Is f(z) Constant in Domain D When f(z) - conj(f(z)) Equals Zero?

[squaremeplz]

Homework Statement

Suppose that f(z) and f(z) - conj(f(z))

proof f(z) is constant on D

Homework Equations

The Attempt at a Solution

If I write the equations as f(x,y) = u(x,y) + i*v(x,y)

then f(z) - conj(f(z))

= u(x,y) + i*v(x,y) - (u(x,y) - i*v(x,y))

and the u(x,y) cancel out and we are left with

f(z) - conj(f(z)) = 2*i*v(x,y)

and by the Cauchy Riemann eq

\frac {du}{dx} = \frac {dv}{dy}

\frac {du}{dy} = - \frac {dv}{dx}

since u(x,y) = 0 for f(z) - conj(f(z)) = 2*i*v(x,y)

in order for \frac {du}{dy} u(x,y) = - 2*i*\frac {dv}{dx}v(x,y)

f(z) has to be constand on D

 

[Mark44]

Homework Statement

Suppose that f(z) and f(z) - conj(f(z))

I don't understand what you are given here. The things that you are supposing are statements that are assumed to be true. All you have here are two expressions, f(z) and f(z) - conj(f(z)). Is there some relationship between these two expressions, such as an equation?

proof f(z) is constant on D

Homework Equations

The Attempt at a Solution

If I write the equations as f(x,y) = u(x,y) + i*v(x,y)

then f(z) - conj(f(z))

= u(x,y) + i*v(x,y) - (u(x,y) - i*v(x,y))

and the u(x,y) cancel out and we are left with

f(z) - conj(f(z)) = 2*i*v(x,y)

and by the Cauchy Riemann eq

\frac {du}{dx} = \frac {dv}{dy}

\frac {du}{dy} = - \frac {dv}{dx}

since u(x,y) = 0 for f(z) - conj(f(z)) = 2*i*v(x,y)

in order for \frac {du}{dy} u(x,y) = - 2*i*\frac {dv}{dx}v(x,y)

f(z) has to be constand on D

 

[squaremeplz]

Im sorry, the problem is: assume that t f(z) and f(z) - conj(f(z)) are both analytic in a domain D. Prove thet f(z) is constant on D

 

[Mark44]

OK, now I follow what you're trying to do.

in order for \frac {du}{dy} u(x,y) = - 2*i*\frac {dv}{dx}v(x,y)
f(z) has to be constand on D

I think you're waving your hands here. You have established from the given conditions that u(x, y) = 0, so what you have is (with derivatives changed to partial derivatives):
- 2*i*\frac {\partial}{\partial x}v(x,y) = 0
This says that v(x, y) = g(x) + C, where g is a function of x alone.

From your Cauchy-Riemann equations, what can you say about
\frac {\partial}{\partial y}v(x,y)?

 

[squaremeplz]

thanks for the reply.

\frac {\partial}{\partial y} v(x,y) = 0 by the riemann eq.

so this says that

v(x,y) = g(y) + C

which is a contradiction to your first statement and hence

v(x,y) = C

does this make sense?

thanks!

 

[Mark44]

Yes, makes sense.

 

[squaremeplz]

ok but doesn't that show that f(z) - conj(f(z)) is constant on D. how do I tie this into f(z) alone?

 

[Mark44]

ok but doesn't that show that f(z) - conj(f(z)) is constant on D. how do I tie this into f(z) alone?

No, it shows that f(z) is constant on D. Recall that you defined f(z) = f(x,y) = u(x,y) + i*v(x,y) back in your original post.