GATE Mathematics Complex Analysis Syllabus

How does Liouville's Theorem prove the Fundamental Theorem of Algebra?

If a non-constant polynomial $p(z)$ had no zero, then $1/p(z)$ would be entire and bounded, hence constant by Liouville — a contradiction. Thus every non-constant polynomial has at least one root in $\mathbb{C}$.

State the Maximum Modulus Principle.

If $f$ is analytic and non-constant in a bounded domain $D$, then $|f(z)|$ attains its maximum only on the boundary $\partial D$, never at an interior point of $D$.

State the Minimum Modulus Principle.

If $f$ is analytic and non-constant in a domain and $f(z)\neq 0$ throughout, then $|f(z)|$ cannot attain a minimum at an interior point; the minimum occurs on the boundary.

State Morera's Theorem (converse of Cauchy's theorem).

If $f$ is continuous on a domain $D$ and $\oint_C f(z)\,dz=0$ for every closed contour $C$ in $D$, then $f$ is analytic on $D$.

State Gauss's Mean Value Theorem for analytic functions.

If $f$ is analytic inside and on a circle $|z-z_0|=R$, then $$f(z_0)=\frac{1}{2\pi}\int_0^{2\pi} f(z_0+Re^{i\theta})\,d\theta,$$ the value at the center equals the average over the circle.

State Taylor's Theorem for an analytic function about $z_0$.

If $f$ is analytic in a disk $|z-z_0|<R$, then $$f(z)=\sum_{n=0}^{\infty}\frac{f^{(n)}(z_0)}{n!}(z-z_0)^{n},$$ valid throughout that disk.

How is the radius of convergence $R$ of $\sum a_n (z-z_0)^n$ found via the root test?

By the Cauchy–Hadamard formula: $$\frac{1}{R}=\limsup_{n\to\infty}|a_n|^{1/n}.$$

Give the ratio-test formula for the radius of convergence of $\sum a_n (z-z_0)^n$.

$$R=\lim_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right|,$$ provided the limit exists.

What is the relationship between the radius of convergence of a Taylor series and the nearest singularity?

The radius of convergence equals the distance from the center $z_0$ to the nearest singularity of $f$ in the complex plane.

State the Laurent series expansion of $f$ in an annulus.

If $f$ is analytic in the annulus $r<|z-z_0|<R$, then $$f(z)=\sum_{n=-\infty}^{\infty} a_n (z-z_0)^{n},\quad a_n=\frac{1}{2\pi i}\oint_C \frac{f(z)}{(z-z_0)^{n+1}}\,dz.$$

In a Laurent series, what are the principal part and the analytic part?

The principal part is $\sum_{n=1}^{\infty} a_{-n}(z-z_0)^{-n}$ (negative powers); the analytic (regular) part is $\sum_{n=0}^{\infty} a_n (z-z_0)^{n}$ (non-negative powers).

How does the structure of the principal part of a Laurent series classify an isolated singularity?

Removable: no negative powers. Pole of order $m$: finitely many negative powers, lowest being $(z-z_0)^{-m}$. Essential singularity: infinitely many negative powers.

Define the residue of $f$ at an isolated singularity $z_0$.

The residue $\operatorname{Res}_{z=z_0} f$ is the coefficient $a_{-1}$ of $(z-z_0)^{-1}$ in the Laurent expansion of $f$ about $z_0$. Equivalently $a_{-1}=\frac{1}{2\pi i}\oint_C f\,dz$.

Give the formula for the residue at a simple pole $z_0$.

$$\operatorname{Res}_{z=z_0} f=\lim_{z\to z_0}(z-z_0)\,f(z).$$ If $f=p/q$ with simple pole, $\operatorname{Res}=\frac{p(z_0)}{q'(z_0)}$.

Give the formula for the residue at a pole of order $m$.

$$\operatorname{Res}_{z=z_0} f=\frac{1}{(m-1)!}\lim_{z\to z_0}\frac{d^{\,m-1}}{dz^{\,m-1}}\big[(z-z_0)^{m} f(z)\big].$$

State the Residue Theorem.

If $f$ is analytic inside and on a positively oriented simple closed contour $C$ except for isolated singularities $z_1,\dots,z_k$ inside, then $$\oint_C f(z)\,dz=2\pi i\sum_{j=1}^{k}\operatorname{Res}_{z=z_j} f.$$

How is a real integral $\int_0^{2\pi} R(\cos\theta,\sin\theta)\,d\theta$ converted to a contour integral?

Substitute $z=e^{i\theta}$, so $\cos\theta=\frac{1}{2}(z+z^{-1})$, $\sin\theta=\frac{1}{2i}(z-z^{-1})$, $d\theta=\frac{dz}{iz}$, turning it into $\oint_{|z|=1}$ evaluated by residues inside the unit circle.

Describe the contour method to evaluate $\int_{-\infty}^{\infty} \dfrac{P(x)}{Q(x)}\,dx$ for rational functions.

Close the contour with a large semicircle in the upper half-plane. If $\deg Q\ge \deg P+2$, the arc contribution vanishes, so the integral equals $2\pi i$ times the sum of residues at poles in the upper half-plane.

State Jordan's Lemma and its use.

For $m>0$, if $f(z)\to 0$ uniformly as $|z|\to\infty$ on the upper semicircle $C_R$, then $\int_{C_R} f(z)e^{imz}\,dz\to 0$. It justifies dropping the arc when evaluating $\int_{-\infty}^{\infty} f(x)e^{imx}\,dx$ by residues.

How are integrals of the form $\int_{-\infty}^{\infty} f(x)\cos(mx)\,dx$ or $\int f(x)\sin(mx)\,dx$ evaluated?

Consider $\int_{-\infty}^{\infty} f(x)e^{imx}\,dx=2\pi i\sum(\text{residues in upper half-plane})$, then take the real part for $\cos$ and the imaginary part for $\sin$.

What is the contribution of a simple pole lying on the real axis (contour indentation)?

Indenting around a simple pole $x_0$ on the real axis with a small semicircle contributes $\pm i\pi\,\operatorname{Res}_{z=x_0} f$ (half the full $2\pi i$ residue), the sign depending on orientation; the integral is taken as a Cauchy principal value.

Classify the singularity of $f(z)=\dfrac{\sin z}{z}$ at $z=0$.

Removable singularity. The Laurent series has no negative powers ($\frac{\sin z}{z}=1-\frac{z^2}{3!}+\cdots$), and $f$ can be defined as $1$ at $z=0$ to make it analytic.

Classify the singularity of $e^{1/z}$ at $z=0$ and state the relevant theorem.

Essential singularity (Laurent series $\sum_{n=0}^\infty \frac{1}{n!\,z^n}$ has infinitely many negative powers). By the Casorati–Weierstrass / Picard theorem, $f$ comes arbitrarily close to every complex value near $z=0$.

Compare a pole and an essential singularity in terms of $\lim_{z\to z_0}|f(z)|$.

At a pole, $|f(z)|\to\infty$ as $z\to z_0$. At an essential singularity, the limit does not exist and $f$ takes values arbitrarily close to every complex number (Casorati–Weierstrass).

State the Cauchy product / uniqueness fact: how do the Taylor and Laurent coefficients relate when $f$ is analytic at $z_0$?

If $f$ is analytic at $z_0$, the Laurent series reduces to the Taylor series, so $a_n=\frac{f^{(n)}(z_0)}{n!}$ for $n\ge0$ and $a_n=0$ for $n<0$. The expansion is unique.

What does it mean for a series $\sum a_n(z-z_0)^n$ when $z$ is exactly on the circle $|z-z_0|=R$?

The behaviour on the circle of convergence is indeterminate by the radius test alone: the series may converge at some boundary points and diverge at others; it must be examined separately.

Give the Taylor (Maclaurin) series of $e^z$, $\sin z$, and $\cos z$ with their radii of convergence.

$e^z=\sum_{n=0}^\infty \frac{z^n}{n!}$, $\sin z=\sum_{n=0}^\infty \frac{(-1)^n z^{2n+1}}{(2n+1)!}$, $\cos z=\sum_{n=0}^\infty \frac{(-1)^n z^{2n}}{(2n)!}$. Each has $R=\infty$ (entire).

Give the geometric series and its radius of convergence.

$$\frac{1}{1-z}=\sum_{n=0}^{\infty} z^{n},\quad |z|<1,\ R=1.$$ It diverges for $|z|\ge1$; the singularity at $z=1$ sets $R$.

Define the winding number (index) of a closed contour $C$ about a point $z_0$.

$$n(C,z_0)=\frac{1}{2\pi i}\oint_C \frac{dz}{z-z_0},$$ an integer counting the net number of times $C$ winds counterclockwise around $z_0$.

State the Argument Principle.

If $f$ is meromorphic inside and on a simple closed contour $C$ (no zeros/poles on $C$), then $$\frac{1}{2\pi i}\oint_C \frac{f'(z)}{f(z)}\,dz=Z-P,$$ where $Z$ and $P$ are the numbers of zeros and poles inside $C$, counted with multiplicity.