Verify Complex Derivative Cauchy-Riemann Eqn.

[mattmns]

The question is as follows.

Let w = 1/z. Check that for z \neq 0 the Cauchy-Riemann equations are satisfied and verify thatdw/dz = -1/z^2
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So I let z = x + iy

Then, \frac{1}{z} = \frac{1}{x + iy} = \frac{1}{x + iy}\frac{x - iy}{x - iy} = \frac{x - iy}{x^2 + y^2}

let w = u + iv
so u = \frac{x}{x^2 + y^2} and v = \frac{-y}{x^2 + y^2}

\frac{\partial u}{\partial x} = \frac{-x^2 + y^2}{(x^2 + y^2)^2} = \frac{\partial v}{\partial y}

\frac{ - \partial u}{\partial y} = \frac{2xy}{(x^2 + y^2)^2 } = \frac{\partial v}{\partial x}

I really don't want to write everything out, but the Cauchy-Riemann equations are satisfied, and thus we can find dw/dz = \frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x}

My problem is right here when I am trying to manipulate this \frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x} to get \frac{1}{z^2} = \frac{1}{(x + iy)^2} = \frac{1}{x^2 + i2xy - y^2}

Is there some trick I am missing or did I take the partial derivatives wrong? I just seem to be stuck here with no ideas. Thanks

 

[quasar987]

You're going to say "duh!". The first thing you found is that

\frac{1}{z} = \frac{x - iy}{x^2 + y^2}

remember? Square that, and you get exactly

-\frac{\partial u}{\partial x} - i\frac{\partial v}{\partial x}

 

[mattmns]

Duh!

Thank you!