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Put $w=z-\pi$. The identity ${\cosh }^2(1/w)=[1+\cosh (2/w)]/2$ gives a Laurent series containing a constant and only even negative powers $w^{-2},w^{-4},\dots$. There is no $w^{-1}$ coefficient, soThe singularity is an essential singularity, not a pole: infinitely many negative Laurent coefficients are nonzero. The residue is still defined as the coefficient of $w^{-1}$.

The Cauchy-Riemann equations are $u_x=v_y$ and $u_y=-v_x$. A holomorphic function is smooth, so equality of mixed partial derivatives givesThus its real and imaginary parts are harmonic functions.

An entire function is holomorphic at every finite point of the complex plane. The globally convergent power series for the exponential implies that $e^z$ and $e^{-z}$ are entire. Their difference therefore givesat every complex $z$, proving that the hyperbolic sine is entire. For powers, the base case $f_1(z)=z$ has derivative one everywhere. If $z^n$ is entire with derivative $n{z}^{n-1}$, the product rule gives $(z^{n+1}{)}^{\prime }=z^n+zn{z}^{n-1}=(n+1)z^n$ everywhere. Induction proves that every positive integer power is entire.

A function has a pole of order $n$ at $z_0$ when it is holomorphic in a punctured neighbourhood and $g(z)=(z-z_0{)}^{n}f(z)$ extends holomorphically to $z_0$ with $g(z_0)\ne 0$. An isolated essential singularity is neither a removable singularity nor a pole; equivalently its Laurent series has infinitely many nonzero negative-power terms. In particularhas an essential singularity at zero.

For $f_N$, the Laurent series about infinity isTo classify the point correctly, use $w=1/z$. Then $f_N(1/w)=w^{-N}{e}^w$, and $e^w$ is holomorphic with value one at zero. ThereforeThe pole at infinity is compatible with an essential singularity at the finite point zero. Negative powers of $z$ are regular powers of the local coordinate $w$ near infinity, so they do not make that point essential.

On a positively oriented contour encircling $z_0$ once inside an annulus of analyticity, the Laurent series coefficients areThis follows by inserting the convergent Laurent expansion and integrating term by term: only the term with exponent $-1$ has nonzero contour integral. Uniform convergence on compact circles in the annulus justifies that step.

For the function at hand,The geometric series converges for $|z|<1$, and the pole at zero excludes the centre. ThusThe coefficients are $a_n=-1$ for $n\ge -1$ and zero for $n<-1$. The coefficient formula gives the same result on $|z|=\rho <1$: $a_n=(2\pi i{)}^{-1}\oint z^{-n-2}/(z-1)dz$, and expansion of $(z-1{)}^{-1}$ selects exactly those coefficients.

The annulus is essential. For $|z|>1$, expand instead in inverse powers:These are different Laurent expansions about the same centre on the two annuli separated by the singularity at $z=1$. The series requested in the PDF is the inner one, not a global expansion.

Let $f$ be meromorphic on and inside a positively oriented simple closed contour $C$, with finitely many isolated poles $z_k$ inside and none on $C$. The residue theorem statesThe positive orientation is counterclockwise. A clockwise contour reverses the sign, and the residue is the coefficient of $(z-z_k{)}^{-1}$ in the local Laurent series.

A simple zero of an analytic function has a local factorization $f(z)=(z-z_0)u(z)$, where $u$ is an analytic function and $u(z_0)=f^{\prime }(z_0)\ne 0$. Since $1/u$ is also an analytic function near $z_0$, its Taylor series givesThe nonzero coefficient of $(z-z_0{)}^{-1}$ proves that this is a simple pole, withMultiplying by the Taylor series $g(z)=g(z_0)+O(z-z_0)$ similarly givesThe assumption $g(z_0)\ne 0$ ensures that the pole is not canceled. In either formula the residue is precisely the coefficient of $(z-z_0{)}^{-1}$ in the Laurent series.

Expand the exponential in inverse powers of $z$:This is a Laurent series, not a Taylor series, and converges throughout $0<|z|<\infty$. Every displayed coefficient of a negative power is nonzero, so its principal part has infinitely many terms. ThusThe residue is the coefficient of $z^{-1}$:

In the Laurent seriesabout a pole of order $m$, the residue is the coefficientwhere the circle is traversed anticlockwise. Equivalently, the residue is $\frac{1}{(m-1)!}{\frac{d^{m-1}}{d{z}^{m-1}}[(z-z_0{)}^{m}f(z)]|}_{z=z_0}$.

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.

If $z_0\ne 0$, factor out $z_0$ and use the geometric series:ThusThe radius of convergence is the distance to the nearest singularity, the simple pole at $z=-z_0$. The case $z_0=0$ must be treated separately: the function is $1/z$, has no Taylor series about zero and is already its own Laurent series on $0<|z|<\infty$.

At every $z=n\pi$, the inner function has a simple pole, with $\cot z=1/(z-n\pi )+h(z)$ for analytic $h$. ThusThe first factor is analytic and nonzero, while the second has infinitely many negative powers in its Laurent series. Therefore, by the exponential of a pole is an essential singularity result,At all other finite points the composition is analytic, so there are no additional finite singularities.

Multiply the exponential power series by the two prefactors:Collecting powers yields the Laurent seriesIts first terms are $z+1+3/(2z)+7/(6z^2)+13/(24z^3)+\cdots$. Infinitely many negative-power coefficients are nonzero, so $\boxed{z=0\text{ is an essential singularity}}$. The residue is the coefficient of $z^{-1}$:The phrase "power series" here refers to a Laurent series; no ordinary Taylor series about an essential singularity exists.

A simple pole at $z_0$ means that $f$ is holomorphic in a punctured neighbourhood and $(z-z_0)f(z)$ extends holomorphically to a nonzero value at $z_0$. Equivalently its Laurent series begins with $a_{-1}/(z-z_0)$ and has no higher negative powers. The residue isIn particular, if $f=g/h$ with $g(z_0)\ne 0$, $h(z_0)=0$ and $h^{\prime }(z_0)\ne 0$, this becomes $\mathrm{Res}(f,z_0)=g(z_0)/h^{\prime }(z_0)$. The associated residue theorem isfor a positively oriented simple contour with no poles on it.

Put $w=z-1$ and $a=1-i$. Partial fraction decomposition givesBoth terms admit a geometric series at $w=0$. The local Laurent series is actually a Taylor series, since the center is regular:The nearest pole is $z=i$, a distance $|1-i|=\sqrt{2}$ from the center; the other pole is at distance $4$. Thus the requested local radius of convergence is $\boxed{R=\sqrt{2}}$.

For completeness, specifying a Laurent series without a region can also mean either of the other maximal annuli. Expanding the first fraction in inverse powers and the second in positive powers givesExpanding both fractions in inverse powers instead givesThese are different Laurent series on different annuli, rather than extensions of the local Taylor series beyond its radius of convergence.

Write the Laurent series about $a$ as ${\sum }_{k\in \mathbb{Z}}{c}_k(z-a{)}^k$. The isolated singularity is:

These cases exhaust the possible principal parts. The residue isFor the first example, $\sin z=z+O(z^3)$ givesThus zero is a double pole, with residue zero. For the other example,so the ratio is $-z/2+O(z^2)$. Its singularity is removable, with extended value zero and residue zero. The numerator's zero has higher order than the denominator's, so there is no pole.

One standard form of Cauchy's residue theorem is the following. Let $\Gamma$ be a positively oriented contour that is a simple closed curve, and let $f$ be a holomorphic function on an open set containing $\Gamma$ and its interior except for finitely many interior isolated singularities $z_1,\dots ,z_m$. There are no singularities on $\Gamma$. ThenThe residue $\mathrm{Res}(f,z_j)$ is the coefficient of $(z-z_j{)}^{-1}$ in the Laurent series at $z_j$. Each interior isolated singularity contributes once because the winding number of this positively oriented contour about each such point is one. Reversing the orientation of a contour reverses the sign. This form of Cauchy's residue theorem applies to the rectangle below.

The residue $\mathrm{res}(f;a)$ is the coefficient $c_{-1}$ of $(z-a{)}^{-1}$ in the Laurent series at the isolated singularity. Equivalently it is $(2\pi i{)}^{-1}$ times the integral around a sufficiently small counterclockwise circle enclosing $a$.

The residue theorem states that if $f$ is holomorphic on a neighborhood of a positively oriented simple closed contour and its interior except for finitely many isolated singularities $a_j$ inside, thenNo singularity may lie on the integration contour. A more general contour weights each residue by its winding number.

The Liouville theorem states that every bounded entire function is constant.

By the Laurent theorem, $f$ has a unique Laurent series $f(z)={\sum }_{n=-\infty }^{\infty }{c}_n{z}^n$ on $R<|z|<S$. Split it intoHere $g$ is holomorphic on $D_S$ and $h$ is holomorphic on $A_{R,\infty }$. To check their extended regions of convergence explicitly, fix $r\in (R,S)$ and put $M_r={\max }_{|z|=r}|f(z)|$. The Laurent coefficient formula givesThe first bound ensures convergence uniformly on every closed disk $|z|\le r^{\prime }<r$; letting $r$ range up to $S$ covers $D_S$. The second gives convergence uniformly on $|z|\ge r^{\prime \prime }>r$; letting $r$ range down to $R$ covers $A_{R,\infty }$. Locally uniform convergence of these power series, in $z$ and $1/z$ respectively, proves the claimed holomorphy. On the original annulus their sum is $f$.

The second estimate also proves $h(z)\to 0$ as $|z|\to \infty$: for any fixed $r\in (R,S)$ and $|z|>r$,

For uniqueness under this condition, suppose $f=g_1+h_1=g_2+h_2$. Define $F=g_1-g_2$ on $D_S$ and $F=h_2-h_1$ on $A_{R,\infty }$. The two definitions agree on the overlap, and their domains cover $\mathbb{C}$, so they define an entire function. Moreover, $F(z)\to 0$ at infinity. It is bounded outside a large disk by this limit, and inside by continuity and the extreme value theorem. The Liouville theorem makes $F$ constant, and its limit makes that constant zero. Hence $g_1=g_2$ and $h_1=h_2$. Without the limiting condition, adding a constant to $g$ and subtracting it from $h$ would preserve the decomposition.

Use the local factorization for a pole of order $n$:Shrink the disc so that the holomorphic function $h$ has no zeros there. Differentiating and dividing gives the logarithmic derivativeThe second term is holomorphic across $a$. Since $n\ge 1$, the first term cannot cancel. Hence $f^{\prime }/f$ has a simple pole, and the coefficient of $(z-a{)}^{-1}$ in its Laurent series isThis is the pole case of the integer residue of a logarithmic derivative.

Let $F=e^g$. It is holomorphic and nowhere zero on a punctured neighbourhood of $a$, and the chain rule givesIf $F$ had a pole of order $n\ge 1$, part (c) would imply ${\mathrm{Res}}_a(g^{\prime })=-n$. But the derivative of a holomorphic function on a punctured disc has residue zero. Indeed, write its Laurent seriesTermwise differentiation is valid on every smaller closed annulus. The only possible $(z-a{)}^{-1}$ term comes from $k=0$ and has coefficient zero. Thusa contradiction. Therefore $\boxed{e^g\text{ cannot have a pole at }a}$. This proves an exponential has no isolated pole, including when $g$ has an essential singularity; no choice of a holomorphic logarithm on the entire punctured disc was assumed.

For $n\ge 1$, $f=z^n{e}^{n/z}$ is holomorphic at every finite nonzero point. At zero its Laurent series has infinitely many negative powers, so zero is an essential singularity. If the Riemann sphere is included, $f(1/w)=w^{-n}{e}^{nw}$ has a pole of order $n$ at $w=0$, so infinity is a pole of order $n$.