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Let $G$ be a region and let $f$ and $g$ be analytic functions on $G$ such that $f(z)g(z)=0$ for all $z \in G$. Show that either $f$ or $g$ is identically zero on $G$.

Here is how I do it: Assume $f$ is non zero on $B(a,R)$, then $fg=0$ implies $a$ is a root of $g$ (of order $n>0$). Therefore $g(z)=(z-a)^nF(z)$ for some non zero analytic function $F(z)$. And then I can't move any more...

The second question is to determine the image of $\{z=x+iy : -\pi < x < \pi, y=3\}$ under the mapping $u+iv = w = \sin z$.

Ishigami
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1 Answers1

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  1. This is a consequence of the identity theorem: Assume $f \not= 0$, then $$N := \{z \in G; f(z) = 0 \}$$ has no accumulation point (otherwise, the identity theorem would imply $f=0$ since $G$ is a region). Clearly, $g|_{G \setminus N} = 0$ since $f \cdot g=0$. Since $G \setminus N$ has an accumulation point, the identity theorem states $g=0$.
  2. By definition, $$\sin z = \frac{1}{2\imath} (e^{\imath \, z} - e^{-\imath \, z})$$ Now set $z=x+3\imath$, $x \in (-\pi,\pi)$, to calculate the image of the given set.
saz
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  • thank you! But I wanna ask why $G\N$ has an accumulation point? – Ishigami May 22 '13 at 07:52
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    @Excelsior $G$ is open and $N$ has no accumulation point in $G$. This implies that $G \setminus N$ has an accumulation point in $G$. (Choose $x \in G,r>0$ such that $B(x,r) \subseteq G$, then $B(x,\delta) \subseteq G \setminus N$ for $\delta<r$ sufficiently small.) – saz May 22 '13 at 08:29