# Mapping of multivalued complex function.

1. Jan 20, 2010

### leoneri

A complex function $$f\left(z\right)=\sqrt{z}$$ can be splitted into two branches:
1. Principal branch: $$f_{1}\left(z\right)=\sqrt{r} e^{i \left(\theta/2\right)}$$
2. Second branch: $$f_{2}\left(z\right)=\sqrt{r} e^{i \left[\left(\theta+2\pi\right) /2\right]}$$

My question is, is there a way to show/proof that the principal branch only map the z-plane to the right half plane of w-plane (Re w > 0) to which positive imaginary semiaxis is added, and that the second branch only map the z-plane to the left half plane of w-plane (Re w < 0) to which negative imaginary semiaxis is added??

Then, how exactly is to plot these branches in z-plane and w-plane? I am confused, because r and theta could be anywhere in the z-plane, and hence it could be mapped to any point in the w-plane. Please tell me if what I said is wrong.

2. Jan 20, 2010

### elibj123

Well,
if you parametrize the entire complex plane by $$( z=re^{i\theta} : r\in R, r\geq0, -\pi\leq\theta\leq\pi)$$

Then the range of $$\sqrt{r}$$ from r>0 remains $$r\in R, r\geq0$$
But the range of $$\theta /2$$ from $$-\pi\leq\theta\leq\pi$$ is
$$-\frac{\pi}{2}\leq \theta /2 \leq \frac{\pi}{2}$$

So the range of the first branch has a parametrization:
$$( w=re^{i\theta}: r\in R, r\geq0, -\frac{\pi}{2}\leq \theta \leq \frac{\pi}{2} )$$

Which is off course a parametrization of the right half plane Re(w)>0

3. Jan 26, 2010

### leoneri

I see it now. Thanks a lot.