Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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 $D$).

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.