[Math] Finding the real and imaginary parts of $\frac {z}{(1-e^z)^2}$

Question

Could anyone help me find the real and imaginary parts of this

$$ \frac {z}{(1-e^{z})^{2}} $$

where $z$ is complex? I can brute force it out but I'm worried that I'm missing an easier way, as I will be partially differentiating the two parts.

Answer 1

Multiply the numerator and denominator of the expression by the complex conjugate of the denominator, with the intent to make the denominator real: $$ \frac{z}{(1-\mathrm{e}^z)^2} = \frac{z}{(1-\mathrm{e}^z)^2} \frac{(1-\mathrm{e}^{z^\ast})^2}{(1-\mathrm{e}^{z^\ast})^2} $$ Now, with $z = x + i y$: $$ (1-\mathrm{e}^z)^2 (1-\mathrm{e}^{z^\ast})^2 = ( 1 - 2 \operatorname{Re}(\mathrm{e}^z) + \mathrm{e}^{2 \operatorname{Re} z } )^2 = ( 1 - 2 \mathrm{e}^x \cos y + \mathrm{e}^{2 x})^2 $$ Using $\operatorname{Re}(a b^\ast) = \operatorname{Re}(a) \operatorname{Re}(b^\ast) - \operatorname{Im}(a) \operatorname{Im}(b^\ast) = \operatorname{Re}(a) \operatorname{Re}(b) + \operatorname{Im}(a) \operatorname{Im}(b) $: $$ \begin{eqnarray} \operatorname{Re}( z (1-\mathrm{e}^{z^\ast})^2 ) &=& x ( 1 - 2 \mathrm{e}^x \cos y + \mathrm{e}^{2x} \cos( 2 y)) + y ( 2 \mathrm{e}^x \sin y + \mathrm{e}^{2x} \sin( 2 y) ) \\ & = & x - 2 \mathrm{e}^x \left( x \cos y - y \sin y \right) + \mathrm{e}^{2x} \left( x \cos (2 y) + y \sin(2 y) \right) \end{eqnarray} $$ The final result is the quotient of these two: $$ \operatorname{Re}\left( \frac{z}{(1-\mathrm{e}^z)^2} \right) = \frac{x - 2 \mathrm{e}^x \left( x \cos y - y \sin y \right) + \mathrm{e}^{2x} \left( x \cos (2 y) + y \sin(2 y) \right)}{ ( 1 - 2 \mathrm{e}^x \cos y + \mathrm{e}^{2 x})^2 } $$