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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...

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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.

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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}

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