- #1

ralqs

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So, my complex analysis professor defined [tex] \partial f / \partial z^*[/tex] as

[tex]\frac {\partial f}{\partial z^*} = \frac {1}{2} \left( \left(\frac {\partial u}{\partial x}-\frac {\partial v}{\partial y}\right) + i\left(\frac {\partial u}{\partial y} + \frac {\partial v}{\partial x}\right)\right)[/tex]

where [tex]z = x + iy[/tex] and [tex]f(z) = u(x,y) + iv(x,y)[/tex]. My prof then showed that [tex] \partial f / \partial z^* = 0[/tex] when f is differentiable.

So my question is, where does this definition come from? It behaves exactly as you would expect it to by the notation (ie [tex] (\partial / \partial z^*) zz^* = z[/tex]), but the definition doesn't make this obvious.

[tex]\frac {\partial f}{\partial z^*} = \frac {1}{2} \left( \left(\frac {\partial u}{\partial x}-\frac {\partial v}{\partial y}\right) + i\left(\frac {\partial u}{\partial y} + \frac {\partial v}{\partial x}\right)\right)[/tex]

where [tex]z = x + iy[/tex] and [tex]f(z) = u(x,y) + iv(x,y)[/tex]. My prof then showed that [tex] \partial f / \partial z^* = 0[/tex] when f is differentiable.

So my question is, where does this definition come from? It behaves exactly as you would expect it to by the notation (ie [tex] (\partial / \partial z^*) zz^* = z[/tex]), but the definition doesn't make this obvious.

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