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 that$dw/dz = -1/z^2$
-------------

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!