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