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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

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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.2012-05-21

2 Answers 2

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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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    Marking this as accepted, conclusion is that the book is most likely wrong.2012-05-21
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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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    You often find 1 or 2 mistakes in exercise books2012-05-21