The number of solutions for n raised to a complex exponent My understanding is that there is one and only one solution when solving for $z$ when $z = n^s$, where $s$ is a complex number of the form $a + bi$.  However, there are many solutions to $z$ when $z^{1/s} = n$.  It seems the solutions should be the same for $z$ in either case.  My first question: is this correct or is it an incorrect/incomplete understanding?
Assuming this is correct, my second question is on how this differs with real number exponents.  For example, $z$ has $m$ solutions when $z = n^{1/m}$ and there are also $m$ solutions to $z$ when $z^m = n$.  In fact, the solutions are the same for $z$ when solving for either form.  Why do we not have this same symmetry with complex exponents?
 A: Your analogy with the real case isn't exactly correct.
Lets imagine what happens when "solving" (or just calculating) the value of $z=n^s$ when $s$ is complex.  You haven't specified what $n$ is, so I'll just assume it's complex.  If it's real there's just a slight modification in the argument.
Any complex number can be written in the form of $re^{i\theta}$, which is known as the polar form of the complex number.  Writing $z=x+iy$ is known as the rectangular form.  Now, I will write $n$ in polar form as $n=re^{i\theta}$, and $s$ in rectangular form as $s=x+iy$.  Taking the exponential, we have that $$n^s=\left( re^{i\theta}\right)^{x+iy}=r^{x+iy}e^{i\theta(x+iy)}=e^{\ln(r)(x+iy)}e^{i\theta x-\theta y}=e^{(\ln(r)x-\theta y)+i(\ln(r)y+\theta x)}$$
Each of $r$, $x$, $\theta$, and $y$ are just real constants.  So, this simplifies to $e^{a+bi}$ for some positive real constants $a$ and $b$.  This is a unique complex number, and while the complex exponential is $2\pi i$ periodic (so $e^{a+bi}=e^{a+bi+2\pi i}=e^{a+bi+2\pi ik}$ for any $k\in\mathbb{Z}$).  But, the fact remains that this is just some specific complex number.
The reverse situation isn't true at all.  By this, I mean that, given a specific $z_0$, finding the $n$ such that $z_0=n^s$ generally has multiple solutions.  Contrary to your assertion though, this is true in the real case also.  To see this, consider the real function $y=x^2$.  We know beforehand that given some $x$, this outputs a unique $y$.  So, in our case of inverses (given a $y_0$, is $x$ such that $y_0=x^2$ unique?) has multiple answers as well.  Let's consider a specific case of $y_0=9$.  Which $x$ are there so $x^2=9$?  Clearly $x=3$ is an answer, but so is x=-3$.
Often when discussing the square root (the inverse function for $x^2$), one mentions the principle square root.  This just means that the inverse of $y=x^2$ can either be $x=\sqrt{y}$ or $x=-\sqrt{y}$.  This is because the inverse of the function $y=x^2$ is what's known as multi-valued.  When working with multivalued functions, it can often be desirable to specify which of the (potentially infinite) choices for it through a process known as making a branch cut.  I won't go into these too much, I'm just mentioning them in case you want to investigate further.
A: The function that takes $n\in\mathbb{N}$ and $s\in\mathbb{C}$ to $n^s$ is $\exp(s\log n)$, a very useful convention.  For fractional real $s$, such as $s={1\over 2}$, one can I suppose complicate things and say that the function is multivalued, but I am not sure anything is gained.
If you are instead attempting to find all $z$ such that $z^{1/s}=n$, there will often be multiple solutions.  For example, if $s={1\over 2}$, so that the equation is $z^2=n$, obviously if $z$ is a solution so is $-z$.  
If now $n$ is complex (not necessarily in $\mathbb{N}$ or $\mathbb{R}^+$), it turns out that $\log n$ is not uniquely defined, and therefore $n^s$ as defined above has issues.  For example, $\log i$ is any number $u$ such that $\exp(u)=i$, which means $u$ can be $i(\pi/2+2\pi m)$ for any $m\in\mathbb{Z}$. Again, if you were seeking complication you could say "ah, even when $n\in \mathbb{N}$, we could make $\log n$ multivalued," and indeed we could, but not much is gained thereby.
Humpty-Dumpty said "When I use a word, it means just what I choose it to mean -- neither more nor less."  That is how mathematicians think -- remember, the novelist who created Humpty-Dumpty was a mathematics don.  Over the years conventions have grown up which are useful and convenient, such as $\log x$ having exactly one value for $x\in\mathbb{R}^+$, or  the definition of $n^s$ in terms of exponentials given above.  
