Analytic in a Domain D

• Feb 12th 2009, 12:59 PM
universalsandbox
Analytic in a Domain D
if f(z) is analytic/holomorphic in a Domain D, prove f(z) must be constant in that domain if any of these are also constant in D:

Im(f(z)), Re(f(z)), |f(z)|

So, f(x + iy) = u(x, y) + iv(x, y)
u, and v have first partial derivatives and satisfy Cauchy-Riemann equations:

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

How can you go about showing this? Thanks.
• Feb 12th 2009, 05:46 PM
HallsofIvy
Quote:

Originally Posted by universalsandbox
if f(z) is analytic/holomorphic in a Domain D, prove f(z) must be constant in that domain if any of these are also constant in D:

Im(f(z)), Re(f(z)), |f(z)|

So, f(x + iy) = u(x, y) + iv(x, y)
u, and v have first partial derivatives and satisfy Cauchy-Riemann equations:

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

How can you go about showing this? Thanks.

Here, v is Im(f(z)) so if it is constant, the right side of each of those two equations is 0 and we have $\displaystyle \frac{\partial u}{\partial x}= 0$ and $\displaystyle \frac{\partial u}{\partial y}= 0$, the Cauchy-Riemann equations. In other words, that both u and v are constant.

Similarly, if Re(f(z)) is constant, so is u and then you can prove that v is constant.

Finally, if |f|= $\displaystyle (u^2+ v^2)^{1/2}$ is constant, so $\displaystyle u^2+ v^2$ is constant and we have $\displaystyle 2u\frac{\partial u}{\partial x}+ 2v\frac{\partial v}{\partial y}= 0$ and $\displaystyle 2u\frac{\partial u}{\partial y}+2v\frac{\partial v}{\partial y}= 0$. Use those two equations together with the two Cauchy-Riemann equations to show that the four derivatives $\displaystyle \frac{\partial u}{\partial x}$, $\displaystyle \frac{\partial u}{\partial y}$, $\displaystyle \frac{\partial v}{\partial x}$, and $\displaystyle \frac{\partial v}{\partial y}$ are all 0.