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This is a question in Complex Analysis for Mathematical Science and Engineering by Saff and Snider. It's on pg 62.

Question: The Jacobian of a mapping $u = u(x,y)$ $v = v(x,y)$ from the xy-plane to the uv-plane is defined to be the determinant $$ J(x_0,y_0) := \begin{vmatrix} \frac{\partial{u}}{\partial{x}} && \frac{\partial{u}}{\partial{y}} \\ \frac{\partial{v}}{\partial{x}} && \frac{\partial{v}}{\partial{y}} \end{vmatrix}$$

Where the derivatives are all evaluated at $(x_0, y_0)$. Show that if $f = u + iv$ is analytic at $z_0 = x_0 + iy_0$, then $J(x_0,y_0) = |f'(z_0)|^2$.

My attempt to answer: Given f'(z) is analytic $f'(z) = \frac{\partial{u}}{\partial{x}} + i\frac{\partial{v}}{\partial{x}}$ which leads us to $|f'(z_0)|^2 = \frac{\partial{u}}{\partial{x}}^2 - \frac{\partial{v}}{\partial{x}}^2$. We can now use the Cauchy-Riemann equations as substitutions; $$|f'(z_0)|^2 = \frac{\partial{u}}{\partial{x}}\left(\frac{\partial{y}}{\partial{y}} \right)- \frac{\partial{v}}{\partial{x}}\left(-\frac{\partial{v}}{\partial{x}}\right) $$

But now the minus signs in the second term cancel and it is positive. The second term from the determinant should be negative. What did I do wrong?

Jekowl
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1 Answers1

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$$|f'(z_0)|^2 = f'(z_0) \overline{f'(z_0)} = (\frac{\partial{u}}{\partial{x}} + i\frac{\partial{v}}{\partial{x}})(\frac{\partial{u}}{\partial{x}} - i\frac{\partial{v}}{\partial{x}}) = \left(\frac{\partial u}{\partial x}\right)^2 + \left(\frac{\partial v}{\partial y}\right)^2$$