Question 1172367
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If function f = u + iv is analytic when u = Sin(x).Cosh(y) then what is the {{{highlight(cross(value))}}} expression of v?
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<pre>
If the complex-value function  f = u + iv  is analytic, then the Cauchy-Riemann equations are held

    {{{(du)/(dx)}}} = {{{(dv)/(dy)}}}      (1)

and

    {{{(du)/(dy)}}} = - {{{(dv)/(dx)}}}    (2)


(as the reference, see this Wikipedia article https://en.wikipedia.org/wiki/Cauchy%E2%80%93Riemann_equations ).


    I am very sorry, but in these equations the symbol "d" must be a Greek letter ("d rounded"), not a "d" Latin,
    but in this editor I can not reproduce it, unfortunately.  So, concider this "d" as "d rounded" in all my 
    formulas in this post).



From equation (1), after differentiating  {{{(du)/(dx)}}}, we have

    {{{(dv)/(dy)}}} = cos(x)*cosh(y)      (3)


From equation (2), after differentiating  {{{(du)/(dy)}}}, we have

    {{{(dv)/(dx)}}} = -sin(x)*sinh(y)     (4)


So, we need to find function v  as antiderivative from equations  (3) and (4).


It is easy to do:  the function  v(x,y) = cos(x)*sinh(y)  satisfies both equations  (3) and (4).


THEREFORE,  the answer to the problem's question is  v = cos(x)*sinh(y).
</pre>

Solved.


You may add an arbitrary constant value to &nbsp;"v" &nbsp;to provide &nbsp;"a general view" to satisfy your professor :).