How to show that $f_1(z) = -f_2(z)$, where $f_1$ and $f_2$ are the square root functions with different branches?

Question

Let $f_1(z)=\sqrt{r}e^{i\theta/2}, (r>0, 0<\theta<\pi)$ and $f_2(z)=\sqrt{r}e^{i\theta/2}, (r>0, \pi<\theta<5\pi/2)$. How to show $f_1(z)=-f_2(z)$ in the first quadrant?

If we set $z_0=re^{i\theta}, (r>0, 0<\theta<\pi)$, then we can plug it into $f_1(z)$, but what about $f_2(z)$? The angle of $z_0$ is not in $(\pi, 5\pi/2)$. Is it correct to add $\pi$ to the argument of $z_0$, then plug it into $f_2$?

Answer 1

To get $f_2(x)$ let $z_0=re^{i(\theta +2\pi) }$. Then $f_2(z_0)=\sqrt{r}e^{i(\theta/2 +\pi) }=-\sqrt{r}e^{i\theta/2}=-f_1(z_0)$. You can do this because $e^{2n\pi i}=1$ for all $n$, while $e^{\pi i}=-1$.

Answer 2

Let $z=x+iy (x,y\in \mathbb R)$. Then $z=x+iy=re^{i\Theta+2n\pi}$ ($\pi<\Theta \le \pi$, $n\in \mathbb Z$). Hence, $z=re^{i\Theta}$ is only one of an infinite number of possiblities for the exponential form of $x+iy$.