Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Question:

$$z=\frac{a+3i}{2+ai}$$

Show that there is only one value of $a$ for which $\operatorname{arg} z= \frac{\pi}{4}$, and find this value.

My attempt: $$\frac{a+3i}{2+ai}\cdot\frac{2-ai}{2-ai}$$ $$=\frac {5a+(6-a^2)i}{4+a^2}$$ $$=\frac {5a}{4+a^2}+\frac {6-a^2}{4+a^2}i$$ $$\tan(\pi/4)=\frac {\mathrm{opposite}}{\mathrm{adjacent}}$$ $$\tan(\pi/4)=\frac{\frac{6-a^2}{4+a^2}}{\frac{5a}{4+a^2}}$$

My Questions:

1) at the last point of my working $$\tan(\pi/4)=\frac{\frac{6-a^2}{4+a^2}}{\frac{5a}{4+a^2}}$$ the answer of the book shows it the other way around like this: $$\tan(\pi/4)=\frac{\frac{5a}{4+a^2}}{\frac{6-a^2}{4+a^2}}$$ Why? Since I use $\tan(\theta)=\mathrm{opp}/\mathrm{adj}$, opposite would be the y-value or imaginary and adjacent the x-value and real.

2) The book states no reason for why we multiply by the complex conjugate and I would like to learn why

share|improve this question
1  
Not surprising that the person got careless, we will get $5a=6-a^2$ in any case, since $1/\tan(\pi/4)=\tan(\pi/4)$. We multiply by the complex conjugate to get the number in "standard form" $c+di$ from which the argument can be read off. –  André Nicolas May 21 '12 at 14:11

2 Answers 2

up vote 1 down vote accepted

Question 1: Your answer is correct, but technically, the answer in the book is correct as well, since $\tan \pi/4 = 1$.

Question 2: Multiplying by the complex conjugate of the denominator helps you get rid of the imaginary numbers in there, since $(a+ib)(a-ib) = a^2 + b^2$.

Edit: The answer in the book is misleading.

share|improve this answer
    
I understood about the conjugate pair, but I didn't get 'The angle is measured from the real axis'. You mean that x is the imaginary axis and y is the real-axis in that case? In any case where did that come from? –  Panayiotis May 21 '12 at 9:40
    
No, I did not mean that. What I meant was that the angle is measured going anti-clockwise from the $x$ axis, which is also the real axis. If you draw a picture, you will see what I mean. –  Johannes Kloos May 21 '12 at 9:45
    
I'm sorry but I still do not understand, could you elaborate with a graph? (this is what I am currently thinking i47.tinypic.com/2wevvus.png) –  Panayiotis May 21 '12 at 9:55
    
Actually, you are right - I somehow got confused reading your question. –  Johannes Kloos May 21 '12 at 10:00
    
Marking this as accepted, conclusion is that the book is most likely wrong. –  Panayiotis May 21 '12 at 10:09

I hope this image helps clear up the confusion :)

enter image description here

Note: Your book probably made a typo. Your answer is correct.

share|improve this answer
    
Actually, this is what the question describes; the book somehow deviates from this!? –  Johannes Kloos May 21 '12 at 10:01
    
I am afraid so. Your answer is correct. –  E.O. May 21 '12 at 10:06
    
Do you think it is safe to assume that the book is wrong?/ –  Panayiotis May 21 '12 at 10:07
    
It's quite likely. –  Johannes Kloos May 21 '12 at 10:09
    
You often find 1 or 2 mistakes in exercise books –  E.O. May 21 '12 at 10:09

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.