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freckles

  • one year ago

\[\text{ Suppose } f \text{ is an analytic function of the complex variable } z=x+i y \text{ given by } \\ f(z)=(2x+3y)+i g(x,y) \\ \text{ where} g(x,y) \text{ is a real-valued function } \\ \text{ of the real variables } x \text{ and } y . \\ \text{ If } g(2,3)=1\text{, then }g(7,3)=? . \]

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  1. freckles
    • one year ago
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    \[f(2+3i)=4+9+ig(2,3) \\ f(2+3i)=13+i g(2,3) \\ f(2+3i)=13+i \\ f(7+3i)=14+9+ig(7,3) \\ f(7+3i)=23+i g(7,3) \\ \] this as far as I can get so far

  2. freckles
    • one year ago
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    There are choices A) -14 B) -9 C) 0 D) 11 E) 18 This question comes from the mathematics gre practice exam.

  3. xapproachesinfinity
    • one year ago
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    hmm interesting, i have some admiration for complex though i have no skills in such things yet lol

  4. anonymous
    • one year ago
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    I believe it might have something to do with cancelling the complex part.

  5. xapproachesinfinity
    • one year ago
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    i have a feeling that it is not that hard just kind of a trick

  6. freckles
    • one year ago
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    was kind of thinking that too

  7. freckles
    • one year ago
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    somehow we have to show it is -14

  8. freckles
    • one year ago
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    maybe and I could be totally crazy we could do something with magnitudes of complex numbers

  9. freckles
    • one year ago
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    \[|g(2,3)|=1 \\ |f(2,3)|=\sqrt{170} \\ |g(7,3)|=\sqrt{a^2} \text{ if } g(7,3)=a \\ |f(7,3)|=\sqrt{23^2+g^2(7,3)}\]

  10. freckles
    • one year ago
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    \[f^2(7,3)=23^2+a^2 \]

  11. xapproachesinfinity
    • one year ago
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    hmm

  12. freckles
    • one year ago
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    still playing

  13. xapproachesinfinity
    • one year ago
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    you -14 was the answer! did you just tried it directly?

  14. xapproachesinfinity
    • one year ago
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    said*

  15. freckles
    • one year ago
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    no I looked at the answers

  16. xapproachesinfinity
    • one year ago
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    oh ok

  17. freckles
    • one year ago
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    how did you know -14 was right?

  18. xapproachesinfinity
    • one year ago
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    no i just saw your reply above mentioning that

  19. freckles
    • one year ago
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    I wonder what analytic means

  20. freckles
    • one year ago
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    looking up...

  21. xapproachesinfinity
    • one year ago
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    isn't that related to graphs ?

  22. xapproachesinfinity
    • one year ago
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    don't know why wolfram ignores the imaginary stuff for such functions

  23. freckles
    • one year ago
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    https://www3.nd.edu/~atassi/Teaching/ame60612/Notes/analytic_functions.pdf go to page 2

  24. freckles
    • one year ago
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    \[f(x,y)=u(x,y)+i v(x,y) \\ \frac{\partial u}{ \partial x}=\frac{\partial v}{ \partial y} \\ \frac{\partial u}{ \partial y}=-\frac{ \partial v }{\partial x} \\ 2=g_y \\ 3=-g_x\]

  25. anonymous
    • one year ago
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    Analytic functions refers to something else. One important property of analytic functions are that they are indefinitely differentiable.

  26. xapproachesinfinity
    • one year ago
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    you know it crossed my mind that we need to use calculus lol

  27. freckles
    • one year ago
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    \[g(x,y)=2y+C(x) \\ -g(x,y)=3x+K(y) \\ \\ 0=2y+3x+C(x)+K(y) \] where C(x) is a function of x and K(y) is a function of y

  28. xapproachesinfinity
    • one year ago
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    so any analytic function has no differentiability problem is that what you are saying?

  29. freckles
    • one year ago
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    \[2g(x,y)=2y-3x+C(x)-K(y) \\ 2g(2,3)=2(3)-3(2)+C(2)-K(3) \\ g(2,3)=C(2)-K(3)\]

  30. freckles
    • one year ago
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    err...

  31. xapproachesinfinity
    • one year ago
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    oh i see! after checking the theorem in that pdf

  32. freckles
    • one year ago
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    lol still trying to play with this fungus

  33. freckles
    • one year ago
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    and that is equal to 1

  34. freckles
    • one year ago
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    \[1=C(2)-K(3)\]

  35. freckles
    • one year ago
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    \[2g(7,3)=2(3)+3(7)+C(7)-K(3)\]

  36. freckles
    • one year ago
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    \[2g(7,3)=27+C(7)-K(3)\]

  37. freckles
    • one year ago
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    \[2g(7,3)-1=27+C(7)-C(2)\]

  38. anonymous
    • one year ago
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    Oh yea, use the Cauchy–Riemann equations to solve this.

  39. freckles
    • one year ago
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    \[2g(7,3)=28+C(7)-C(2) \\ g(7,3)=14+\frac{C(7)-C(2)}{2}\]

  40. freckles
    • one year ago
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    oops i made a type-o somewhere above

  41. freckles
    • one year ago
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    \[g(x,y)=2y+C(x) \\ -g(x,y)=3x+K(y) \\ \\ 0=2y+3x+C(x)+K(y)\] \[2g(x,y)=2y-3x+C(x)-K(y) \\ 2g(2,3)=2(3)-3(2)+C(2)-K(3) \\ g(2,3)=C(2)-K(3)\] \[C(2)-K(3)=1 \\ 2g(7,3)=6-21+C(7)-K(3) \\ 2g(7,3)=-15+C(7)-K(3) \\ 2g(7,3)-1=-15+C(7)-K(3)-C(2)+K(3) \\ 2g(7,3)=-14+C(7)-C(2) \\ g(7,3)=-7+\frac{C(7)-C(2)}{2}\]

  42. freckles
    • one year ago
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    I think I'm still making a mistake somewhere

  43. anonymous
    • one year ago
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    Oh I get it now.

  44. anonymous
    • one year ago
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    You simply use the Cauchy-Riemann equations above, and calculate the partials. There, you had \[2 = g_{y}\]

  45. anonymous
    • one year ago
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    The C-R equations are your best bet. It becomes a problem similar to finding potential functions.

  46. anonymous
    • one year ago
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    and \[3 = -g_{x}\]

  47. anonymous
    • one year ago
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    Yes, and then you find the potential function by adding the constant term.

  48. anonymous
    • one year ago
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    So your potential function g(x,y) becomes -3x + 2y + C

  49. anonymous
    • one year ago
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    But you are given an initial condition, g(2,3)=1. Plugging it back into our potential function, you will find that C = 1.

  50. anonymous
    • one year ago
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    Therefore, you get \[g(x,y) = -3x+2y+1\]

  51. anonymous
    • one year ago
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    So just plug in, g(7,3) and you will get -14.

  52. freckles
    • one year ago
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    @Loser66 check this out

  53. freckles
    • one year ago
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    I will still need to look at that theorem thing because I think I was using a little wrong. lol like I thought I was supose to write g_x=-3 implies g(x,y)=-3x+K(y) where K is a function of y and g_y=2 implies g(x,y)=2y+C(x) where C is a function of x but I kept getting ugly stuff

  54. freckles
    • one year ago
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    thanks for your help guys

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