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A zero of order $n\ge 1$ at $a$ means that $f$ is holomorphic near $a$ andEquivalently its Taylor series starts with the nonzero coefficient of $(z-a{)}^n$, soUnlike a pole, a zero is a regular point of the function itself; its reciprocal has a pole of the corresponding order.

With the same factors, if $n\le m$, thenhas no negative power and extends holomorphically across $a$. Thus its apparent singularity is a removable singularity, with extension valueFor $m>n$ the extended quotient has a zero of order $m-n$; for $m=n$ the extension is nonzero.

By the finite-order zero factorizations,with $F,G$ holomorphic near $a$. Shrinking the disk if necessary, continuity ensures $G$ never vanishes there, so $F/G$ is holomorphic and nonzero at $a$. ConsequentlyIf $n>m$, the exponent is $-(n-m)$, with a holomorphic nonvanishing factor. Hence

A zero of order $n\ge 1$ at $a$ means that $f$ is holomorphic near $a$ andEquivalently its Taylor series starts with the nonzero coefficient of $(z-a{)}^n$, soUnlike a pole, a zero is a regular point of the function itself; its reciprocal has a pole of the corresponding order.

With the same factors, if $n\le m$, thenhas no negative power and extends holomorphically across $a$. Thus its apparent singularity is a removable singularity, with extension valueFor $m>n$ the extended quotient has a zero of order $m-n$; for $m=n$ the extension is nonzero.

By the finite-order zero factorizations,with $F,G$ holomorphic near $a$. Shrinking the disk if necessary, continuity ensures $G$ never vanishes there, so $F/G$ is holomorphic and nonzero at $a$. ConsequentlyIf $n>m$, the exponent is $-(n-m)$, with a holomorphic nonvanishing factor. Hence

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.

Using the exponential form of the complex cosine, $\cos z=0$ is equivalent to $e^{2iz}=-1$. Thus the zeros are exactlyAt each point ${\cos }^{\prime }(z_n)=-\sin z_n\ne 0$, so the cosine has a simple zero. Cubing it givesThe order of a zero of a holomorphic function is therefore three at every zero. There are no finite singularities because this is an entire function.

To classify infinity, examine ${\cos }^3(1/w)$ at $w=0$. The identity ${\cos }^{3}z=(3\cos z+\cos 3z)/4$ shows that its Laurent series contains the nonzero negative-power coefficientsThere are infinitely many terms in the principal part, the criterion for an essential singularity. Therefore $\boxed{\infty \text{ is an essential singularity, not a pole.}}$ This is also the general characterization of a nonpolynomial entire function at infinity.

A holomorphic function has a zero of order $n$ at $a$ when it is holomorphic on a neighbourhood of $a$ and factors asEquivalently, $f(a)=f^{\prime }(a)=\cdots =f^{(n-1)}(a)=0$ and $f^{(n)}(a)\ne 0$. The order of a zero of a holomorphic function is the degree of the first nonzero term of its Taylor series.