Do you know about the geometric interpretation of complex numbers and square roots? Have you heard about the $r{e}^{i\theta }$ or polar coordinate $(r,\theta )$ notation for complex numbers? Get back to me and I'll see if I can help.

OK, here's a very quick, rough and ready summary:

I described above the problem that you can't specify complex square roots so that the surd rules apply. Clearly, for each complex number you can specify a unique square root (for example, you could always choose the complex number whose imaginary part is positive if it's nonzero, or whose real part is positive if the imaginary part is zero), however the surd rules won't apply. In fact, there is another problem here.

If you think of taking the complex square root as a function $f(z)$ (so that $f(z{)}^2=z$ for all $z$) then there is no way to make sure that $f(z)$ is continuous (no sudden jumps in the value of $f(z)$) and single valued. Here's why. Suppose $f(z{)}^2=z$ for all $z$. Consider $g(t)=f(e^{2\pi it})$ for $0\le t\le 1$. $g(0)=f(1)=\pm 1$. Let's assume that $f(1)=1$ (the same argument works if $f(1)=-1$). If the value of $g(t)$ doesn't "jump" for any values of $t$ (i.e. the value of $f(z)$ doesn't jump for any $z$ on a unit circle) then we must have that for $0\le t\le 1/2$ $g(t)=e^{i\pi t}$ (since if at any point $g(t)=-e^{i\pi t}$ then the value of $g(t)$ would jump at that point). So $g(1/2)=e^{i\pi /2}=i$. Continue this argument for $1/2\le t\le 1$ to show that $g(t)=e^{i\pi t}$ and so $g(1)=e^{i\pi }=-1$. But $g(1)=f(e^{2\pi i})=f(1)=1$. In other words, $f(z)$ is not single valued.

That was quite complicated, I hope you followed it. Basically, as we go round the unit circle anticlockwise starting at 1 (taking square roots along the way), when we get back to 1 the square root has changed from 1 to -1. In other words, we can't choose a unique, continuous square root function.

There are lots of other functions like this, for example the complex logarithm function and the complex nth root function.

There are two major ways of dealing with this problem, one is mostly used by applied mathematicians and one is mostly used by pure mathematicians. The applied mathematicians tend to use "branch cuts" and the pure mathematicians tend to use "Riemann surfaces". I'll quickly describe both.

Let's start off with branch cuts. Suppose we said, I want a function $f(z)$ such that $f(z{)}^2=z$ for all $z$ except for negative real values of $z$. In this case, we can have a single valued, continuous value of the complex square root. Here is one way of doing it: $f(r{e}^{\mathrm{it}})=r^0.5e^{\mathrm{it}/2}$ where $-\pi <t<\pi$. Can you see why the argument I used above doesn't apply here?

If we do this, we say that the we have made a "branch cut" from 0 to -infinity along the negative real axis. As long as there is a line of points from 0 to infinity (a branch cut from 0 to infinity) which we do not want the value of sqrt(z) on, we can make sqrt(z) single valued and continuous.

For more complicated functions, you might have to make more than one branch cut, but I won't go into that now.

The other way of solving the problem is Riemann surfaces, but this is even more difficult. Imagine, if you will, a 4 dimensional space. In fact, it's a 2 dimensional space where each dimension is complex. In other words, it's the space of points $(x+iy,u+iv)$ for all real values of $x$, $y$, $u$, $v$ . Two complex dimensions, four real dimensions. The "concrete Riemann surface" of a function $f(z)$ is the set of points $(z,f(z))$ in this 4 dimensional space ( $z$ and $f(z)$ are both complex numbers). So, for example, if $f(z)=z^2$ then the concrete Riemann surface of $f(z)=z^2$ is the set of points $(z,z^2)$. What does this look like? Well, it's a surface which means that it is a 2 dimensional object sitting in a 4 dimensional space. It turns out that we can study the square root "function" by studying this surface in this 4 dimensional space. It turns out that if you take this surface of points $(z,z^2)$ and you twist it and deform it in the right way, you can make it look like the surface of a doughtnut (the mathematical word is a torus) with one point removed.

I'm not sure if that will have made any sense to you, it's an extremely difficult subject to understand (we weren't taught it until the 3rd year at Cambridge). If there were any bits you didn't follow, just post a note here and I'll see if I can explain a bit more fully.