Complex multi-valued function concept So this multi-valued function $f(z) = \sqrt{r}e^{i \frac{\theta}{2}}$ is multi-valued because it can output multiple results with a single input.
First of all, I think this is because it's totally dependent on the fact that a complex number CANNOT be written uniquely using the polar form. Like the fact that any complex number the angle will always have the $\theta +2\pi k$. Like its the angle that results in the multi-valued function I believe.  
This is something I would like clarification on, thanks.
Precisely, for example $z=i$ in polar form it will be written as $z=e^{i(\frac{\pi}{2} +2\pi k)}$ and $f(z)=f(e^{i(\frac{\pi}{2} +2\pi k)})=e^{i(\frac{\pi}{4} +\pi k)}=e^{i(\frac{\pi}{4})}e^{i\pi k} = \pm e^{i\frac{\pi}{4}}$.
And in the last step it is a plus or minus because $e^{i\pi k}$ is $1$ if $k$ is even and $-1$ if $k$ is odd. 
But for like a cube root function would have three outputs instead of two like for the square root complex function right?
I want to confirm if my reasoning with the particular example is correct as to why $f(z)$ is a multi-valued function.
 A: Yes you are correct in what you are saying to some degree. Using the $z = a + bi, (a,b\in \mathbb{R})$ form then like you have said, functions are defined uniquely since for 
$$z_1 = a_1 + b_1 i, \quad z_2 = a_2 + b_2 i$$
$$z_1 = z_2 \iff a_1 = a_2 \text{ and } b_1=b_2.$$
For the polar form on the other hand, no function (other than $f(z)=0$, or not depending on who you ask) can produce a unique result. When considering analytic (or regular) functions $f$, it is important when defining the function to also define a corresponding branch cut. 
For example, $f(z)=\log z$ must be defined with a corresponding branch cut so that a given $z$ cannot produce multiple different outputs of $f(z)$ with different arguments. Recall the definition of the complex natural logarithm, $\log w = \log\lvert w \rvert + i(\arg w + 2\pi k),$ for $k\in \mathbb{Z}$. Therefore, for $f(z)=\log {z}$ we can produce an infinite number of outputs by choosing any integer value $k$. Unfortunately, this multivaluedness defies the definition of differentiability (regularity) of a complex function - the limit definitions do not hold. This problem is solved by accompanying $f(z)$ with a branch cut such as the principle branch $\arg{z} \in (-\pi, \pi]$ - this restricts our arguments to produce single value output. 
There are an infinite number of choices for a branch cut, but it must be chosen to lie on a boundary where the function $f(z)$ is otherwise discontinuous (therefore defying traditional differentiability). 
To have a look at these more, here is a university resource on the topic of multivaluedness of complex functions (of a single variable): https://people.maths.bris.ac.uk/~maavm/mathmethods_files/branch_cuts.pdf
