Thread: Complex function C-R stuff

Hey,

I've been struggling with this tute question,



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,

Originally Posted by Daniiel

Hey,

I've been struggling with this tute question,



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 , then is...

(1)

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



(2)

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



(3)

Now the linear equations system (3) is nonsingular because its determinant is , so that the (3) has the only solution ...

Kind regards

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) ?

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...



Kind regards

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

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?