Proof: if f holomorphic then f(z)=λz+c

Hello. I need a bit of help or a tip maybe..

I am to show that if f is a holomorphic function that is of the form f(x+iy) = u(x) + i*v(y) where u and v are real functions,

then f(z) = λz+c where λ is a real number and c is a complex one.

How would I begin to prove this?

Thanks to everyone in advance.

Re: Proof: if f holomorphic then f(z)=λz+c

Quote:

Originally Posted by

**seijo** I am to show that if f is a holomorphic function that is of the form f(x+iy) = u(x) + i*v(x) where u and v are real functions,

then f(z) = λz+c where λ is a real number and c is a complex one.

How would I begin to prove this?

You can't prove it, because it isn't true (unless $\displaystyle \lambda = 0$).

Notice that, when considered as f(z) = f(x,y), the assumption is that f(x,y) is holomorphic and $\displaystyle \partial{f}/\partial{y} = 0$.

But your f(x+iy) = λ(x+iy)+c varies with y (unless $\displaystyle \lambda = 0$).

The thing to prove is that the function must be constant (i.e. it actually is true, but only with $\displaystyle \lambda = 0$).

You can prove that by a straighforward application of the C-R equations.

Re: Proof: if f holomorphic then f(z)=λz+c

I see your point. But I made a typo. It's meant to be f(x+iy) = u(x) + i*v(y) not f(x+iy) = u(x) + i*v(x).

I suppose in this case it actually does make sense? Would the C-R equations still be the way to go?

Re: Proof: if f holomorphic then f(z)=λz+c

Yes, it drops straight out: