Complex Analysis Proof: Derivative I'm not quite sure what the question is asking me to show or even to show this. Please provide some guidance, because I think I need some kind of proof:

Fix $c \in \mathbb{C} \setminus \{0\}$. Find the derivative of $f (z) = z^c$.

I would normally just use the power rule. $f'(z)=cz^{c-1}$, but that seems too easy. What should be done?
 A: Part of the problem is that $z\mapsto a^c$ is "multiple-valued".  The simplest example is when $c=1/2$.  You learned at your mother's knee that each positive real number $z$ has two square roots, called $\pm\sqrt{z}$, and each negative real number $z$ has two square roots, called $\pm i\sqrt{-z}$ (where of course $-z$ is a positive number, and so is $\sqrt{-z}$.
Now picture what happens when $z$ goes around the circle of unit radius centered at $0$.  Suppose you single out the positive value of the square root of $1$, and then let $z$ run counterclockwise from $1$.  As $z$ moves along the circle from $1$ to (say) $\cos22^\circ+i\sin22^\circ$, then $\sqrt{z}$ moves counterclockwise from $1$ to $\cos11^\circ+i\sin11^\circ$, i.e. it goes just half as far.  By the time $z$ reaches $i$, $\sqrt{z}$ reaches $\cos45^\circ+i\sin45^\circ$, and as $z$ keeps going counterclockwise from $i$ to $-1$, $\sqrt{z}$ continues counterclockwise from $\cos45^\circ+i\sin45^\circ$ to $\cos90^\circ+i\sin90^\circ = 0+i\cdot1=i$, and we get the square root of $-1$ equal to $i$.  But then $z$ continues counterclockwise along the bottom half of the circle, going from $-1$ back to $1$, and $\sqrt{z}$ continues counterclockwise, going from $i$ to $-1$, and then we have the other square root of $1$.  Then $z$ continues going counterclockwise, going around the circle for the second time, as $\sqrt{z}$ also continues counterclockwise, along the bottom half of the circle.  By the time $z$ reaches $-1$, then $\sqrt{z}$ reaches $-i$, the other square root of $-1$.  And then $z$ continues counterclockwise, going from $-1$ back to $1$, as $\sqrt{z}$ continues counterclockwise, going from $-i$ to $1$.
How, then, do we define $z^c$, when $c$ isn't necessarily even a real number?  The complex exponential $z\mapsto e^z$ can be defined by a power series, and is periodic with period $2\pi i$, so its inverse is multiple-valued, this time with infinitely many values rather than just two.  We can say $z^c = e^{c\log z}$, and this is multiple-valued because $\log$ is multiple-valued.
Any two values of $\log z$ differ from each other by $2\pi i n$, for some integer $n$, and therefore the derivative of $\log z$ is single-valued because $(d/dz)(2\pi i n)=0$.
So if one uses the chain rule to say $\dfrac{d}{dz} e^{c\log z} = e^{c\log z}\cdot\dfrac c z$, then the multiple-valued part is just the function you had before differentiating, an whichever one of its multiple values you're looking at, use that same one on both sides of the equality.
So you do get $cz^{c-1}$, another multiple-value function, but any one of the multiple values of $z^c$ has a corresponding one of the multiple values of $cz^{c-1}$.
