Argument of a general Complex number

Question

How to prove that for large $k>N$, $\frac{1}{2} \le a \le 3\pi$, $0 <b \le 3\pi$, the argument of $z$ satisfy \begin{align*} \frac{\pi}{2} \le \mathrm{arg}z<\frac{3\pi}{4}, \end{align*} where $$z=\frac{i(k-2 \pi b)+2\pi a}{2 \sqrt{a-i b}}$$ I tried to split $z=x+i y$ and try to prove that $y>-x$. The inequality turns to be much complicated.

Answer 1

Let $\sqrt {a+ib} =c+id$ with $c,d$ real. The the real part of $z$ is $2\pi ac-d (k-2 \pi b)/|\sqrt {a-ib}|^{2}$ which is clearly negative for large values of $k$. Now note that the argument of a complex number lies between $\frac {\pi} 2$ and $\frac {3\pi} 4$ iff its real part is negative.