# Trouble with Cauchy Riemann…not sure which law to use?

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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I assume you are talking about finding a harmonic conjugate (so you are given the real part of a holomorphic function and you're supposed to find the imaginary part)? It's not entirely clear. In that case, I'm pretty sure you would need both Cauchy-Riemann equations. – Daan Michiels May 11 '12 at 18:39
Hi, yes. Although from my solution here, I've only used one: u=−3xy^2 −2y^2 +x^3 +2x^2 ￼￼￼￼∂u/∂x = −3 y2 + 3 x2 + 4 x = ∂v/∂y by C-R ￼￼Hence v = −y3 + 3 x2 y + 4 x y – Flo May 11 '12 at 18:46
Oh! I think I've got you now...thanks for the help! – Flo May 11 '12 at 18:51
Once you know $\partial v/\partial y$, you can find $v$ by integrating with respect to $y$ (I assume this is what you did). However, this gives a constant of integration that still depends on $x$. To find the integration constant, you need the other equation. – Daan Michiels May 11 '12 at 18:54
Are you sure this is the correct expression for $u$? It is not harmonic. – Daan Michiels May 11 '12 at 19:43

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 $D$).
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. – Daan Michiels May 11 '12 at 19:13