What Does the Branch of the Cube Root Function Mean?

[alexmahone]

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".

 

[Euge]

Hi Alexmahone,

Recall that a nonzero complex number has infinitely many arguments. In fact, any two arguments of a complex number differ by an integer multiple of $2\pi$. This is why we consider branches of the cube root; the cube root function is generally multi-valued, and the branch cuts allows us to define the cube root so that it is single-valued (so for every $z$ in the domain there is a unique argument of $z$).

In your situation, $f$ is defined on the complement of the $[0,\infty)$, and is determined by the equation

$$f(z) = e^{(1/3)\operatorname{arg}(z)} \qquad (0 < \operatorname{arg}(z) < 2\pi)$$

Evaluate $\arg(-i)$ first, then plug it into the formula for $f(z)$ to get the answer.