Complex function C-R stuff

Hey,

I've been struggling with this tute question,

http://img200.imageshack.us/img200/6955/unledcjb.jpg

I really don't understand where im supposed to be going,

I've been trying to show the u derivatives are zero because the hint says to do that,

So far ive just been messing around with

f(z) = u + iy, |f(z)|=u^2+y^2 =k

then i took d/dx partial derivative

2u(du/dx) + 2v(dv/dx) = 0 all partial derivatives

Now im stuck plugging in C-R equations trying to find a way to show du/dx and du/dy are zero,

I don't really understand what to do after that also,

Thanks in advanced,

Re: Complex function C-R stuff

Quote:

Originally Posted by

**Daniiel** Hey,

I've been struggling with this tute question,

http://img200.imageshack.us/img200/6955/unledcjb.jpg
I really don't understand where im supposed to be going,

I've been trying to show the u derivatives are zero because the hint says to do that,

So far ive just been messing around with

f(z) = u + iy, |f(z)|=u^2+y^2 =k

then i took d/dx partial derivative

2u(du/dx) + 2v(dv/dx) = 0 all partial derivatives

Now im stuck plugging in C-R equations trying to find a way to show du/dx and du/dy are zero,

I don't really understand what to do after that also,

Thanks in advanced,

If $\displaystyle |f(z)|=k= \text{constant}$, then is...

$\displaystyle f(z)= k\ e^{i\ \theta(x,y)}$ (1)

Now if we write the C-R equations in 'nonstandard' form...

$\displaystyle u_{x}-v_{y}=0$

$\displaystyle u_{y}+v_{x}=0$ (2)

... from (1) and (2) we obtain...

$\displaystyle \sin \theta\ \frac{d \theta}{dx} + \cos \theta\ \frac{d \theta}{d y}=0$

$\displaystyle \cos \theta\ \frac{d \theta}{dx} - \sin \theta\ \frac{d \theta}{d y}=0$ (3)

Now the linear equations system (3) is nonsingular because its determinant is $\displaystyle \delta=-1 \ne 0$, so that the (3) has the only solution $\displaystyle \frac{d \theta}{d x} = \frac{d \theta}{d y} =0$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

Re: Complex function C-R stuff

Thanks alot,

So you let u = cos(theta) v=sin(theta) and f(z) was = k(u+iv)

im a little confused though,

If |f(z)|=k=|u+iv|

how can f(z) = k(u+iv) ?

Re: Complex function C-R stuff

Quote:

Originally Posted by

**Daniiel** Thanks alot,

So you let u = cos(theta) v=sin(theta) and f(z) was = k(u+iv)...

Not exactly!... the correct procedure is...

$\displaystyle |f(z)|=k \implies f(z)= k\ (\cos \theta + i\ \sin \theta) \implies u= k\ \cos \theta\ ,\ v=k\ \sin \theta $

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

Re: Complex function C-R stuff

Sorry i'm still a little confused,

I don't understand how f(z)=k(u+iv) and how its not just f(z) = k

because k = u^2 + v^2 or 1 if using cos and sin

k=|f(z)|=|u+iv|

so f(z) = k(u+iv) = k only if u = 1 and v = 0

Re: Complex function C-R stuff

Ok that parts all sweet now,

But how do I apply this in answering the question about |z|<100,

do i just say as it is true for |f(z)| and that = k

it is also true for f(z) since it also = k?