# Is this correct? - showing analytic function is constant.

• Feb 8th 2010, 05:57 PM
platinumpimp68plus1
Is this correct? - showing analytic function is constant.
for analytic function f:A-->C, where A is an open connected set. If f(A) is completely contained in the line {x+iy | x=2y}, show that f is constant.

My answer: i said that the analytic function f(x,y)=u(x,y)+iv(x,y) has to be in the form u(x,y)=x, v(x,y)=2x. Then Cauchy-Riemann relations:

du/dx=1 =/= 0=dv/dy
du/dy=0 =/= -2=-dv/dx

So the only way they can hold is if x=c, c some constant, so that all the partial derivatives are equal to 0.

Is this correct? (I feel like its a little too simple...)
• Feb 8th 2010, 06:09 PM
Bruno J.
Are you sure the function must be of that form? What about $\displaystyle u(x,y)=xg(x,y)$ and $\displaystyle v(x,y)=2xg(x,y)$ where $\displaystyle g: \mathbb{R}^2 \rightarrow \mathbb{R}$ is an arbitrary function?
• Feb 8th 2010, 06:34 PM
platinumpimp68plus1
Oh hm, well that would work too I suppose. Do I use the same process (ie. show that their partials don't satisfy C-R)?
• Feb 8th 2010, 07:00 PM
Bruno J.
Give it a try! (Cool)
• Feb 9th 2010, 04:46 PM
platinumpimp68plus1
okay, so i plugged it in, equated terms for g(x,y), and found dg/dy must be equal to 0 (because the dg/dx's cancelled out). consequently, du/dy and dv/dy are also equal to 0, hence for C-R to hold, du/dx and dv/dx are 0 as well --> f is constant.

yay?