Complex conjugates

1. Oct 19, 2004

Complex conjugates (new question in 1st post)

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?

Last edited: Oct 20, 2004
2. Oct 19, 2004

Tide

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

3. Oct 19, 2004

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

4. Oct 19, 2004

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

5. Oct 19, 2004

Thanks, that makes a lot of sense.

6. Oct 20, 2004

7. Oct 20, 2004

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

Last edited: Oct 20, 2004