Find the holomorphic function f(x+iy) such that Re f(x+iy)= cosh(3y)sin(3x) and f(0)=0

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- anonymous

Find the holomorphic function f(x+iy) such that Re f(x+iy)= cosh(3y)sin(3x) and f(0)=0

- katieb

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- anonymous

ugh i havent taken calc yet so i cant do this

- anonymous

cauchy reimann yes?

- anonymous

yes satellite

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## More answers

- anonymous

so the derivative of this wrt x must be the derivative of the imaginary part wrt y right? the derivative is
\[3\cos(3x)\cosh(3y)\] integrate wrt y and get
\[\cos(3x)\sinh(3y)\] i think.

- anonymous

it's been a while. am i on the right track?

- anonymous

you also need the partial wrt y is minus the partial wrt x of the imaginary part.
if we are lucky it already is
oh and you also need that f(0)=0 i forgot the +C when i integrated

- anonymous

how to you like that? i miracle! it works

- anonymous

so i guess unless i totally screwed this up the answer is
\[f(x+iy)=\cosh(3y)\sin(3x)+\cos(3x)\sinh(3y)\]

- anonymous

think we lost raheen maybe he will be back

- anonymous

rather
\[f(x+iy)=\cosh(3y)\sin(3x)+i\cos(3x)\sinh(3y)\]

- anonymous

i forgot the i part

- anonymous

@raheen look ok?

- anonymous

it does not come with a money guarantee because it has been several years but i think it looks good. you can check cauchy reimann equations for this and see if they work

- anonymous

satellite you are about to reach , great work I think you did
how about using the 2 conditions of C-R
then you need to integrate and don forget to find the constant

- anonymous

well i integrated wrt y and got the answer. then i checked that the second equation and it worked.
and by inspection you can see that
\[f(0)=0\]

- anonymous

i mean if the C-R equations are going to work, then after i take the derivative wrt x and integrate wrt y , then the second condition
\[\frac{\delta u}{\delta y}=-\frac{\delta v}{\delta x}\] had better work or else we are screwed. both conditions must hold.
but in any case i checked and they do

- anonymous

that's so great satellite, thank you.

- anonymous

yw

- Zarkon

the only thing I would add is that when you integrated with respect to y you don't get C you get some function of x

- anonymous

Zarkon, thank you, you are right it's C(x)

- Zarkon

It doesn't look like it will change the final answer though

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