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By taking the partial derivitives and second partialy derivative with respect to x and y I have found that b=-3d and c=-3a but I don't understand the last bit of the question or how to do it.

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    $f(z)=(a+id)z^3 +i*c$ where c is a constant2012-11-26

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Using what you have so far, write $u(x,y) = ax^3 -3d x^2y -3a xy^2 + dy^3.$

The Cauchy-Riemann equations give $\frac{\partial v}{\partial y} = \frac{\partial u}{\partial x} = 3ax^2 -6dxy-3ay^2$

and $ \frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y}= 3dx^2+6axy -3dy^2.$

The first equation then tells you that $v(x,y) = 3ax^2 y-3dxy^2-ay^3 +F(x)$ while the second gives $ v(x,y) = dx^3+ 3ax^2y-3dxy^2 + G(y).$

Putting these together yields $v(x,y) = 3ax^2y-3dxy^2-ay^3+dx^3$ up to a additive constant.

Thus the function $f(x+iy) = (ax^3 -3d x^2y -3a xy^2 + dy^3) + i(3ax^2y-3dxy^2-ay^3+dx^3) +K$ is analytic with the required real part. For a nicer form, we can let $y=0$, which gives $ f(x) = ax^3+idx^3+K.$ Since that is the function represented on the real line, our function can be seen to be $f(z) = (a+id)z^3+K.$

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$u(x,y)=ax^3 -3dx^2y-3axy^2+dy^3$ from $CR$ $equation $ you get $v(x,y)=3ax^2y -ay^3 -3dxy^2+dx^3 +k$ where k is a constant from this you can get $f(z)$

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    Apologies for the downvote - I accidentally pressed downvote when I meant upvote, and thought I corrected it, only to find it is still registering my vote as a downvote (and it won't let me correct it 8 hours later.) Something is broken about the way it deals with votes from the iPad, I think.2012-11-27