If $u+v = \frac{2 \sin 2x}{e^{2y}+e^{-2y}-2 \cos 2x}$ then find corresponding analytical function $f(z)=u+iv$

Question

If $u+v = \frac{2 \sin 2x}{e^{2y}+e^{-2y}-2 \cos 2x}$ then find corresponding analytical function $f(z)=u+iv$

My attempt: we have $(1+i)f(z) = u-v+i(u+v) = U+iV$ (say)

then

$V = u+v =\frac{ \sin 2x}{\cosh 2y- \cos 2x} $ (from given equation u+v)

$dU=U_xdx+U_ydy = V_ydx-V_xdy$(from cauchy reiman equations)

Now i got

$V_x= \frac{2 \cos 2x \cosh 2y - 2}{(\cosh 2y-\cos 2x)^2}, V_y=\frac{-2 \sin 2x \sinh 2y}{(\cosh 2y-\cos 2x)^2}$

$dU = V_ydx-V_xdy$

Substituting above $V_x, V_y$ values in $dU$ i am not able to proceed to find U. This is getting messy. Can you provide me any hint?

Answer 1

Let us use the Milne-Thomson method to continue.

You have computed correctly $\frac{\partial V} {\partial y} = \psi_1(x,y)$ and $\frac{\partial V} {\partial x} = \psi_2(x,y)$ where $$V = \frac{\sin 2x}{\cosh 2y - \cos 2x}$$ Cheers!

Now, we have $$\psi_1(z,0) = - \frac{\sin 2z(2\sinh 0)}{(\cosh 0-\cos 2z)^2} =0$$ and $$\psi_2(z,0) = \frac{2\cos 2z \cosh 0 - 2} {(\cosh 0 - \cos 2z) ^2}$$ $$ = \frac{2\cos 2z - 2}{(1-\cos 2z)^2}$$ $$= - \frac2{1- \cos 2z} = - \csc^2 z$$

Then an application of the Milne-Thomson method results in: $$(1+i)f(z) = \int \left[\psi_1(z,0) + i\psi_2(z,0) \right] \, dz +C $$ $$=i \cot z + C $$ $$\implies \boxed{f(z) = \frac12(1+i)\cot z + c} $$