Let f be the branch of the function $ \sqrt {z^2 -4} $ defined by f(z)=$ \sqrt{r_1 r_2} e^{i \frac{ \theta_1 +\theta_2}{2} }$ where $\ r_1=|z-2|,r_2=|z+2|, \theta_1=arg(z-2), \theta_2=arg(z+2) $ and $\frac{- \pi}{2}<\theta_1,\theta_2 \le \frac{3\pi}{2}$. I have to show that if y $ \in\Bbb R$, then f(yi)=-$ \sqrt {4+y^2 } $.
Here is my attempt: We know that z=x+yi and as I have to evaluate f(yi) then x=0. So I make the substitution in f(z)=$ \sqrt{r_1 r_2} e^{i \frac{ \theta_1 +\theta_2}{2} }$ and I get
$ \sqrt {4+y^2 } e^{i\frac{arg(yi-2)+arg(yi_2)}{2}}$.So I need this $ \frac{arg(yi-2)+arg(yi_2)}{2}$ to be equal to $\pi$. Therefore i'll get the result $f(yi)=- \sqrt {4+y^2 } $.
But I don't know how to do it :(
Please somebody help.
Let $f(z)=\sqrt{z^2-4}$. The plane is cut such that $-\pi/2<\arg(z\pm 2)\le 3\pi/2$ to give
$$f(z)=\sqrt{|z^2-4|}e^{i\frac12(\arg(z-2)+\arg(z+2))}$$
Noting that for $z=iy$, $\arg(z-2)+\arg(z+2)=\pi$ for all $y\in \mathbb{R}$, we can write
$$\begin{align} f(iy)&=\sqrt{y^2+4}\,\,e^{i \frac{\pi}{2}}\\\\ &=i\sqrt{y^2+4}\\\\ \end{align}$$