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I am trying to decipher what this means:
[tex]F(z) = \overline{f(\bar{z})}[/tex]
Thanks for the help.
[tex]F(z) = \overline{f(\bar{z})}[/tex]
Thanks for the help.
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.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 understand what you are saying but I still don't quite see it.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.
This is exactly what needs proving. Where you use the fact that the function is entire.[itex] f(z) = f(x,y) = u(x,y) + iv(x,y) \rightarrow f( \overline{z}) =u(x,y) - iv(x,y)[/itex]
Yes.Then [itex]\overline{f(\bar{z})} = u(x,-y)-iv(x,-y)[/itex]??
Since f is holomorphic, we know:Yes.
How do I check that?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...
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)##?How do I check that?
I wrote that in terms of u and v in the post just above yours.
I need to prove the second part. So can I say [itex]y=-y[/itex]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)##?
Hmm, no you can't say y=-y.I need to prove the second part. So can I say [itex]y=-y[/itex]
I am not sure what you are attempting to allude to from the ellipses.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...
Thanks. Sorry for not being able to decipher that myself.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.