Define a Principal Branch in Complex Analysis The actual problem is to define a principal branch for $a^z=e^{zlog(a)}$ and to give a general formula. From what I understand about principal branches, it is already in that form? I'm missing something key about that. But I also am having a hard time with the general formula. The transition from real numbers to complex is confusing me, since "log" is just defined as the inverse of $e^x$, and $e^x$ is defined by the power series. I'm just not sure where to begin.
 A: The issue of branches of log can be seen this way: as you said, $e^z$ and $logz$ are not global inverses of each other, still, they are local inverses of each other, (e.g., by the inverse function theorem. ) A branch of $logz$ is a local inverse of $e^z$.
Remember that the exponential $e^z$ is periodic, with period $2\pi$ This means you can tile the complex plane with strips $[y,y+i2\pi)$ , in which you can define local inverses of $e^z$. Any of these local inverses (you get one for each value y ) is a branch of the log.
As an example, the main log, aka, Logz, is the invertible map $e^z$ between the complex plane with the negative real axis removed, and the strip $(0,0+i2\pi]$=$(0,i2\pi$]. The inverse of this function is then what is called $Logz$ 
A: In the complex numbers, unlike in the case of real numbers, the function $e^z$ is not one-to-one.  Therefore it does not have an inverse that we can call "log".  Essentially, the difficulty arises from the fact that $e^{2 \pi i n} = 1$ for every integer $n$.  Thus, "$\log 1$" could be any of the numbers $2 \pi i n$.  In general, there will be infinitely many possible choices for the log of any nonzero complex number, all of which differ from each other by an integer multiple of $2 \pi i$.  So when we talk about "$\log z$" over the complex numbers, we have to specify which choice we are making for each value of $z$.
To define $a^z$, you just need to choose a value of $\log{a}$, and then define $a^z = e^{z \log{a}}$, as you said.  Given a nonzero value of $a$, there are many choices for $\log{a}$, but once you fix it, then $a^z$ is a perfectly well-defined analytic function for all $z$ (since you can use the same choice of $\log{a}$ for all $z$.)  No principal branches involved.
There is a problem, however, with defining $z^a$.  (Is this what you meant?)  Now you have $z^a = e^{a \log{z}}$, and you have to make a choice of $\log{z}$ for each nonzero $z$.  Making a random choice for each nonzero $z$ would produce a badly behaved function; the interesting question is whether we can make the choices so that the resulting function is continuous for all nonzero $z$.
It turns out, unfortunately, that we cannot make a continuous choice of $\log{z}$ for all $z \ne 0$.  The main difficulty can be described as follows.  As you probably know, every $z \ne 0$ can be written in the polar form $z = r e^{i \theta}$ for real numbers $r, \theta$, and $r>0$.  Geometrically, $r$ is the distance of $z$ from the origin, and $\theta$ is the angle from the positive $x$-axis to vector from 0 to $z$.  Thus we can "define" $\log{z}$ by $$"\log{z} = (\log{r}) + i \theta "$$
where I have put the equation in quotes because we have to be careful how we interpret it.
First of all, what is $\log{r}$?  This is no problem: $r$ is a positive real number, which therefore has a unique real logarithm.  We define $\log{r}$ to be the real logarithm of $r$.  The problem comes with the imaginary part: we have lots of choices for $\theta$, because $0, 2\pi, 4\pi, \ldots, -2\pi, -4\pi, \ldots$ all represent the same angle.  So the question boils down to: Can we make a continuous choice of $\theta$ for all complex numbers $z \ne 0$?
We can't.  Let's say we choose the value $\theta=0$ at the point $z=1$.  Now follow the circle of radius 1 centered at 0, walking counterclockwise along the circle.  Now if you think about it carefully, the choice of $\theta$ at each point along this circle is completely determined by the choice of $\theta=0$ at the point $z=1$, if we want $\theta$ to be continuous.  Thus, on the line $y=x$ (for positive $x$), we have to choose $\theta=\pi/4$, not something else like $9\pi/4$; on the positive $y$-axis, we have to choose $\theta=\pi/2$, not $5\pi/2$, etc. But now when we walk all the way around the circle, back to the point $z=1$, we find that our original choice 0 is now forced to be $2\pi$.  Thus $\theta$ cannot be defined continuously everywhere away from 0.
What can we salvage?  We can try to define $\theta$ continuously on a smaller subset of the nonzero complex numbers, instead of all of them.  Essentially, the problem was that looping around the origin once increases $\theta$ by $2 \pi$.  If we remove enough complex numbers so that it is impossible to loop around the origin without hitting one of these removed points, then $\theta$ can be defined continuously on the remaining points.  These removed points are called a "branch cut".
A common choice of branch cut is the negative real numbers.  It's not hard to convince yourself that you can't make a loop around the origin without hitting a negative real number at some point.  And you can define $\theta$ continuously for all $z$ excluding the real numbers $z \le 0$.  If you choose $\theta=0$ for $z=1$, as before, then for points $z$ slightly "above" the negative real axis, we choose $\theta$ positive and having magnitude slightly less than $\pi$; for points $z$ slightly "below" the negative real axis, we choose $\theta$ negative and having magnitude slightly less than $\pi$.  All other points fill in naturally.  Note that there is a discontinuity as you cross the branch cut (the negative real numbers), as expected.
Once we define $\theta$, we can define $\log z = (\log r) + i \theta$, and thus define $z^a = e^{a \log{z}}$, which was our original goal.
Having said all that, there is one case where all these choices don't matter, and we can define $z^a$ continuously over all nonzero $z$.  What case is that?  That is the case where $a$ is an integer: For in that case, making a different choice of $\log{z}$ changes the value of $a \log{z}$ by an integer multiple of $2 \pi i$ and thus the value of $e^{a \log{z}}$ isn't affected at all.  So the value of $z^a$ is independent of the choice of $\log{z}$ when $a$ is an integer.  And if you think about it, there's a good reason for that: for example, when $a=3$, the value of $z^3$ had better not be any number other than $z \cdot z \cdot z$ !
