Complex comparison

$z \overline z = (1+i) \cdot (1-i) = 1^2-i^2 = 1-(-1) = 2 = 2e^{i\cdot 0}$

This means that the angle is 0 (radians of degrees doesn't matter) and because the angles are in the range $[0,2\pi) = [0^\circ,360^\circ)$ , no other option can have a smaller angle.

---

Here are the other angles

$z = 1+i = \sqrt{2} e^{\color{#D61F06} \frac {\pi}{4}}$

$z^2 = (1+i)^2 = 1^2+2i+i^2 = 2i = 2e^{\color{#D61F06} \frac {\pi}{2}}$

$\frac 1z = \frac 1{1+i} = \frac {1 (1-i)}{(1+i)(1-i)} = \frac {1-i}{2} = \frac 12 - \frac i2 = \frac 1{\sqrt{2}} e^{\color{#D61F06} \frac {7\pi}{4}}$