complex numberscomplex-analysis
Let $f(z)=z^{1/3}$ be the branch of the cube root whose domain of definition is given by $0<\theta<2\pi$, $z\neq 0$ (i.e. the branch cut is along the ray $\theta=0$.) Find $f(-i)$.
Could someone please help me understand the question? I'm not too clear on "branches" and "branch cuts".
This is explained in Wikipedia :
case $n=2$ : draw the two branches $\ z\mapsto z^{1/2}$ and $\ z\mapsto z^{1/2}e^{2\pi i/2}$ (we display the imaginary part only for $\ x+iy\,\mapsto z:=\Im (x+iy)^{1/2}\,$)
case $n=3$ : draw the three branches $\ z\mapsto z^{1/3}$ , $\ z\mapsto z^{1/3}e^{2\pi i/3}$ and $\ z\mapsto z^{1/3}e^{4\pi i/3}$
Hoping this helped to imagine to generalize,
Given a point $z=re^{i\theta}$, not on the parabola, with $0\le\theta\lt2\pi$, imagine starting at $(r,0)$ and going around the circle of radius $r$ counterclockwise until you get to $z$. If you get to $z$ without crossing the parabola, define $\log z=\log r+i\theta$. If you cross the parabola before you get to $z$, define $\log z=\log r+i(\theta-2\pi)$.
Best Answer
When $\theta \in [0,2 \pi]$, $-i=e^{i 3 \pi/2}$. Thus,
$$f(-i) = \left ( e^{i 3 \pi/2} \right )^{1/3} = e^{i \pi/2} = i$$
If the branch cut were along the negative real axis, then $\theta \in [-\pi,\pi]$ and $-i = e^{-i \pi/2}$ so that $f(-i) = e^{-i \pi/6}$.