Solving F(z): Complex Analysis Explained

In summary: 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...u_x(x,w)=v_y(x,w) if and only if x=w and y=0.
  • #1
fauboca
158
0
I am trying to decipher what this means:

[tex]F(z) = \overline{f(\bar{z})}[/tex]

Thanks for the help.
 
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  • #2
Welcome to PF, fauboca! :smile: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).
 
  • #3
I like Serena said:
Welcome to PF, fauboca! :smile:


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

I am not sure how to do this problem. So we know f is differentiable everywhere. How does this help?
 
  • #4
You'll need to prove that:

[tex] f(z) = \overline{f({\overline{z}}) }, [/tex]

if you write out f in a real and imaginary part and apply the CR conditions you'll find this quick and easy.
 
  • #5
dirk_mec1 said:
You'll need to prove that:

[tex] f(z) = \overline{f({\overline{z}})}, [/tex]

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 [itex]u(x,y)[/itex] be the real and [itex]v(x,y)[/itex] be the imaginary parts of F. How do I define them for [itex]\overline{f({\overline{z}})}[/itex]
 
  • #6
[tex] f(z) = f(x,y) = u(x,y) + iv(x,y) \rightarrow f( \overline{z}) =u(x,y) - iv(x,y)[/tex]
 
  • #7
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.
 
  • #8
dirk_mec1 said:
[itex] f(z) = f(x,y) = u(x,y) + iv(x,y) \rightarrow f( \overline{z}) =u(x,y) - iv(x,y)[/itex]

This is exactly what needs proving. Where you use the fact that the function is entire.
 
  • #9
Then [itex]\overline{f(\bar{z})} = u(x,-y)-iv(x,-y)[/itex]??
 
  • #10
fauboca said:
Then [itex]\overline{f(\bar{z})} = u(x,-y)-iv(x,-y)[/itex]??

Yes.
 
  • #11
I like Serena said:
Yes.

Since f is holomorphic, we know:

[tex]u_x = v_y \ \text{and} \ u_y = -v_x[/tex]

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

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

[tex]u_x(x,-y) = v_y(x,-y) \ \text{and} \ -u_y(x,-y) = v_x(x,-y)[/tex]

Now, what should I do or how should I approach it?
 
  • #12
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...
 
  • #13
I like Serena said:
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.
 
  • #14
fauboca said:
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)##?
 
  • #15
I like Serena said:
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 [itex]y=-y[/itex]
 
  • #16
fauboca said:
I need to prove the second part. So can I say [itex]y=-y[/itex]

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...
 
  • #17
I like Serena said:
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.
 
  • #18
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.
 
  • #19
I like Serena said:
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.
 

1. What is complex analysis?

Complex analysis is a branch of mathematics that deals with the study of functions of complex numbers. It involves the manipulation and study of complex numbers, which are numbers of the form a+bi, where a and b are real numbers and i is the imaginary unit (√-1).

2. What is F(z) in complex analysis?

F(z) is a function of a complex variable z. It represents a relationship between a complex number z and another complex number w, where w = F(z). F(z) can be a polynomial, trigonometric function, exponential function, or any other type of function involving complex numbers.

3. How do you solve F(z) in complex analysis?

To solve F(z), you can use various techniques such as the Cauchy-Riemann equations, contour integration, or the residue theorem. These techniques involve manipulating and evaluating complex numbers and functions to find the value of F(z) for a given complex number z.

4. Why is complex analysis important?

Complex analysis has many applications in mathematics, physics, and engineering. It is used to solve problems in fields such as fluid dynamics, electromagnetism, quantum mechanics, and signal processing. It also provides a deeper understanding of the behavior of functions and their properties.

5. Is complex analysis difficult to learn?

Complex analysis can be challenging to learn, especially for those without a strong background in mathematics. It requires a good understanding of calculus, algebra, and geometry. However, with practice and dedication, it is a fascinating and rewarding subject to study.

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