Solving F(z): Complex Analysis Explained

[fauboca]

I am trying to decipher what this means:

$$F(z) = \overline{f(\bar{z})}$$

Thanks for the help.

 

[I like Serena]

Welcome to PF, fauboca! The overbar notation indicates the complex conjugate.

Apparently the function F(z) is defined by taking the complex conjugate of z, applying the function f, followed by taking the conjugate of the result.

Note that for regular functions F(z) is equal to f(z).

 

[fauboca]

Welcome to PF, fauboca!


The overbar notation indicates the complex conjugate.

Apparently the function F(z) is defined by taking the complex conjugate of z, applying the function f, followed by taking the conjugate of the result.

Note that for regular functions F(z) is equal to f(z).

I was hoping with this information I would have been able to solve the problem, but I am still unsure on how to work it.

The question is:

If $f:\mathbb{C}\to\mathbb{C}$ is entire, i.e. holomorphic at every point, then $F(z) = \overline{f(\bar{z})}$.

I am not sure how to do this problem. So we know f is differentiable everywhere. How does this help?

 

[dirk_mec1]

You'll need to prove that:

$$f(z) = \overline{f({\overline{z}}) },$$

if you write out f in a real and imaginary part and apply the CR conditions you'll find this quick and easy.

 

[fauboca]

You'll need to prove that:

$$f(z) = \overline{f({\overline{z}})},$$

if you write out f in a real and imaginary part and apply the CR conditions you'll find this quick and easy.

I understand what you are saying but I still don't quite see it.

So if we let $u(x,y)$ be the real and $v(x,y)$ be the imaginary parts of F. How do I define them for $\overline{f({\overline{z}})}$

 

[dirk_mec1]

$$f(z) = f(x,y) = u(x,y) + iv(x,y) \rightarrow f( \overline{z}) =u(x,y) - iv(x,y)$$

 

[conquest]

I suppose the question is actually to show that F(z)=f(z) ?

Anyway suppose you write z = x+ iy then the conjugate is x - iy

so take u(x,y) to u(x,-y) and v(x,y) to v(x,-y). Then (and of course taking the conjuagte of the function also means another minus sign. The the Cauchy Riemann relation will provide the answer.

 

[conquest]

$f(z) = f(x,y) = u(x,y) + iv(x,y) \rightarrow f( \overline{z}) =u(x,y) - iv(x,y)$

This is exactly what needs proving. Where you use the fact that the function is entire.

 

[fauboca]

Then $\overline{f(\bar{z})} = u(x,-y)-iv(x,-y)$??

 

[I like Serena]

Then $\overline{f(\bar{z})} = u(x,-y)-iv(x,-y)$??

Yes.

 

[fauboca]

Yes.

Since f is holomorphic, we know:

$$u_x = v_y \ \text{and} \ u_y = -v_x$$

Now, how do I use $F(z)=\overline{f(\bar{z})}=u(x,-y)-iv(x,-y)$.

I read that I need to use C.R. equations. I need to show:

$$u_x(x,-y) = v_y(x,-y) \ \text{and} \ -u_y(x,-y) = v_x(x,-y)$$

Now, what should I do or how should I approach it?

 

[I like Serena]

With just f(z) being holomorphic, that is not enough to show that F(z)=f(z).

So I assume you're supposed to show that F(z) is holomorphic?
If so, according to C.R. you should check that $\Re(F_x)=\Im(F_y)$ and that $\Re(F_y)=-\Im(F_x)$, starting from the fact that f(z) is holomorphic...

 

[fauboca]

With just f(z) being holomorphic, that is not enough to show that F(z)=f(z).

So I assume you're supposed to show that F(z) is holomorphic?
If so, according to C.R. you should check that $\Re(F_x)=\Im(F_y)$ and that $\Re(F_y)=-\Im(F_x)$, starting from the fact that f(z) is holomorphic...

How do I check that?

I wrote that in terms of u and v in the post just above yours.

 

[I like Serena]

How do I check that?

I wrote that in terms of u and v in the post just above yours.

Well, if $u_x(x,y)=v_y(x,y)$ for all x and y, don't you also have $u_x(x,-y)=v_y(x,-y)$?

 

[fauboca]

Well, if $u_x(x,y)=v_y(x,y)$ for all x and y, don't you also have $u_x(x,-y)=v_y(x,-y)$?

I need to prove the second part. So can I say $y=-y$

 

[I like Serena]

I need to prove the second part. So can I say $y=-y$

Hmm, no you can't say y=-y.

But since $u_x(x,y)=v_y(x,y)$ is true for all x and y, you could also say $u_x(x,w)=v_y(x,w)$ is true for all x and w.
Now define y=-w...

 

[fauboca]

Hmm, no you can't say y=-y.

But since $u_x(x,y)=v_y(x,y)$ is true for all x and y, you could also say $u_x(x,w)=v_y(x,w)$ is true for all x and w.
Now define y=-w...

I am not sure what you are attempting to allude to from the ellipses.

 

[I like Serena]

Since f(z) is holomorphic, $u_x(x,w)=v_y(x,w)$ is true for all x and w.
In particular, if we define y=-w, that means that $u_x(x,-y)=v_y(x,-y)$ is true for all x and y.

Put otherwise, the relation holds true for any point in the complex plane.
It does not matter if we put a minus sign in front of the imaginary part of a point, as long as we do so consistently.
The resulting points (x,-y) are all in the same complex plane for which the relation holds true.

 

[fauboca]

Since f(z) is holomorphic, $u_x(x,w)=v_y(x,w)$ is true for all x and w.
In particular, if we define y=-w, that means that $u_x(x,-y)=v_y(x,-y)$ is true for all x and y.

Put otherwise, the relation holds true for any point in the complex plane.
It does not matter if we put a minus sign in front of the imaginary part of a point, as long as we do so consistently.
The resulting points (x,-y) are all in the same complex plane for which the relation holds true.

Thanks. Sorry for not being able to decipher that myself.