s3.wordpress.com/latex.php?latex=%28x%2C+y%29+%3D+%286%2C+5%29&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='(x, y) = (6, 5)' title='(x, y) = (6, 5)' class='latex' /> is the only solution. </p>
<p>Problem(2) Let <img src='http://s1.wordpress.com/latex.php?latex=p+%5Cnot%3D+2&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='p \not= 2' title='p \not= 2' class='latex' /> be a prime. Suppose <img src='http://s2.wordpress.com/latex.php?latex=x%5E2+%5Cequiv+a+%5Cpmod%7Bp%7D&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='x^2 \equiv a \pmod{p}' title='x^2 \equiv a \pmod{p}' class='latex' /> is solvable. Prove that <img src='http://s3.wordpress.com/latex.php?latex=x%5E2+%5Cequiv+a+%5Cpmod%7Bp%5E2%7D&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='x^2 \equiv a \pmod{p^2}' title='x^2 \equiv a \pmod{p^2}' class='latex' /> is also solvable. </p>
<p>Proof: First of all, we may assume <img src='http://s1.wordpress.com/latex.php?latex=a+%5Cnot+%5Cequiv+0+%5Cpmod%7Bp%7D&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='a \not \equiv 0 \pmod{p}' title='a \not \equiv 0 \pmod{p}' class='latex' />, otherwise the problem is trivial. Let <img src='http://s2.wordpress.com/latex.php?latex=y&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='y' title='y' class='latex' /> be a solution to <img src='http://s3.wordpress.com/latex.php?latex=x%5E2+%5Cequiv+a+%5Cpmod%7Bp%7D.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='x^2 \equiv a \pmod{p}.' title='x^2 \equiv a \pmod{p}.' class='latex' /> Write <img src='http://s1.wordpress.com/latex.php?latex=x+%3D+y+%2B+pk&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='x = y + pk' title='x = y + pk' class='latex' />. Then <img src='http://s2.wordpress.com/latex.php?latex=x%5E2+%3D+y%5E2+%2B+2ypk+%2B+p%5E2k%5E2+%5Cequiv+y%5E2+%2B+2ypk+%5Cpmod%7Bp%5E2%7D.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='x^2 = y^2 + 2ypk + p^2k^2 \equiv y^2 + 2ypk \pmod{p^2}.' title='x^2 = y^2 + 2ypk + p^2k^2 \equiv y^2 + 2ypk \pmod{p^2}.' class='latex' /> Now, since <img src='http://s3.wordpress.com/latex.php?latex=y%5E2+%5Cequiv+a+%5Cpmod%7Bp%7D&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='y^2 \equiv a \pmod{p}' title='y^2 \equiv a \pmod{p}' class='latex' />, we may write <img src='http://s1.wordpress.com/latex.php?latex=y%5E2+%3D+a+%2B+pt&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='y^2 = a + pt' title='y^2 = a + pt' class='latex' /> for some <img src='http://s2.wordpress.com/latex.php?latex=t.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='t.' title='t.' class='latex' /> Therefore, <img src='http://s3.wordpress.com/latex.php?latex=x%5E2+%5Cequiv+a+%5Cpmod%7Bp%5E2%7D&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='x^2 \equiv a \pmod{p^2}' title='x^2 \equiv a \pmod{p^2}' class='latex' /> if and only if <img src='http://s1.wordpress.com/latex.php?latex=pt+%2B+2ypk+%5Cequiv+0+%5Cpmod%7Bp%5E2%7D.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='pt + 2ypk \equiv 0 \pmod{p^2}.' title='pt + 2ypk \equiv 0 \pmod{p^2}.' class='latex' /> i.e. <img src='http://s2.wordpress.com/latex.php?latex=t+%2B+2yk+%5Cequiv+0+%5Cpmod%7Bp%7D.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='t + 2yk \equiv 0 \pmod{p}.' title='t + 2yk \equiv 0 \pmod{p}.' class='latex' /></p>
<p>Finally, since <img src='http://s3.wordpress.com/latex.php?latex=%282y%2C+p%29+%3D+1&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='(2y, p) = 1' title='(2y, p) = 1' class='latex' />, the linear congruence equation <img src='http://s1.wordpress.com/latex.php?latex=%282y%29k+%5Cequiv+-t+%5Cpmod%7Bp%7D&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='(2y)k \equiv -t \pmod{p}' title='(2y)k \equiv -t \pmod{p}' class='latex' /> is solvable for all <img src='http://s2.wordpress.com/latex.php?latex=t.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='t.' title='t.' class='latex' /> Therefore, we are done. The number <img src='http://s3.wordpress.com/latex.php?latex=x+%3D+y+%2B+pk&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='x = y + pk' title='x = y + pk' class='latex' /> is the appropriate solution. </p>
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		<title>Challenging problem in Linear Algebra</title>
		<link>http://koopakoo.wordpress.com/2009/08/19/challenging-problem-in-linear-algebra/</link>
		<comments>http://koopakoo.wordpress.com/2009/08/19/challenging-problem-in-linear-algebra/#comments</comments>
		<pubDate>Wed, 19 Aug 2009 11:53:06 +0000</pubDate>
		<dc:creator>koopakoo</dc:creator>
				<category><![CDATA[HKAL]]></category>
		<category><![CDATA[Linear Algebra]]></category>
		<category><![CDATA[Pure Mathematics]]></category>
		<category><![CDATA[Teaching]]></category>

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		<description><![CDATA[Let 
Solve the following matrix equation  where 
       <img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=koopakoo.wordpress.com&blog=4676682&post=663&subd=koopakoo&ref=&feed=1" />]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p>Let <img src='http://s1.wordpress.com/latex.php?latex=%5Cmathcal%7BM%7D+%3D+%5Cleft%5C%7B+%5Cbegin%7Bpmatrix%7D+a+%26+-b+%5C%5Cb+%26+a+%5Cend%7Bpmatrix%7D+%5C%2C+%7C+%5C%2C+a%2C+b%2C+%5Cin+%5Cmathbb%7BR%7D+%5Cright%5C%7D.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\mathcal{M} = \left\{ \begin{pmatrix} a &amp; -b \\b &amp; a \end{pmatrix} \, | \, a, b, \in \mathbb{R} \right\}.' title='\mathcal{M} = \left\{ \begin{pmatrix} a &amp; -b \\b &amp; a \end{pmatrix} \, | \, a, b, \in \mathbb{R} \right\}.' class='latex' /></p>
<p>Solve the following matrix equation <img src='http://s2.wordpress.com/latex.php?latex=X%5E4+%2B+X%5E2+%2B+I+%3D+0&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='X^4 + X^2 + I = 0' title='X^4 + X^2 + I = 0' class='latex' /> where <img src='http://s3.wordpress.com/latex.php?latex=X+%5Cin+%5Cmathcal%7BM%7D.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='X \in \mathcal{M}.' title='X \in \mathcal{M}.' class='latex' /></p>
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		<title>Answers to Linear Algebra Concept Check</title>
		<link>http://koopakoo.wordpress.com/2009/08/19/answers-to-linear-algebra-concept-check/</link>
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		<pubDate>Wed, 19 Aug 2009 11:50:10 +0000</pubDate>
		<dc:creator>koopakoo</dc:creator>
				<category><![CDATA[Teaching]]></category>

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		<description><![CDATA[Let (E) be the following system: 


Let  be the augmented matrix. 
Determine whether the following statements are True or False.
1) (E) can have exactly 3 solutions. 
Ans: False
2) (E) has a unique solution iff 
Ans: True
3) A is invertible iff 
Ans: True
4) The system:  is consistent if  or 
Ans: True
5) If , [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=koopakoo.wordpress.com&blog=4676682&post=661&subd=koopakoo&ref=&feed=1" />]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p>Let (E) be the following system: </p>
<p><img src='http://s2.wordpress.com/latex.php?latex=a_%7B11%7Dx+%2B+a_%7B12%7Dy+%2B+a_%7B13%7Dz+%3D+b_1%5C%5Ca_%7B21%7Dx+%2B+a_%7B22%7Dy+%2B+a_%7B23%7Dz+%3D+b_2+%5C%5Ca_%7B31%7Dx+%2B+a_%7B32%7Dy+%2B+a_%7B33%7Dz+%3D+b_3+&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='a_{11}x + a_{12}y + a_{13}z = b_1\\a_{21}x + a_{22}y + a_{23}z = b_2 \\a_{31}x + a_{32}y + a_{33}z = b_3 ' title='a_{11}x + a_{12}y + a_{13}z = b_1\\a_{21}x + a_{22}y + a_{23}z = b_2 \\a_{31}x + a_{32}y + a_{33}z = b_3 ' class='latex' /></p>
<p><span id="more-661"></span></p>
<p>Let <img src='http://s3.wordpress.com/latex.php?latex=%28A+%5C%3B+%7C+%5C%3B+b%29&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='(A \; | \; b)' title='(A \; | \; b)' class='latex' /> be the augmented matrix. </p>
<p>Determine whether the following statements are True or False.</p>
<p>1) (E) can have exactly 3 solutions. </p>
<p>Ans: False</p>
<p>2) (E) has a unique solution iff <img src='http://s1.wordpress.com/latex.php?latex=%5Cdet%28A%29+%5Cnot%3D+0.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\det(A) \not= 0.' title='\det(A) \not= 0.' class='latex' /></p>
<p>Ans: True</p>
<p>3) A is invertible iff <img src='http://s2.wordpress.com/latex.php?latex=%5Cdet%28A%29+%5Cnot%3D+0.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\det(A) \not= 0.' title='\det(A) \not= 0.' class='latex' /></p>
<p>Ans: True</p>
<p>4) The system: <img src='http://s3.wordpress.com/latex.php?latex=%5Cbegin%7Bpmatrix%7D+3+%26+%5Cast+%26+%7C+%26+%5Cast+%5C%5C+0+%26+a+%26+%7C+%26+b+%5Cend%7Bpmatrix%7D&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\begin{pmatrix} 3 &amp; \ast &amp; | &amp; \ast \\ 0 &amp; a &amp; | &amp; b \end{pmatrix}' title='\begin{pmatrix} 3 &amp; \ast &amp; | &amp; \ast \\ 0 &amp; a &amp; | &amp; b \end{pmatrix}' class='latex' /> is consistent if <img src='http://s1.wordpress.com/latex.php?latex=a+%5Cnot+%3D0&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='a \not =0' title='a \not =0' class='latex' /> or <img src='http://s2.wordpress.com/latex.php?latex=a+%3D+b+%3D+0.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='a = b = 0.' title='a = b = 0.' class='latex' /></p>
<p>Ans: True</p>
<p>5) If <img src='http://s3.wordpress.com/latex.php?latex=%5Cdet%28A%29+%3D+0&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\det(A) = 0' title='\det(A) = 0' class='latex' />, then <img src='http://s1.wordpress.com/latex.php?latex=A+%3D+0.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='A = 0.' title='A = 0.' class='latex' /></p>
<p>Ans: False</p>
<p>6) If <img src='http://s2.wordpress.com/latex.php?latex=A%5E2+%3D+0&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='A^2 = 0' title='A^2 = 0' class='latex' />, then <img src='http://s3.wordpress.com/latex.php?latex=A+%3D+0.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='A = 0.' title='A = 0.' class='latex' /></p>
<p>Ans: False</p>
<p>7) If (E) is consistent and <img src='http://s1.wordpress.com/latex.php?latex=%5Cdet%28A%29+%3D+0&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\det(A) = 0' title='\det(A) = 0' class='latex' />, then (E) has infinitely many solutions. </p>
<p>Ans: True</p>
<p> <img src='http://s.wordpress.com/wp-includes/images/smilies/icon_cool.gif' alt='8)' class='wp-smiley' /> If (E) has infinitely many solutions, then <img src='http://s2.wordpress.com/latex.php?latex=%5Cdet%28A%29+%3D+0.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\det(A) = 0.' title='\det(A) = 0.' class='latex' /></p>
<p>Ans: True</p>
<p>9) If <img src='http://s3.wordpress.com/latex.php?latex=A%5ET+%3D+-A&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='A^T = -A' title='A^T = -A' class='latex' />, then <img src='http://s1.wordpress.com/latex.php?latex=%5Cdet%28A%29+%3D+0.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\det(A) = 0.' title='\det(A) = 0.' class='latex' /></p>
<p>Ans: True (be careful, this is true because A is a 3 by 3 matrix.) This is false if A is a 2 by 2 matrix. </p>
<p>10) If <img src='http://s2.wordpress.com/latex.php?latex=A%5ET+%3D+-A&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='A^T = -A' title='A^T = -A' class='latex' />, then (E) is consistent. </p>
<p>Ans: False. </p>
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		<title>Mental exercises for my Pure Complex Class (1)</title>
		<link>http://koopakoo.wordpress.com/2009/08/19/mental-exercises-for-my-pure-complex-class-1/</link>
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		<pubDate>Wed, 19 Aug 2009 11:35:15 +0000</pubDate>
		<dc:creator>koopakoo</dc:creator>
				<category><![CDATA[HKAL]]></category>
		<category><![CDATA[Pure Mathematics]]></category>
		<category><![CDATA[Teaching]]></category>

		<guid isPermaLink="false">http://koopakoo.wordpress.com/?p=656</guid>
		<description><![CDATA[Dear Students,
Please avoid as much calculations as possible. In ideal situation, you should not do any calculation at all. 
Problem 1
Let  be a complex number.
(a) Find 
(b) Find 
(c) Find 
(d) Find 
(e) Find 
(f) What is 
(g) Suppose ,  write  in terms of 
Problem 2
(a) Write  in the form 
(b) [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=koopakoo.wordpress.com&blog=4676682&post=656&subd=koopakoo&ref=&feed=1" />]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p>Dear Students,</p>
<h1><span style="color:#ff0000;">Please avoid as much calculations as possible. In ideal situation, you should not do any calculation at all. </span></h1>
<p>Problem 1<br />
Let <img src='http://s1.wordpress.com/latex.php?latex=z+%3D+3+%2B+4i&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='z = 3 + 4i' title='z = 3 + 4i' class='latex' /> be a complex number.<br />
(a) Find <img src='http://s2.wordpress.com/latex.php?latex=Re%28z%29.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='Re(z).' title='Re(z).' class='latex' /><br />
(b) Find <img src='http://s3.wordpress.com/latex.php?latex=Im%28z%29.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='Im(z).' title='Im(z).' class='latex' /><br />
(c) Find <img src='http://s1.wordpress.com/latex.php?latex=%5Coverline%7Bz%7D.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\overline{z}.' title='\overline{z}.' class='latex' /><br />
(d) Find <img src='http://s2.wordpress.com/latex.php?latex=%7Cz%7C%5E2.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='|z|^2.' title='|z|^2.' class='latex' /><br />
(e) Find <img src='http://s3.wordpress.com/latex.php?latex=Re%281%2Fz%29.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='Re(1/z).' title='Re(1/z).' class='latex' /><br />
(f) What is <img src='http://s1.wordpress.com/latex.php?latex=z+%5Coverline%7Bz%7D%3F&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='z \overline{z}?' title='z \overline{z}?' class='latex' /><br />
(g) Suppose <img src='http://s2.wordpress.com/latex.php?latex=z+%3D+re%5E%7Bi+%5Ctheta%7D&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='z = re^{i \theta}' title='z = re^{i \theta}' class='latex' />,  write <img src='http://s3.wordpress.com/latex.php?latex=r&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='r' title='r' class='latex' /> in terms of <img src='http://s1.wordpress.com/latex.php?latex=z&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='z' title='z' class='latex' /></p>
<p>Problem 2<br />
(a) Write <img src='http://s2.wordpress.com/latex.php?latex=z+%3D+i&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='z = i' title='z = i' class='latex' /> in the form <img src='http://s3.wordpress.com/latex.php?latex=re%5E%7Bi+%5Ctheta%7D.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='re^{i \theta}.' title='re^{i \theta}.' class='latex' /><br />
(b) Find the two square roots of <img src='http://s1.wordpress.com/latex.php?latex=i&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='i' title='i' class='latex' /> and represent the two square roots on the Gaussian Plane.</p>
<p><span id="more-656"></span></p>
<p>Problem 3<br />
(a) State Euler&#8217;s Formula<br />
(b) State a friend of Euler&#8217;s Formula<br />
(c) Write <img src='http://s2.wordpress.com/latex.php?latex=%5Csin%28n+%5Ctheta%29&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\sin(n \theta)' title='\sin(n \theta)' class='latex' /> and <img src='http://s3.wordpress.com/latex.php?latex=%5Ccos%28n+%5Ctheta%29&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\cos(n \theta)' title='\cos(n \theta)' class='latex' /> in terms of powers of <img src='http://s1.wordpress.com/latex.php?latex=e.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='e.' title='e.' class='latex' /></p>
<p>Problem 4<br />
(a) Write done the n-th roots of unity in terms of <img src='http://s2.wordpress.com/latex.php?latex=%5Czeta.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\zeta.' title='\zeta.' class='latex' /><br />
(b) What is <img src='http://s3.wordpress.com/latex.php?latex=%5Czeta&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\zeta' title='\zeta' class='latex' /> as a power of <img src='http://s1.wordpress.com/latex.php?latex=e&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='e' title='e' class='latex' />?<br />
(c) Write out one nth root of <img src='http://s2.wordpress.com/latex.php?latex=i&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='i' title='i' class='latex' /> as a power of <img src='http://s3.wordpress.com/latex.php?latex=e.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='e.' title='e.' class='latex' /><br />
(d) Write out all the nth roots of <img src='http://s1.wordpress.com/latex.php?latex=i&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='i' title='i' class='latex' /> in terms of your answer in (c) and <img src='http://s2.wordpress.com/latex.php?latex=%5Czeta.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\zeta.' title='\zeta.' class='latex' /></p>
<p>Problem 5<br />
(a) Describe geometrically the set of complex numbers <img src='http://s3.wordpress.com/latex.php?latex=z&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='z' title='z' class='latex' /> satisfying <img src='http://s1.wordpress.com/latex.php?latex=%7Cz+-+i%7C+%3D+2.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='|z - i| = 2.' title='|z - i| = 2.' class='latex' /><br />
(b) Describe geometrically the set of complex numbers <img src='http://s2.wordpress.com/latex.php?latex=z&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='z' title='z' class='latex' /> satisfying <img src='http://s3.wordpress.com/latex.php?latex=%7Cz+-+1%7C+%3D+%7Cz+-+i%7C.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='|z - 1| = |z - i|.' title='|z - 1| = |z - i|.' class='latex' /><br />
(c) Sketch on the Argand diagram, the set of points satisfying both <img src='http://s1.wordpress.com/latex.php?latex=%7Cz+-+1%7C+%3D+%7Cz+-+i%7C&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='|z - 1| = |z - i|' title='|z - 1| = |z - i|' class='latex' /> and <img src='http://s2.wordpress.com/latex.php?latex=Arg%28z%29+%3D+%5Cfrac%7B%5Cpi%7D%7B4%7D.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='Arg(z) = \frac{\pi}{4}.' title='Arg(z) = \frac{\pi}{4}.' class='latex' /><br />
(d) Without doing any calculation, determine the number of points <img src='http://s3.wordpress.com/latex.php?latex=z&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='z' title='z' class='latex' /> satisfying both <img src='http://s1.wordpress.com/latex.php?latex=%7Cz+-+2%7C+%3D+%7Cz+-+3i%7C&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='|z - 2| = |z - 3i|' title='|z - 2| = |z - 3i|' class='latex' /> and <img src='http://s2.wordpress.com/latex.php?latex=Arg%28z+%2B+i%29.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='Arg(z + i).' title='Arg(z + i).' class='latex' /><br />
(e) Consider the circle C: <img src='http://s3.wordpress.com/latex.php?latex=%7Cz+%7C+%3D+1&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='|z | = 1' title='|z | = 1' class='latex' />, let <img src='http://s1.wordpress.com/latex.php?latex=w&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='w' title='w' class='latex' /> be the center of any circle that touches <img src='http://s2.wordpress.com/latex.php?latex=C&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='C' title='C' class='latex' /> with radius 2. Describe the locus of <img src='http://s3.wordpress.com/latex.php?latex=w.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='w.' title='w.' class='latex' /></p>
<p>Problem 6<br />
(a) Suppose <img src='http://s1.wordpress.com/latex.php?latex=w&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='w' title='w' class='latex' /> is a non-zero complex number. Let <img src='http://s2.wordpress.com/latex.php?latex=z_1%2C+z_2%2C+%5Cdots%2C+z_5&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='z_1, z_2, \dots, z_5' title='z_1, z_2, \dots, z_5' class='latex' /> be roots of the equation $z^5 &#8211; 1 = 0.$ Describe geometrically, the shape formed by complex numbers <img src='http://s3.wordpress.com/latex.php?latex=wz_1%2C+wz_2%2C+wz_3%2C+wz_4%2C+wz_5.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='wz_1, wz_2, wz_3, wz_4, wz_5.' title='wz_1, wz_2, wz_3, wz_4, wz_5.' class='latex' /><br />
(b) Let <img src='http://s1.wordpress.com/latex.php?latex=w+%5Cnot+%3D+1&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='w \not = 1' title='w \not = 1' class='latex' /> be a cube root of unity. What is <img src='http://s2.wordpress.com/latex.php?latex=1+%2B+w+%2B+w%5E2%3F&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='1 + w + w^2?' title='1 + w + w^2?' class='latex' /><br />
How about <img src='http://s3.wordpress.com/latex.php?latex=1+%2B+w%5E5+%2B+w%5E%7B10%7D%3F&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='1 + w^5 + w^{10}?' title='1 + w^5 + w^{10}?' class='latex' /></p>
<p>Problem 7<br />
Let <img src='http://s1.wordpress.com/latex.php?latex=%5Czeta+%3D+e%5E%7B%5Cfrac%7B2+%5Cpi+i%7D%7Bn%7D%7D&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\zeta = e^{\frac{2 \pi i}{n}}' title='\zeta = e^{\frac{2 \pi i}{n}}' class='latex' />.<br />
(a) What is <img src='http://s2.wordpress.com/latex.php?latex=1+%2B+%5Czeta+%2B+%5Czeta%5E2+%2B+%5Cdots+%2B+%5Czeta%5E%7Bn-1%7D.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='1 + \zeta + \zeta^2 + \dots + \zeta^{n-1}.' title='1 + \zeta + \zeta^2 + \dots + \zeta^{n-1}.' class='latex' /><br />
(b) What is <img src='http://s3.wordpress.com/latex.php?latex=%5Csum_%7Bk+%3D+1%7D%5E%7Bn%7D+%5Ccos%7B%5Cfrac%7B2k+%5Cpi%7D%7Bn%7D%7D%3F&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\sum_{k = 1}^{n} \cos{\frac{2k \pi}{n}}?' title='\sum_{k = 1}^{n} \cos{\frac{2k \pi}{n}}?' class='latex' /><br />
(c) What is <img src='http://s1.wordpress.com/latex.php?latex=%5Csum_%7Bk+%3D+1%7D%5E%7Bn%7D+%5Csin%7B%5Cfrac%7B2k+%5Cpi%7D%7Bn%7D%7D%3F&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\sum_{k = 1}^{n} \sin{\frac{2k \pi}{n}}?' title='\sum_{k = 1}^{n} \sin{\frac{2k \pi}{n}}?' class='latex' /></p>
<p>Problem 8<br />
Suppose you are asked to evaluate the two sums:<br />
(i) <img src='http://s2.wordpress.com/latex.php?latex=%5Csum_%7Bk+%3D+1%7D%5E%7Bn%7D+%5Ccos%283k+%5Ctheta%29&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\sum_{k = 1}^{n} \cos(3k \theta)' title='\sum_{k = 1}^{n} \cos(3k \theta)' class='latex' /> and<br />
(ii) <img src='http://s3.wordpress.com/latex.php?latex=%5Csum_%7Bk+%3D+1%7D%5E%7Bn%7D+%5Csin%283k+%5Ctheta%29&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\sum_{k = 1}^{n} \sin(3k \theta)' title='\sum_{k = 1}^{n} \sin(3k \theta)' class='latex' />.<br />
What is your first step? And you next step? Do you see the entire solution lay out in your head?</p>
<p>Problem 9<br />
Let <img src='http://s1.wordpress.com/latex.php?latex=p%28x%29&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='p(x)' title='p(x)' class='latex' /> be a polynomial of degree 4 with real coefficients. Given that <img src='http://s2.wordpress.com/latex.php?latex=1+%2B+2i&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='1 + 2i' title='1 + 2i' class='latex' /> and <img src='http://s3.wordpress.com/latex.php?latex=3+%2B+4i&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='3 + 4i' title='3 + 4i' class='latex' /> are roots of <img src='http://s1.wordpress.com/latex.php?latex=p%28x%29.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='p(x).' title='p(x).' class='latex' /><br />
(a) What is the sum of roots?<br />
(b) What is the product of the roots?<br />
(c) Do you know how to do the problem if the condition &#8220;<img src='http://s2.wordpress.com/latex.php?latex=p%28x%29&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='p(x)' title='p(x)' class='latex' /> has real coefficients is removed?</p>
<p>Problem 10<br />
Expand the following in your head:<br />
(a) (1 + x)(1 + y)(1 + z)<br />
(b) (1 &#8211; x)(1 &#8211; y)(1 &#8211; z)</p>
<p>Problem 11<br />
How can you tell whether <img src='http://s3.wordpress.com/latex.php?latex=%5Csqrt%7B2%7D+%2B+%5Csqrt%7B3%7D&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='\sqrt{2} + \sqrt{3}' title='\sqrt{2} + \sqrt{3}' class='latex' /> is rational?</p>
<p>Problem 12<br />
Let C be the set of points satisfying <img src='http://s1.wordpress.com/latex.php?latex=%7Cz+-+1%7C+%3D+2.&#038;bg=ffffff&#038;fg=61636a&#038;s=0' alt='|z - 1| = 2.' title='|z - 1| = 2.' class='latex' /><br />
(a) Describe C geometrically.<br />
(b) Let P = 4 + 3i. What is the shortest distance from P to C?<br />
(c) Find the complex number w on C such that the distance between P and w is the answer you found in (b). [Remember, no calculation, you can work this out in your brain.]</p>
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