You want to show that if you write $re^{i\theta} + se^{i\phi}$ as $te^{i\chi}$, then $\chi$ lies between $\theta$ and $\phi$. Without loss of generality, assume $\theta\leq\phi$. Factoring out $e^{i\theta}$ you get $re^{i\theta}+se^{i\phi} = e^{i\theta}(r + se^{i(\phi-\theta)})$. Since multiplying by $e^{i\theta}$ is just a rotation by an angle $\theta$, it is enough to consider the case where $\theta=0$ and $0\lt \phi\lt \pi$.
In that case, you have $r + se^{i\phi} = r + s(\cos(\phi)+i\sin(\phi))$, and you want to express it in the form $t(\cos(\chi) + i \sin(\chi))$. Looking at real and complex parts, you see that $t\sin(\chi)=s\sin(\phi)$, and $r+s\cos(\phi) = t\cos(\chi)$.
Assume first that $0\lt \phi\lt \frac{\pi}{2}$. Then $\chi$ must also lie in the first quadrant, since we need both $\sin(\chi)$ and $\cos(\chi)$ to be positive (since $t$, $s\sin(\phi)$, and $r+s\cos(\phi)$ are all positive). If, on the other hand, $\frac{\pi}{2}\lt \phi\lt \pi$, then $\cos(\phi)$ is negative. If $r+s\cos(\phi)\gt 0$, then we need $\cos(\chi)\gt 0$, so $\chi$ is in the first quadrant and automatically smaller than $\phi$ and we are done. If $r+s\cos(\phi)$ is negative then we need $\chi$ in the second quadrant.
In the former case, $0\lt \chi,\phi\lt \frac{\pi}{2}$, then from $\sin(\chi)=\frac{s}{t}\sin(\phi)$ and since $\sin(x)$ is increasing on $0\leq x\leq\frac{\pi}{2}$, then $0\lt\chi\lt\phi$ if and only if $\frac{s}{t}\lt 1$, if and only if $s\lt t$.
Now note that $t^2 = ||r+se^{i\phi}||^2 = r^2 + s^2 + 2rs\cos(\phi) \gt s^2$, since all of $r$, $s$, and $\cos(\phi)$ are positive. Since $s$ and $t$ are both positive, $s\lt t$, which shows that $0\lt\chi\lt\phi$, as desired.
In the other case, where $\frac{\pi}{2}\lt\phi\lt \pi$ and $r+s\cos(\phi)\lt 0$, then we know $\chi$ is also in the second quadrant where $\sin(x)$ is decreasing, so from $\sin(\chi)=\frac{s}{t}\sin(\phi)$ we get that $\frac{\pi}{2}\lt \chi\lt \phi$ if and only if $\frac{s}{t}\gt 1$, if and only if $s\gt t$. Here, since $\cos(\theta)\lt 0$, then you have again $t^2 = ||r+se^{i\phi}||^2 = s^2 + r(r + 2s\cos(\phi)) \lt s^2$ (since $r+s\cos(\phi)\lt 0$ in this situation), so you get $t\lt s$ and hence $\frac{\pi}{2}\lt\chi\lt\phi$, as desired
$$|z|=\frac1{|z|}\Rightarrow |z|=1,$$
$$|z|=|1-z|\Rightarrow \sqrt{x^2+y^2}=\sqrt{(1-x)^2+y^2}\Rightarrow x=\frac12,$$
where $z=x+iy$. Next, we will determine the imaginary part
$$\sqrt{\left(\frac12\right)^2+y^2}=1\Rightarrow y=\pm \frac{\sqrt{3}}2.$$
Therefore solutions are
$$z_1=\frac12+i\frac{\sqrt{3}}2,$$
$$z_2=\frac12-i\frac{\sqrt{3}}2.$$
Best Answer
If $z^{2018}=\overline z$,$$|z|^{2018}=\bigl|\overline z\bigr|=|z|$$and therefore $|z|=0$ or $|z|=1$. But $|z|=1\implies\overline z=\frac1z$. So, the solutions are $0$ and the complex numbers $z$ such that $z^{2019}=1$. And the solutions of this last equation are the numbers of the form$$\exp\left(\frac{2k\pi i}{2019}\right)\text{, with }k\in\{0,1,\ldots,2018\}.$$