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October 20th, 2011, 09:16 AM   #1
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2011 complex numbers examination

[color=#000000]This is a test I put, for the third high school class.
[attachment=1:3klpwlsq]2011comptest.pdf[/attachment:3klpwlsq]
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 2011comptest.pdf (67.7 KB, 109 views)

 October 20th, 2011, 03:01 PM #2 Senior Member   Joined: Jul 2011 Posts: 245 Thanks: 0 Re: 2011 complex numbers examination Ah.. Beautiful. I thank you, good man. I shall give it my best shot. Understandably, I am likely to fail.
 October 21st, 2011, 11:27 AM #3 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: 2011 complex numbers examination That looks good! It would be a tough 120 minutes for me, even at my finest
 October 21st, 2011, 02:29 PM #4 Senior Member   Joined: Jul 2011 Posts: 245 Thanks: 0 Re: 2011 complex numbers examination Surprisingly, these aren't as bad as I thought. I've worked through question 1 and the majority of 2. Here are my solutions: Question 1: A. i. \begin{align} \text{Define:}&\\ z_{1}&=a+bi\\ z_{2}&=c+di\\ \overline{z_{1}+z_{2}}&=\overline{(a+c)+(b+d)i}\\ &=a+c-(b+d)i\\ &=(a-bi)+(c-di)\\ &=\overline{z_1}+\overline{z_2} \end{align} ii. $lt;a,b=>+==a+bi+c+di=(a+c)+(b+d)i=<(a+c),(b+d)=>=$ B (I went ahead and proved these 1. T: \begin{align} z&=\frac{-b\pm \sqrt{\Delta}}{2a}\\ \Delta&=-n \, \, \{n: n>0\}\\ \therefore z&=\frac{-b\pm \sqrt{-n}}{2a}\\ &=\frac{-b\pm \sqrt{n}i}{2a}\\ &=\frac{-b}{2a}\pm\frac{\sqrt{n}}{2a}i &=\{p+qi,p-qi\} \end{align} 2. F: \begin{align} z+\overline{z}&=(a+bi)+(a-bi)\\ &=2a\\ &=2\Re(z)\\ &\neq 2\Im(z) \end{align} 3. T: \begin{align} |z|^2&=(\sqrt{a^2+b^2})^2\\ &=a^2+b^2\\ &=(a+bi)(a-bi)\\ &=z\cdot \overline{z} \end{align} 4. F: $|z|=\sqrt{\Re(z)^2+\Im(z)^2}$ i.e. The value of z is its distance from the origin, which is derived through pythagorean theorem (or, equivalently, distance formula). 5. I don't know how to do this as I've never looked at trig seriously (just a freshman [first year] in highschool right now). Question 2: A. To answer this, I proved that $z$ is a complex number in the form $a+bi$ with $b=0$. The proof is as follows: \begin{align} |z^2-2iz-1|+|z^2+2iz-1|&=2|z^2+1|\\ |(z-i)^2|+|(z+i)^2|&=2|(z^2+1|\\ z=a+bi \Rightarrow |(a+bi-i)^2|+|(a+bi+i)^2|&=2|(a+bi)^2+1|\\ |a^2+2 i a b-2 i a-b^2+2 b-1|+|a^2+2 i a b+2 i a-b^2-2 b-1|&=2|a^2+2abi-b^2+1|\\ |(a^2-b^2+2b-1)+(2ab-2a)i|+|(a^2-b^2-2b-1)+(2ab+2a)i|&=2|(a^2-b^2+1)+(2ab)i|\\ \sqrt{(a^2-b^2+2b-1)^2+(2ab-2a)^2}+\sqrt{(a^2-b^2-2b-1)^2+(2ab+2a)^2}&=2\sqrt{(a^2-b^2+1)^2+(2ab)^2} \text{We see that this equation is not true unless } b&=0\\ \therefore z &\in \mathbb{R} \end{align} i. \begin{align}z \in \mathbb{R} \Rightarrow z&=a\\ |z-i|&=|a-i|\\ &=\sqrt{a^2+(-1)^2}\\ &=\sqrt{a^2+1^2}\\ &=|z+i|\end{align} ii. See above. Part B troubles me because I have yet to take trigonometry. -_- I need to do some studying and come back to this.
 October 25th, 2011, 02:42 AM #5 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: 2011 complex numbers examination The test paper attached above has been updated and now gives correct solutions for Question 3.
October 25th, 2011, 05:44 AM   #6
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Re: 2011 complex numbers examination

Quote:
 Originally Posted by CherryPi Part B troubles me because I have yet to take trigonometry. -_- I need to do some studying and come back to this.
The fun thing is that I never understood trig very well until I learned complex numbers, which make many of the mysterious parts of trig make sense.

 October 25th, 2011, 08:40 AM #7 Global Moderator   Joined: Dec 2006 Posts: 20,919 Thanks: 2203 4 a. $z_a\,=\,\frac{a}{1\,+\,ai}\,=\,\frac{a(1\,-\,ai)}{1\,+\,a^{\small2}}\,=\,\frac{a}{1\,+\,a^{\s mall2}}\,-\,\frac{a^{\small2}}{1\,+\,a^{\small2}}\,i$ $\text{If }a\,=\,\tan(t/2),\text{ where }-\pi\,<\,t\,<\,\pi, \\ \,z_a\,=\,x\,+\,yi,\text{ where }x\,=\,\frac{\small1}{\small2}sin(t)\text{ and }y\,=\,(\frac{\small1}{\small2}\cos(t)\,-\,\frac{\small1}{\small2})i, \\ \text{so the locus has equation }x^{\small2}\,+\,(y\,+\,\frac{\small1}{\small2})^{ \small2}\,=\,\frac{\small1}{\small4},\text{ but excludes the point (0, -1).} \\ \text{The locus is circular, with t relating to the direction of a radius.}$ 4 b. $|z_a|^{\small2}\,=\,x^{\small2}\,+\,y^{\small2}\,= \,\frac{\small1}{\small2}(1\,-\,\cos(t)),\text{ so }|z_a|\text{ has minimum 0 (for z_0\,=\,0) and no maximum.}$ 4 c. $z_a,\,z_{{\small{-1}}/a\,},\text{ where }a\,\ne\,0,\text{ correspond to values of }t\text{ that differ by \pi,} \\ \text{so the corresponding points are diametrically opposed.}$

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