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Homework Help: Find f(z) if possible from u(x,y) and v(x,y)

  1. Nov 4, 2012 #1
    1. The problem statement, all variables and given/known data

    a. u(x,y)=(x+2)2-y2, v(x,y)=2(x+2)y
    b. u(x,y)=cosxcoshy, v(x,y)=-sinxsinhy
    c. u(x,y)=(x2+y2+y)/(x2+y2), v(x,y)=x/(x2+y2)
    d. u(x,y)=eycosx, v(x,y)=eysinx

    3. The attempt at a solution

    a. f(x+iy)=(x+2)2-y2+i2(x+2)y

    b. f(x-iy)=cosxcoshy-isinxsinhy
    Am I allowed to answer this question as f([itex]\bar{z}[/itex])=cos[itex]\bar{z}[/itex]?

    Same as b... f([itex]\bar{z}[/itex])=([itex]\bar{z}[/itex]-i)/[itex]\bar{z}[/itex]?

    d. du/dx != dv/dy, fails Cauchy-Reimann?

    Edit: How do I go from x-iy to x+iy? This escapes me. :(
    Last edited: Nov 4, 2012
  2. jcsd
  3. Nov 5, 2012 #2


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    Homework Helper

    Welcome to PF!

    Hi IHazAName! Welcome to PF! :smile:

    Is the question to find an f such that f = u + iv ?

    a and b look ok

    (for b, just replace y by -y :wink:)

    I haven't checked c

    For d, isn't cauchy-riemann only relevant to differentiability? Try again :wink:
  4. Nov 5, 2012 #3


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    If [itex]f(\bar{z})= cos(\bar{z})[/itex], then f(z)= cos(z).

    And, as in b, if [itex]f(\bar{z})= (\bar{z}- i)/\bar{z}[/itex] then f(z)= (z- i)/z.

    The problem asked you to find a function- it did not say the function had to be analytic.

    If z= x- iy, then [itex]\bar{z}= x+ iy[/itex].
  5. Nov 5, 2012 #4
    Thanks guys!

    ...Oops. I forgot to mention that...these are supposed to be analytic functions as that is what we're learning. :(

    So, just to make extra sure: for f(x-iy)=cos(x)cosh(y)-isin(x)sinh(y) that is still f(z)=cos(z)? Even though using x+iy grants f(x+iy)=cos(x)cosh(y)+isin(x)sinh(y)

    I guess my confusion with that in our notes we only solve it one way...where we're given a f(z)...say f(z)=z and plug in x+iy so that f(z) becomes f(x+iy)=x+iy, and solve for u(x,y) and v(x,y). In these problems I was just trying to go in the reverse direction, so getting a result that needed x-iy was confusing to me.
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