Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

As a function on $\mathbb R^2$, I want to compute the Jacobian of $f(z)=z^n$. Is there an easy way to this? Write $z=x+iy$ .. and compute real part and imaginary part of $f$ and differentiate with respect to $x,y$ seems to be very tedious work...

share|cite|improve this question
up vote 3 down vote accepted

For a holomorphic function $f$, its Jacobian matrix is $\left(\begin{array}{cc}\mathrm{Re}~f' & -\mathrm{Im}~f'\\\mathrm{Im}~f' &\mathrm{Re}~f' \end{array}\right)$, and its Jacobian determinant is $|f'|^2$. Please see Cauchy-Riemann equations for reference.

share|cite|improve this answer
I thought he meant the Jacobian matrix, not the determinant. – Harald Hanche-Olsen Oct 31 '12 at 17:34
oh... I'm enough for determinant. – Detectives Oct 31 '12 at 17:38
I thought I found some way. since power function can easily be seen if we see in polar coordinate, try to see f as functions of r,theta and then try to apply chain rule – Detectives Oct 31 '12 at 17:39
@HaraldHanche-Olsen: I have updated. – 23rd Oct 31 '12 at 17:49
@mathlover: have you learned some complex analysis? If so, then compared with using polar coordinate, compute the complex derivative is a little more convenient in this case. – 23rd Oct 31 '12 at 17:54

Let $z=x+iy, \quad f(x,\ y)=u(x,\ y)+iv(x,\ y).$ Using polar coordinates \begin{gather} x=\rho \cos{\varphi}\\ y=\rho \sin{\varphi} \end{gather} and the chain rule, Jacobian matrix \begin{gather} \dfrac{\partial(u,\, v)}{\partial(x, \, y)}=\pmatrix{\dfrac{\partial u}{\partial x} & \dfrac{\partial u}{\partial y} \\ \dfrac{\partial v}{\partial x} & \dfrac{\partial v}{\partial y}}=\dfrac{\partial(u,\, v)}{\partial(\rho, \, \varphi)}\cdot \dfrac{\partial(\rho, \, \varphi)}{\partial(x, \, y)}. \end{gather}

share|cite|improve this answer

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.