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I'm unsure of which Cauchy-Riemann law to use when I'm given either a real or imaginary function. For instance. I might be given a real function and asked to work out the imaginary part.

For instance, if I'm given the real part: $-3xy^2-2y^2+x^3+2x$ and asked to work out the imaginary, then I'd need to use the $\frac{du}{dx}=\frac{dv}{dy}$ rule rather than the $-\frac{du}{dy}=\frac{dv}{dx}$ rule before finding the imaginary part. Why is this?

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    Are you sure this is the correct expression for $u$? It is not harmonic.2012-05-11

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You need both. Let us take $u(x,y)=-3xy^2+x^3+2x+y.$ Then we get $\frac{\partial u}{\partial x} = -3y^2+3x^2+2 = \frac{\partial v}{\partial y}.$ Integrating with respect to $y$ leaves us with $v(x,y) = -y^3+3x^2y+2y + C(x),$ noting that the integration constant could be different for different $x$. To find $C(x)$, you would use the other Cauchy-Riemann equation: $\frac{\partial v}{\partial x} = 6xy + C'(x) $ and $-\frac{\partial u}{\partial y} = 6xy-1 $ and these should be equal, so $C'(x)=-1$. This implies $ C(x) = -x+D $ for some constant (really constant, this time) $D$. The final result is then $ v(x,y) = -y^3+3x^2y+2y-x+D .$ Note that a harmonic conjugate is only defined up to a constant (in this case it's called $D$).

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    You may wonder why I chose this $u$, and not the one you mentioned. There is a good reason for this: the $u$ you mentioned is not a harmonic function (did you type it correctly?), so it cannot be the real part of a holomorphic function. I could have just left out the term $-2y^2$, but then we would have gotten $C'(x)=0$, which by coincidence defeats the point I was trying to make.2012-05-11