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The hypothesis says that $f$ has a zero of order $k$ at $a$, so writewith $g$ holomorphic near $a$. Differentiating givesChoose $\delta >0$ so small that the closed disc $\overline{D}(a,\delta )$ lies in the domain, $g$ never vanishes there, and the parenthesized factor never vanishes there. Thus $a$ is the only critical point of $f$ in that disc.

On the boundary circle, $f$ is nonzero. SetIf $0<|b|<\epsilon$, then $|b|<|f(z)|$ on the boundary. Rouché's theorem applied to $f$ and $f-b$ says that $f-b$ has the same number of zeros in $D(a,\delta )$ as $f$, namely $k$, counted with multiplicity.

None of these zeros is $a$ because $b\ne 0$, and none is a critical point because $f^{\prime }$ has no other zero in the disc. Every zero of $f-b$ is therefore simple. Hence there are exactly

Because $\cos 0=1$, continuity gives a disc $U$ about zero on which $\cos z\ne 0$. The function $-\tan z=(\log \cos z{)}^{\prime }$ has a holomorphic primitive $G$ on a sufficiently small disc, normalized by $G(0)=0$. Thenis holomorphic and satisfies $f(0)=1$ and $f(z{)}^2=\cos z$ on $U$.

No such square root exists throughout $|z|<2$. The point $z=\pi /2$ lies in that disc and is a simple zero of $\cos z$, since $-\sin (\pi /2)=-1$. Every zero of the square of a holomorphic function has even order of a zero of a holomorphic function, contradicting this simple zero.