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Given the real part of an analytic function $f:\mathbb{C}\rightarrow \mathbb{C}$

$$\text{Re}[f] = \frac{\sin(2x)}{\cosh(2y)-\cos(2x)},$$

I have to find the imaginary part of $f$.

I tried using the Cauchy-Riemann equations, but $\frac{\partial\text{Re}[f]}{\partial x}=\frac{\partial\text{Im}[f]}{\partial y}$ gives me the following integral:

$$\text{Im}[f]=2\int\frac{\cos(2x)\cosh(2y)-1}{\left(\cosh(2y)-\cos(2x)\right)^2}dy$$

and here is where the difficulties start. I put it in Wolfram Alpha and yeah, I don't think this is the way to go.

Is there another way to calculate Im$[f]$ without having such crazy integrals in your way?

TwoStones
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    An idea, don't know if it helps. The denominator is $2\sin(z)\sin(\bar z)$. Edit, yes it helps, try developing $1/\tan(z)$ – zwim Jul 01 '19 at 21:02
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    duplicate https://math.stackexchange.com/questions/3208652/integrating-int-frac-cosh2y-cos2x-1-cosh2y-cos2x2dy/3208728#3208728 – logo Jul 01 '19 at 21:19

1 Answers1

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Using addition formulas you have:

$\cosh(2y)-\cos(2x)=\cos(2iy)-\cos(2x)=2\sin(x+iy)\sin(x-iy)=2\sin(z)\sin(\bar z)$

Applying same kind of transformation to a numerator that would be symmetric, gives:

$\sin(2x)-i\sinh(2y)=\sin(2x)-\sin(2iy)=2\cos(x+iy)\sin(x-iy)=2\cos(z)\sin(\bar z)$

Thus $f(z)=\dfrac 1{\tan(z)}$ is a candidate and since it is analytic, it is the one searched for (have a look at the post below).

Is it Possible to have Two Distinct Analytic Functions with the Same Real Part?

zwim
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