• Support PF! Buy your school textbooks, materials and every day products Here!

Cauchy riemann equations and constant functions

  • Thread starter reb659
  • Start date
  • #1
64
0

Homework Statement



Let f(z) be analytic on the set H. Let the modulus of f(z) be constant. Does f need be constant also? Explain.

Homework Equations



Cauchy riemann equations

Hint: Prove If f and f* are both analytic on D, then f is constant.




The Attempt at a Solution



I think f need be constant.
Let f*=conjugate operator

Let f = U+iV Then f* = U-iV
Since F is analytic we can use CR equations and we get
1) Ux = Vy
and
2) Uy = -Vx .
Applying CR to f* gives
3) Ux = -Vy
and
4) Uy = Vx

1) and 3) imply Vy = -Vy and 2) and 4) imply Vx = -Vx.
But the only function that can equal its negative is zero, and thus Vx = Vy = 0, and so V = constant.
Likewise, the same argument for Ux and Uy gives Ux = Uy = 0 and so U = a constant. And both U and V constant implies that f is constant. So we have the hint proven.


Let f = u+iv. We’re given |f| = c.

Stuck.
 
Last edited:

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
Hint: f times f* is |f|^2 which is a constant.
 
  • #3
64
0
But what about the function f(z)=z=x+iy? Isn't that function analytic on the say, unit disk with a constant modulus but f is not constant?
 
  • #4
Dick
Science Advisor
Homework Helper
26,258
618
But what about the function f(z)=z=x+iy? Isn't that function analytic on the say, unit disk with a constant modulus but f is not constant?
If you mean the unit disk, it doesn't have constant modulus. If you mean the unit circle, yes, it does. They should have stated assumptions on the domain H. Let's say it needs to be open and connected.
 
  • #5
64
0
Ah yes, I meant unit circle. I was thinking the domain would need to have restrictions because I thought I proved it yet found a counterexample at the same time. But for the original problem:

Let f=U + iV. We have |f|=c for some constant c. |z|^2=zz*, so we have |f|^2=c^2 which implies f*=c^2/f. f is analytic by assumption and constants are always analytic over any D, so by the above hint I proved f must be constant.
 
  • #6
Dick
Science Advisor
Homework Helper
26,258
618
Ah yes, I meant unit circle. I was thinking the domain would need to have restrictions because I thought I proved it yet found a counterexample at the same time. But for the original problem:

Let f=U + iV. We have |f|=c for some constant c. |z|^2=zz*, so we have |f|^2=c^2 which implies f*=c^2/f. f is analytic by assumption and constants are always analytic over any D, so by the above hint I proved f must be constant.
Sure, that's it.
 

Related Threads on Cauchy riemann equations and constant functions

Replies
7
Views
706
Replies
4
Views
828
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
5
Views
4K
  • Last Post
Replies
14
Views
4K
  • Last Post
Replies
6
Views
2K
Replies
3
Views
686
Replies
2
Views
5K
Top