Take [itex]x=r\cos\theta[/itex] and [itex]y=r\sin\theta[/itex](adsbygoogle = window.adsbygoogle || []).push({});

If [itex]f(z)=u(r,\theta) + iv(r,\theta)[/itex], is analytic with u and v real, show that the derivative is given by

[tex]f'(z) = \left( \cos\theta \frac{\partial u}{\partial r}- \sin\theta\frac{1}{r}\frac{\partial u}{\partial\theta}\right) + i\left( \cos\theta\frac{\partial v}{\partial r} - \sin\theta\frac{1}{r}\frac{\partial v}{\partial\theta} \right)[/tex]

Since f is analytic, I use the result

[tex]f'(z)=\frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x}[/tex]

Though this seems to give

[tex]f'(z) = \left( (1/\cos\theta) \frac{\partial u}{\partial r}- (1/\sin\theta)\frac{1}{r}\frac{\partial u}{\partial\theta}\right) + i\left( (1/\cos\theta)\frac{\partial v}{\partial r} - (1/\sin\theta)\frac{1}{r}\frac{\partial v}{\partial\theta} \right)[/tex]

Can anyone see why this isn't correct?

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# Analytic Function Derivative

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