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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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@jim How did you work that out? – Adam Nov 26 '12 at 13:58
$f(z)=(a+id)z^3 +i*c$ where c is a constant – jim Nov 26 '12 at 14:07

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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Probably shouldn't use $c$, since it is different from the $c$ in the original problem – Thomas Andrews Nov 26 '12 at 15:20
thanks for the response – jim Nov 27 '12 at 5:47
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. – Thomas Andrews Nov 27 '12 at 14:29

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