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.

 

[math771]

Sure, I can try to help explain the question for you.

In this context, "branches" refer to different possible values for the function $f(z)=z^{1/3}$. For example, the cube root of 8 has three possible values: 2, -1+sqrt(3)i, and -1-sqrt(3)i. These are the three branches of the cube root function.

However, in this problem, we are only interested in one specific branch of the cube root function, denoted by $f(z)=z^{1/3}$. This branch is defined by the condition that $0<\theta<2\pi$, which means that the angle of the complex number $z$ must be between 0 and 2pi (or between 0 and 360 degrees). This branch also excludes the point $z=0$, which is the origin on the complex plane.

Now, the "branch cut" is a line or curve on the complex plane where the function is not defined. In this case, the branch cut is along the ray $\theta=0$, which means that the function is not defined for any complex number with an angle of 0 degrees (or any real number on the positive x-axis).

So, to find $f(-i)$, we need to first determine which branch of the cube root function we are using (in this case, the one with $0<\theta<2\pi$ and $z\neq 0$). Then, we simply plug in $z=-i$ into the function $f(z)=z^{1/3}$ to get our answer.

I hope that helps clarify the question for you! Let me know if you have any other questions.