Show that the function $\phi (x,y)=\arctan{\frac{2x}{x^2+y^2-1}}$ is harmonic by considering $w(z)=\frac{i+z}{i-z}$. Show that $\phi (x,y)=\arctan{\frac{2x}{x^2+y^2-1}}$ is harmonic by considering $w(z)=\frac{i+z}{i-z}$.

I know that if $\phi$ is harmonic then it satisfies Laplace's equation but I don't see how considering $w(z)$ will help me.. I've tried looking at the real and imaginary parts of $z$ but they don't seem to match up. I'm guessing the solutions will have something to do with the Cauchy Riemann equations but I've tried differentiating $w(z)$ with respect to $x$ and $y$, where $z=x+iy$, and that hasn't shed any light either.
Hint:


  $w$ is a fractional linear transformation, hence a conformal mapping (of what domains?). Therefore its inverse is also conformal. Composing a harmonic function with a conformal map gives you another harmonic function. So it remains to see what function you get by composing $\phi$ and $w$.
suppose $f(z) = Re^{i\Theta}$ and $f$ is differentiable. then, it can be shown that $\Theta$ is harmonic. 

now, $$w(z)=\frac{i+z}{i-z}=\frac{x+i(y+1)}{-x - i(y-1)}= \frac{[x+i(y+1)][-x+i(y-1)]}{[-x - i(y-1)][-x+i(y-1)]}=\frac{(1-x^2-y^2)-2ix}{x^2+(y-1)^2}=Re^{i\Theta}$$

where $\Theta = \tan^{-1}\left(\frac{2x}{x^2+y^2-1}\right)$ and therefore is harmonic.