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If $f(z)= u - iv$ is an analytic function of $z = x + iy$ and $u-v=\Large\frac{e^y-\cos x+\sin x}{\cosh y - \cos y}$, find $f(z)$ subject to condition, $f\left(\frac{\pi}2\right)=\large \frac{3 - i}2$.

I haven't solved such problems before, I tried to use $\cosh y = \frac{e^y + e^{-y}}2$ but it didn't help.

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Here is a strategy outline, it seems like no fun to execute due to the complicated function. We know (because $f$ is analytic and written funny) that $u_x = -v_y$ and $u_y = v_x$. We can compute $u_x - v_x$ and $u_y - v_y$ using the explicit formula. So we have four linear equations in the four "variables" $u_x, u_y, v_x, v_y$ (whose coefficients are functions), so you can solve for them.

Now if you know $u_x$ and $u_y$ you can recover $u$ up to a constant (integrate $u_x$ with respect to $x$, you'll get something well-defined up to a function of $y$, now differentiate with respect to $y$ to figure out that function of $y$ up to a constant); same for $v$. To solve for those two unknown constants, you can use your explicit value of $f(\pi/2)$ much as you would when integrating in a first calculus course.

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