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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

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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
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.

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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? – εν οίδα ότι ουδέν οίδα 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 – εν οίδα ότι ουδέν οίδα 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. – εν οίδα ότι ουδέν οίδα 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.

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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?/ – εν οίδα ότι ουδέν οίδα 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

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