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.