Complex conjugates

[Hypnotoad]

I have a couple questions on complex variables:

1.)If you have a complex function defined as follows:
f(z)=u(x,y)+iv(x,y)
with x,y real, what do you get if you take the complex conjugate of the variable z?
f(z^*)=?

I was thinking that it wouldn't change since the complex variable has been replaced with two real variables, but that doesn't seem right.

If I take the conjugate of the entire function, is this what I should get:

f^*(z)=u(x,y)-iv(x,y)


2.) If you are trying to prove that a function is not analytic at a specific point, is it sufficient to show that the Cauchy-Riemann conditions do not hold? I'm trying to show that the derivative of a function at zero is dependant on the direction that you approach zero. I've shown that the C-R conditions are not met, but I'm not sure how to show explicitly for that point that they are not met.


EDIT: NEW QUESTION

I figured I would just edit this topic instead of starting a new one. I'm trying to prove that if a function f(z) is analytic then the function f^{*}(z^{*}) is also analytic.

I'm not sure how to get this started. It makes sense that taking the conjugate wouldn't affect the differentiability, but I don't know how to prove that. Any hints on how to get this problem started?

[Tide]

To get the complex conjugate of f(z) replace y with -y.

[Hypnotoad]

Is that just a general rule, or is that something I should be trying to prove?

[kenhcm]

z is defined by x+iy. Therefore, the function f(z) is equivalent to a function f(x,y) with two arguments. Since z^\ast=x-iy, f(z^\ast) is similarly equivalent to a function f(x,-y).

Hope that this is the proof you are looking for.

Best regards,
Kenneth

[Hypnotoad]

Thanks, that makes a lot of sense.

[Hypnotoad]

I added a question to the first post. I'd appreciate any help you can offer.

[kenhcm]

Given f(z)=u(x,y)+iv(x,y) is analytic, we have

\frac{\partial u(x,y)}{\partial x}=\frac{\partial v(x,y)}{\partial y} and

\frac{\partial u(x,y)}{\partial y}=-\frac{\partial v(x,y)}{\partial x}.

Next, we have f^\ast(z^\ast)=u(x,-y)-iv(x,-y). Rewrite it as f^\ast(z^\ast)=u'(x,y')+iv'(x,y') where u'(x,y')=u(x,-y) and v'(x,y')=-v(x,-y). Now, we have the following:

\frac{\partial u'(x,y')}{\partial x}=\frac{\partial u(x,y)}{\partial x}\,,

\frac{\partial u'(x,y')}{\partial y'}=-\frac{\partial u(x,y)}{\partial y}=\frac{\partial v(x,y)}{\partial x}\,,

\frac{\partial v'(x,y')}{\partial x}=-\frac{\partial v(x,y)}{\partial x}\,,

\frac{\partial v'(x,y')}{\partial y'}=\frac{\partial v(x,y)}{\partial y}=\frac{\partial u(x,y)}{\partial x}\,.

Isn't it the Cauchy-Riemann condition still be hold for f^\ast(z^\ast)?

Best regards,
Kenneth