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Math Help - Complex variables : f(z) = zIm(z^2)?

  1. #1
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    Complex variables : f(z) = zIm(z^2)?

    Complex variables: f(z) = z Im(z^2)?

    If f(z) = z Im(z^2), determine the points in the complex plane such that f is differentiable.
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    Re: Complex variables : f(z) = zIm(z^2)?

    Hey mous99.

    What have you tried for this question?

    (Hint: Something is differentiable if it satisfies the Cauchy-Riemann equations).
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  3. #3
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    Re: Complex variables : f(z) = zIm(z^2)?

    Let z = x+ iy z^2 = (x^2-y^2) + 2ixy or,
    im(z^2 = 2ixy F(z) = z (2ixy))

    Hence F(z) is differentiable at all points in complex plain except at (x,y)=(0,0) that is origin.

    Is this answer right or wrong?
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  4. #4
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    Re: Complex variables : f(z) = zIm(z^2)?

    Quote Originally Posted by mous99 View Post
    Let z = x+ iy z^2 = (x^2-y^2) + 2ixy or,
    im(z^2 = 2ixy F(z) = z (2ixy))

    You have a serious mistake.

    \text{Im}(z^2)=2xy. NOT 2xyi.
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  5. #5
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    Re: Complex variables : f(z) = zIm(z^2)?

    why {Im}(z^2)=2xy?
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    Re: Complex variables : f(z) = zIm(z^2)?

    Quote Originally Posted by mous99 View Post
    why {Im}(z^2)=2xy?
    If indeed you do not know that, then you have no business trying this question.

    If z=x+yi then \text{Re}(z)=x~\&~\text{Im}(z)=y.

    EXAMPLE:
    \text{Re}(xy+x^2i)=xy~\&~\text{Im}(xy+x^2i)=x^2.
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  7. #7
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    Re: Complex variables : f(z) = zIm(z^2)?

    Let z = x+ yi
    Im (z2) = Im (x+yi) ^2
    = Im (x^2- y ^ 2 +2xyi)


    then how to continue for Im (z2) = 2xy???
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    Re: Complex variables : f(z) = zIm(z^2)?

    Quote Originally Posted by mous99 View Post
    Let z = x+ yi
    then how to continue for Im (z2) = 2xy???

    z\text{Im}(z^2)=(2x^2y)+i(2xy^2)
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  9. #9
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    Re: Complex variables : f(z) = zIm(z^2)?

    yes, i have same answer with you.
    but how to determine the points 2x^2 y + 2xy^2 i in the complex plane such that f is differentiable?
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    Re: Complex variables : f(z) = zIm(z^2)?

    Quote Originally Posted by mous99 View Post
    yes, i have same answer with you.
    but how to determine the points 2x^2 y + 2xy^2 i in the complex plane such that f is differentiable?
    (Hint: Something is differentiable if it satisfies the Cauchy-Riemann equations).
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  11. #11
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    Re: Complex variables : f(z) = zIm(z^2)?

    f(z) = z Im (z^2) = (x+iy) (2xy) = 2(x^2)y + 2x(y^2)i



    determine the points in the complex plane such that f is differentiable.

    df/dx=2xy+2y(x+iy), so i*df/dx=2ixy+2iy(x+iy)

    And df/dy=2ixy+3x(x+iy)

    This tell us f(z) is differentiable when x=iy.


    is it right?
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