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raheen

  • 3 years ago

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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  1. hahd
    • 3 years ago
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    ugh i havent taken calc yet so i cant do this

  2. satellite73
    • 3 years ago
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    cauchy reimann yes?

  3. raheen
    • 3 years ago
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    yes satellite

  4. satellite73
    • 3 years ago
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    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.

  5. satellite73
    • 3 years ago
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    it's been a while. am i on the right track?

  6. satellite73
    • 3 years ago
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    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

  7. satellite73
    • 3 years ago
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    how to you like that? i miracle! it works

  8. satellite73
    • 3 years ago
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    so i guess unless i totally screwed this up the answer is \[f(x+iy)=\cosh(3y)\sin(3x)+\cos(3x)\sinh(3y)\]

  9. satellite73
    • 3 years ago
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    think we lost raheen maybe he will be back

  10. satellite73
    • 3 years ago
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    rather \[f(x+iy)=\cosh(3y)\sin(3x)+i\cos(3x)\sinh(3y)\]

  11. satellite73
    • 3 years ago
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    i forgot the i part

  12. satellite73
    • 3 years ago
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    @raheen look ok?

  13. satellite73
    • 3 years ago
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    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

  14. raheen
    • 3 years ago
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    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

  15. satellite73
    • 3 years ago
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    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 \[f(0)=0\]

  16. satellite73
    • 3 years ago
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    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

  17. raheen
    • 3 years ago
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    that's so great satellite, thank you.

  18. satellite73
    • 3 years ago
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    yw

  19. Zarkon
    • 3 years ago
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    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

  20. raheen
    • 3 years ago
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    Zarkon, thank you, you are right it's C(x)

  21. Zarkon
    • 3 years ago
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    It doesn't look like it will change the final answer though

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