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    • Thread Starter

    I'm given that f_1(x+iy) = u_1(x,y)+iv_1(x,y) and f_1 is differentiable everywhere on the complex plane.

    Given that u_1(x,y) = 2x^2 - 2xy - 2y^2 and f_1(1) = 2, I'm supposed to find f_1(z).

    I found v_1(x,y) = 4xy - y^2 using the Cauchy Riemann equations.

    So f_1(z) = u_1(x,y) + iv_1(x,y) = 2x^2 - 2xy - 2y^2 + 4xyi - y^2 i

    Which is really annoying to simplify!

    I've gotten to the point where

    f_1(z) = 2(x+iy)^2 -y^2i -2xy and now I'm stuck.
    • Study Helper

    Study Helper
    (Original post by wanderlust.xx)
    I've gotten to the point where

    f_1(z) = 2(x+iy)^2 -y^2i -2xy and now I'm stuck.
    What about using z+\bar{z}=2x etc. as a basis for building the trailing parts of that formula.

    You have the wrong eqn for v1(x,y)

    When using the Cauchy-Riemann eqns, you will be doing partial integration which means you should have a function of integration rather than a constant of integration to include in your eqn for v1(x,y).

    It then becomes simple to simplify!

    BTW, are you sure that f1(1) = 2 and not f1(1) = 2+i, ?
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