Express $\arg (e^{z - i / z})$ in terms of $x$ and $y$. 
Express $\arg (e^{z - i / z})$ in terms of $x$ and $y$.

I start with $\arg (e^{z - i / z}) = \arg (\frac{ze^z-i}{z})$ and after some algebraic manipulations end up with $\arg (\space\space\space e^x(x\cos y - y \sin y) + i[e^x(x \sin y + y \cos y) - 1]\space\space\space) - \arg(z)$.  At this point is taking inverse tangents appropriate or have I overlooked a simpler method?

For $z=x+iy$ we have:
$$
z-\frac{i}{z}=x+iy-i\left(\frac{x-iy}{x^2+y^2} \right)= x-\frac{y}{x^2+y^2}+i\left( y-\frac{x}{x^2+y^2}\right)= a+ib
$$
with:
$$
a=x-\frac{y}{x^2+y^2} \qquad b= y-\frac{x}{x^2+y^2}
$$
We have $e^{a+ib}= e^a e^{ib}$ and $b$ is the argument ( without the $2 \pi$ periodicity) of $e^{a+ib}$, so the argument of $e^{z-\frac{i}{z}}$ is:
$$
b= y-\frac{x}{x^2+y^2}
$$
