Find the image of $|z|<1$ under the mapping $w=\frac{z}{(z+i)(z-i)}$. 
Describe the image of $|z|<1$ under the mapping 
  $$w=\dfrac{z}{(z+i)(z-i)}$$

I have tried solving for $z$ but I don’t know where to go from there. And even if I get an equation of $w=u+iv$, I don’t know how that changes what the function will look like.
 A: If you are dealing with the image of the unit circle and its interior, use the polar form:
$$z=re^{i\phi}$$
Your expression simplifies to:
$$w=\frac{z}{(z+i)(z-i)}=\frac{z}{z^2+1}=\frac{re^{i\phi}}{r^2e^{2i\phi}+1}$$
The edge of your domain (the unit circle $r=1$) maps to:
$$w=\frac{e^{i\phi}}{e^{2i\phi}+1}=\frac{1}{e^{i\phi}+e^{-i\phi}}=\frac{1}{2\cos\phi}$$
which is always a real number, but it goes off to infinity and comes back to the other side. When $z$ starts at $\phi=0$ ($z=1$), we go from $\frac{1}{2}$ along the real axis to infinity (at $z=i$), then it comes back from negative infinity, going to $-\frac12$ at $z=-1$, then it backtraces to negative infinity, comes back from positive infinity and returns to the original point. This is where your boundary of the circular domain maps to: a half-line from $-1/2$ to the left and a half-line from $1/2$ to the right. 
Checking that $z=0$ maps to $w=0$, we see that the inside of the circle maps to the domain which includes the point $w=0$ and extends all the way to the boundary, which in this case is the entire complex plane (with the points precisely on the unit circle, mapping onto the discussed lines, twice: lower halfcircle and upper halfcircle map to the same numbers). It seems wrong that the continuous circle splits up into two discontinuous parts, but that's just because we are "closing" back through infinity to the other side (the unit circle includes both poles of the function).
Imagine the upper half-circle "fanning out", stretching its upper point to infinity, like trying to fill an entire upper half-plane by stretching a small piece of rubber. The same with lower part.
