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

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.

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Isn't $z/(1-e^z)^{-2}=z(1-e^z)^2$? That should simplify things unless you meant something different. – Eric Naslund Sep 18 '11 at 11:59
@EricNaslund: Hmm.. thanks for spotting that, typo – Freeman Sep 18 '11 at 12:02
Write $z = x + iy$, multiply top and bottom by the conjugate of the bottom, and use $e^z = e^x(\cos y + i\ \sin y)$ repeatedly. Is this what you had in mind by 'brute force'? – Barinder Banwait Sep 18 '11 at 12:20
Yes, this is what I had in mind, it get's pretty messy. This is half of part c of a 6 part 2002 finals question, that approach seems a little long winded to me. If it is the correct approach then it's a bit of a mean question. – Freeman Sep 18 '11 at 12:25
@Barinder: I've converted it to a comment. – Zev Chonoles Sep 18 '11 at 13:35

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: $$\operatorname{Re}\left( \frac{z}{(1-\mathrm{e}^z)^2} \right) = \frac{x - 2 \mathrm{e}^x \left( x \cos y - y \sin y \right) + \mathrm{e}^{2x} \left( x \cos (2 y) + y \sin(2 y) \right)}{ ( 1 - 2 \mathrm{e}^x \cos y + \mathrm{e}^{2 x})^2 }$$