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?
 A: When $\text{z}\in\mathbb{C}$:


*

*$$\Re\left[\text{z}-\frac{i}{\text{z}}\right]=\Re\left[\text{z}\right]-\frac{\Im\left[\text{z}\right]}{\Re^2\left[\text{z}\right]+\Im^2\left[\text{z}\right]}$$

*$$\Im\left[\text{z}-\frac{i}{\text{z}}\right]=\Im\left[\text{z}\right]-\frac{\Re\left[\text{z}\right]}{\Re^2\left[\text{z}\right]+\Im^2\left[\text{z}\right]}$$


So:


*

*$$e^{\text{z}-\frac{i}{\text{z}}}=e^{\Re\left[\text{z}-\frac{i}{\text{z}}\right]+\Im\left[\text{z}-\frac{i}{\text{z}}\right]i}=e^{\Re\left[\text{z}-\frac{i}{\text{z}}\right]}\cdot
   e^{\Im\left[\text{z}-\frac{i}{\text{z}}\right]i}$$

*$$e^{\Im\left[\text{z}-\frac{i}{\text{z}}\right]i}=\cos\left(\Im\left[\text{z}-\frac{i}{\text{z}}\right]\right)+\sin\left(\Im\left[\text{z}-\frac{i}{\text{z}}\right]\right)i$$


We can write:


*

*$$\Re\left[e^{\text{z}-\frac{i}{\text{z}}}\right]=\exp\left\{\Re\left[\text{z}\right]-\frac{\Im\left[\text{z}\right]}{\Re^2\left[\text{z}\right]+\Im^2\left[\text{z}\right]}\right\}\cdot\cos\left(\Im\left[\text{z}\right]-\frac{\Re\left[\text{z}\right]}{\Re^2\left[\text{z}\right]+\Im^2\left[\text{z}\right]}\right)$$

*$$\Im\left[e^{\text{z}-\frac{i}{\text{z}}}\right]=\exp\left\{\Re\left[\text{z}\right]-\frac{\Im\left[\text{z}\right]}{\Re^2\left[\text{z}\right]+\Im^2\left[\text{z}\right]}\right\}\cdot\sin\left(\Im\left[\text{z}\right]-\frac{\Re\left[\text{z}\right]}{\Re^2\left[\text{z}\right]+\Im^2\left[\text{z}\right]}\right)$$


Now, use the complex argument:


*

*When $\Re\left[e^{\text{z}-\frac{i}{\text{z}}}\right]>0$ and $\Im\left[e^{\text{z}-\frac{i}{\text{z}}}\right]>0$:
$$\arg\left(e^{\text{z}-\frac{i}{\text{z}}}\right)=\arctan\left(\frac{\Im\left[e^{\text{z}-\frac{i}{\text{z}}}\right]}{\Re\left[e^{\text{z}-\frac{i}{\text{z}}}\right]}\right)=\arctan\left(\tan\left(\Im\left[\text{z}\right]-\frac{\Re\left[\text{z}\right]}{\Re^2\left[\text{z}\right]+\Im^2\left[\text{z}\right]}\right)\right)$$

*When $\Re\left[e^{\text{z}-\frac{i}{\text{z}}}\right]>0$ and $\Im\left[e^{\text{z}-\frac{i}{\text{z}}}\right]<0$:
$$\arg\left(e^{\text{z}-\frac{i}{\text{z}}}\right)=-\arctan\left(\frac{\Im\left[e^{\text{z}-\frac{i}{\text{z}}}\right]}{\Re\left[e^{\text{z}-\frac{i}{\text{z}}}\right]}\right)=-\arctan\left(\tan\left(\Im\left[\text{z}\right]-\frac{\Re\left[\text{z}\right]}{\Re^2\left[\text{z}\right]+\Im^2\left[\text{z}\right]}\right)\right)$$

A: 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:
$$
b= y-\frac{x}{x^2+y^2}
$$
