Find all holomorphic functions, $f: \mathbb{C} \rightarrow \mathbb{C}$. so that $f'(0)=1$ and $f(x+iy)=e^{x}f(iy)$

I've been messing with this problem for most of today and haven't managed to get much from it.

I started considering the real and imaginary parts of $f$. So, $f(x+iy) = u(x,y) + i v(x,y)$ and $f(iy) = u(0,y) + iv(0,y)$ Since $f$ is holomorphic we know that the Cauchy Riemann equations hold for both of those.

Since $f$ is holomorphic we also know that $\lim_{x\rightarrow0} \frac{f(x) - f(0)}{x} = \lim_{x\rightarrow0} \frac{e^xf(0) - f(0)}{x} = 1$

Which gives us $\lim_{x\rightarrow0} \frac{e^x - 1}{x} = \frac{1}{f(0)} = 1$ so $f(0) = 1$, so $v(0,0)=0$ and $u(0,0)=1$.

Similarly $\lim_{y\rightarrow0} \frac{f(iy) - f(0)}{iy} = \lim_{y\rightarrow0} \frac{f(iy) - 1}{iy} = 1$, but I don't think this is very useful.

I tried messing a bit with the Cauchy Riemann equations, but I didn't manage to get much from that. Is my approach the wrong one or is there something I'm missing?

Any help would be greatly appreciated!

  • 2
    $\begingroup$ Do you know the identity theorem? $\endgroup$ – Cameron Williams Apr 25 '15 at 2:09
  • $\begingroup$ I do now, but I don't think I'm supposed to use it. I imagine we'll be allowed to use the result soon enough. $\endgroup$ – John Williams Apr 25 '15 at 22:46

On $\mathbb R$, we have $f(x) = f(0) e^x$. So now apply the identity theorem to $F(z)=f(z)-f(0)e^z$, which has all the real numbers as roots. Since $\mathbb R$ definitely has a limit point, we will have $F(z)=0 \quad \forall z \in \mathbb C$, which implies $f(z)=f(0)e^z.$ And the derivative condition would give us $f(0)=1.$


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