Draw solutions to complex number inequality $0 < \arg[(1-i)\overline z ] \le \frac \pi 4$

Question

I have to draw a picture of

$$\{ z\in\mathbb{ C } ; 0 < \arg[ (1-i)\overline z ] \le \frac \pi 4 \}$$

I totally don't get it and i don't even know how to start solving this.

Answer 1

Note that $$0<\arg (1-i)\overline z< \frac{\pi}4$$ implies $$0<-\frac{\pi}4 -\arg z < \frac{\pi}4,$$ since the argument of the product is the sum of the arguments, and the argument of $\overline z$ is the opposite of the argument of $z$. From the last inequality you get directly to $$-\frac{\pi}2 <\arg z<-\frac{\pi}4.$$

Answer 2

Let $z=a+ib$ and evaluate

$$\arg[(1-i)\overline z]=\arg(i-i)+\arg(\overline z)=-\frac\pi4 + \tan^{-1}\left(-\frac ba \right)$$ $$=-\tan^{-1}(1) - \tan^{-1}\frac ba =\tan^{-1}\frac{b+a}{b-a}$$

Given that $0 < arg( (1-i)\overline z$ ) $\le$ $\frac \pi 4$, we have

$$0 < \frac{b+a}{b-a} \le 1$$

which has the solutions,

1) $a \ge 0, \>\>\> b +a < 0$

2) $a\le 0, \>\>\> b + a > 0$.

Check them in the inequality to determine that only the first solution is valid.