title: "Complex Integration Practice: Cauchy Theorem & Residues" metaTitle: "Complex Integration Practice: Cauchy Theorem & Residues" metaDescription: "Master complex integration with 25+ solved problems on Cauchy's theorem, integral formula, Taylor/Laurent series, singularities, residues, and residue theorem." keywords: "complex integration, Cauchy integral theorem, Cauchy integral formula, Taylor series, Laurent series, singularities classification, residues, residue theorem, complex analysis"

Complex Variable — Integration: Practice Questions & Solutions

> Perfect for: Semester exams, GATE, ESE, and competitive exam preparation. Step-by-step solutions with exam-oriented explanations.

Module 1: Complex Integration (Line Integrals)

Q1. Evaluate $\int_C z \, dz$ where $C$ is the straight line from $0$ to $1+i$.

Solution:

Parametrize the line: $z(t) = t(1+i)$, $0 \le t \le 1$.

Then $dz = (1+i) \, dt$ and $z = t(1+i)$.

$\int_C z \, dz = \int_0^1 t(1+i) \cdot (1+i) \, dt = (1+i)^2 \int_0^1 t \, dt$

$= (1+2i+i^2) \cdot \frac{1}{2} = 2i \cdot \frac{1}{2} = i$

$\boxed{\int_C z \, dz = i}$

Q2. Evaluate $\int_C (z^2+1) \, dz$ where $C$ is the unit circle $$ traversed counterclockwise.

Solution:

$f(z) = z^2+1$ is a polynomial, hence entire (analytic everywhere).

By Cauchy's Integral Theorem, the integral of an analytic function over a simple closed contour is zero.

$\boxed{\int_C (z^2+1) \, dz = 0}$

Q3. Evaluate $\oint_C \frac{1}{z} \, dz$ where $C$ is the unit circle .

Solution:

Parametrize: $z = e^{i\theta}$, $0 \le \theta \le 2\pi$, so $dz = ie^{i\theta} d\theta$.

$\oint_C \frac{1}{z} \, dz = \int_0^{2\pi} \frac{1}{e^{i\theta}} \cdot ie^{i\theta} \, d\theta = i\int_0^{2\pi} d\theta = 2\pi i$

$\boxed{\oint_C \frac{1}{z} \, dz = 2\pi i}$

> Note: This is the fundamental result. The function $1/z$ is not analytic at $z=0$ (inside $C$), so Cauchy's theorem does not apply.

Q4. Evaluate $\oint_C \overline{z} \, dz$ where $C$ is the unit circle $|z|=1$.

Solution:

On $|z|=1$, we have $\overline{z} = \frac{1}{z}$.

$\oint_C \overline{z} \, dz = \oint_C \frac{1}{z} \, dz = 2\pi i$

$\boxed{\oint_C \overline{z} \, dz = 2\pi i}$

Q5. Evaluate $\int_C (x^2 + iy^2) \, dz$ along the parabola from to .

Solution:

Parametrize: $x = t$, $y = t^2$, so $z = t + it^2$, .

$dz = (1 + 2it) \, dt$

$f(z) = x^2 + iy^2 = t^2 + it^4$

$\int_C f(z) \, dz = \int_0^1 (t^2 + it^4)(1 + 2it) \, dt$

$= \int_0^1 [t^2 + 2it^3 + it^4 + 2i^2t^5] \, dt$

$= \int_0^1 [t^2 + 2it^3 + it^4 - 2t^5] \, dt$

$= \left[\frac{t^3}{3} + \frac{it^4}{2} + \frac{it^5}{5} - \frac{t^6}{3}\right]_0^1$

$= \frac{1}{3} + \frac{i}{2} + \frac{i}{5} - \frac{1}{3} = i\left(\frac{5+2}{10}\right)$

$\boxed{\int_C (x^2 + iy^2) \, dz = \frac{7i}{10}}$

Module 2: Cauchy's Integral Theorem

Q6. State Cauchy's Integral Theorem and verify it for $f(z) = z^3$ on the circle $|z|=2$.

Solution:

Cauchy's Integral Theorem: If $f(z)$ is analytic inside and on a simple closed contour $C$, then:

$\boxed{\oint_C f(z) \, dz = 0}$

Verification:

$f(z) = z^3$ is a polynomial, hence analytic everywhere.

By the theorem:

$\oint_{|z|=2} z^3 \, dz = 0$

Direct verification (parametrize $z = 2e^{i\theta}$):

$\int_0^{2\pi} (2e^{i\theta})^3 \cdot 2ie^{i\theta} \, d\theta = 16i\int_0^{2\pi} e^{4i\theta} \, d\theta = 16i\left[\frac{e^{4i\theta}}{4i}\right]_0^{2\pi} = 4(e^{8\pi i} - 1) = 0$

$\boxed{\oint_{|z|=2} z^3 \, dz = 0 \quad \checkmark}$

Q7. Evaluate $\oint_C \frac{e^z}{z-1} \, dz$ where is the circle .

Solution:

$f(z) = e^z$ is analytic inside and on $C$.

The point $z = 1$ lies inside $|z|=2$.

By Cauchy's Integral Formula:

$\oint_C \frac{f(z)}{z-a} \, dz = 2\pi i \cdot f(a)$

Here $a = 1$:

$\oint_C \frac{e^z}{z-1} \, dz = 2\pi i \cdot e^1 = 2\pi e i$

$\boxed{\oint_C \frac{e^z}{z-1} \, dz = 2\pi e i}$

Q8. Evaluate $\oint_C \frac{\cos z}{z^2+1} \, dz$ where is .

Solution:

Factor: $z^2+1 = (z-i)(z+i)$.

Singularities at $z = i$ and $z = -i$, both inside $|z|=2$.

Use partial fractions:

$\frac{1}{z^2+1} = \frac{1}{2i}\left(\frac{1}{z-i} - \frac{1}{z+i}\right)$

So:

$\oint_C \frac{\cos z}{z^2+1} \, dz = \frac{1}{2i}\left[\oint_C \frac{\cos z}{z-i} \, dz - \oint_C \frac{\cos z}{z+i} \, dz\right]$

Apply CIF to each:

$= \frac{1}{2i}[2\pi i \cos(i) - 2\pi i \cos(-i)]$

Since $\cos(-i) = \cos(i)$:

$= \frac{1}{2i}[2\pi i \cos i - 2\pi i \cos i] = 0$

$\boxed{\oint_C \frac{\cos z}{z^2+1} \, dz = 0}$

Q9. Evaluate $\oint_C \frac{z^2}{(z-1)^2} \, dz$ where is .

Solution:

Generalized Cauchy Integral Formula:

$f^{(n)}(a) = \frac{n!}{2\pi i} \oint_C \frac{f(z)}{(z-a)^{n+1}} \, dz$

Here $f(z) = z^2$, $a = 1$, $n = 1$.

$\oint_C \frac{z^2}{(z-1)^2} \, dz = \frac{2\pi i}{1!} f'(1)$

$f'(z) = 2z \Rightarrow f'(1) = 2$

$= 2\pi i \cdot 2 = 4\pi i$

$\boxed{\oint_C \frac{z^2}{(z-1)^2} \, dz = 4\pi i}$

Q10. Evaluate $\oint_C \frac{e^{2z}}{(z+1)^3} \, dz$ where is .

Solution:

Here $f(z) = e^{2z}$, $a = -1$, $n = 2$.

$f'(z) = 2e^{2z}, \quad f''(z) = 4e^{2z}$

$f''(-1) = 4e^{-2}$

By generalized CIF:

$\oint_C \frac{e^{2z}}{(z+1)^3} \, dz = \frac{2\pi i}{2!} f''(-1) = \pi i \cdot 4e^{-2} = \frac{4\pi i}{e^2}$

$\boxed{\oint_C \frac{e^{2z}}{(z+1)^3} \, dz = \frac{4\pi i}{e^2}}$

Module 3: Taylor's and Laurent's Series

Q11. Find the Taylor series of $f(z) = e^z$ about $z = 0$ and determine its radius of convergence.

Solution:

$f^{(n)}(z) = e^z \Rightarrow f^{(n)}(0) = 1$

Taylor series:

$e^z = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} z^n = \sum_{n=0}^{\infty} \frac{z^n}{n!}$

$= 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \dots$

Radius of convergence: $R = \infty$ (since $e^z$ is entire).

\boxed{e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!}, \quad |z| < \infty}

Q12. Find the Taylor series of $f(z) = \frac{1}{1-z}$ about $z = 0$ and its radius of convergence.

Solution:

This is a geometric series:

$\frac{1}{1-z} = \sum_{n=0}^{\infty} z^n = 1 + z + z^2 + z^3 + \dots$

The series converges when |z| < 1.

\boxed{\frac{1}{1-z} = \sum_{n=0}^{\infty} z^n, \quad |z| < 1}

Q13. Find the Laurent series of $f(z) = \frac{1}{z(z-1)}$ valid in the annulus 0 < |z| < 1.

Solution:

For 0 < |z| < 1, we have |z| < 1, so:

$\frac{1}{z-1} = -\frac{1}{1-z} = -\sum_{n=0}^{\infty} z^n$

Therefore:

$\frac{1}{z(z-1)} = -\frac{1}{z}\sum_{n=0}^{\infty} z^n = -\sum_{n=0}^{\infty} z^{n-1} = -\sum_{n=-1}^{\infty} z^n$

$= -\frac{1}{z} - 1 - z - z^2 - \dots$

\boxed{\frac{1}{z(z-1)} = -\sum_{n=-1}^{\infty} z^n = -\frac{1}{z} - 1 - z - z^2 - \dots, \quad 0 < |z| < 1}

Q14. Find the Laurent series of $f(z) = \frac{1}{z-1}$ valid for |z| > 1.

Solution:

For |z| > 1, we have \left|\frac{1}{z}\right| < 1.

$\frac{1}{z-1} = \frac{1}{z}\cdot\frac{1}{1-\frac{1}{z}} = \frac{1}{z}\sum_{n=0}^{\infty} \frac{1}{z^n} = \sum_{n=0}^{\infty} \frac{1}{z^{n+1}} = \sum_{n=1}^{\infty} \frac{1}{z^n}$

$= \frac{1}{z} + \frac{1}{z^2} + \frac{1}{z^3} + \dots$

\boxed{\frac{1}{z-1} = \sum_{n=1}^{\infty} z^{-n}, \quad |z| > 1}

Q15. Find the Laurent series of $f(z) = e^{1/z}$ about $z = 0$.

Solution:

Replace $z$ by $\frac{1}{z}$ in the Taylor series of $e^z$:

$e^{1/z} = \sum_{n=0}^{\infty} \frac{(1/z)^n}{n!} = \sum_{n=0}^{\infty} \frac{1}{n! \, z^n}$

$= 1 + \frac{1}{z} + \frac{1}{2!z^2} + \frac{1}{3!z^3} + \dots$

$\boxed{e^{1/z} = \sum_{n=0}^{\infty} \frac{1}{n! \, z^n}, \quad z \neq 0}$

Module 4: Singularities and Classification

Q16. Find and classify the singularities of $f(z) = \frac{z-1}{(z+1)(z-2)^2}$.

Solution:

Singularities occur where the denominator is zero:

1. $z = -1$: Simple zero of denominator, non-zero numerator.

   • $\boxed{\text{Simple pole (pole of order 1)}}$

2. $z = 2$: Zero of order 2 in denominator, .

Q17. Classify the singularity of $f(z) = \frac{\sin z}{z}$ at $z = 0$.

Solution:

$\sin z = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \dots$

$\frac{\sin z}{z} = 1 - \frac{z^2}{3!} + \frac{z^4}{5!} - \dots$

The Laurent series has no negative powers of $z$.

$\boxed{z = 0 \text{ is a removable singularity}}$

Defining $f(0) = 1$ makes the function analytic at $z=0$.

Q18. Classify the singularity of $f(z) = e^{1/z}$ at $z = 0$.

Solution:

From Q15, the Laurent series is:

$e^{1/z} = 1 + \frac{1}{z} + \frac{1}{2!z^2} + \frac{1}{3!z^3} + \dots$

There are infinitely many negative powers of $z$.

$\boxed{z = 0 \text{ is an essential singularity}}$

Q19. Determine the nature of singularities of $f(z) = \frac{1}{\sin z}$.

Solution:

Singularities where $\sin z = 0$, i.e., $z = n\pi$ for $n \in \mathbb{Z}$.

At $z = n\pi$:

$\lim_{z \to n\pi} (z-n\pi)\frac{1}{\sin z} = \lim_{z \to n\pi} \frac{z-n\pi}{\sin z}$

Using L'Hôpital's rule or $\sin z \approx (-1)^n(z-n\pi)$ near $z=n\pi$:

$= \frac{1}{\cos(n\pi)} = \frac{1}{(-1)^n} = (-1)^n \neq \infty$

So $z = n\pi$ are simple poles.

At $z = \infty$: The singularities accumulate at $\infty$, so:

$\boxed{z = n\pi \text{ (simple poles)}, \quad z = \infty \text{ is a non-isolated essential singularity}}$

Module 5: Zeros of Analytic Functions

Q20. Find the zeros of $f(z) = z^3 - 1$ and determine their orders.

Solution:

$z^3 - 1 = 0 \Rightarrow z^3 = 1 = e^{2\pi i n}$

$z = e^{2\pi i n/3}, \quad n = 0, 1, 2$

The zeros are:

• $z_0 = 1$
• $z_1 = e^{2\pi i/3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$

Since $f'(z) = 3z^2$ and $f'(z_k) = 3z_k^2 \neq 0$ for all :

$\boxed{\text{Three simple zeros (order 1) at } z = 1, e^{\pm 2\pi i/3}}$

Q21. Show that $z = 0$ is a zero of order 3 for $f(z) = z^3 \sin z$.

Solution:

$f(z) = z^3 \sin z = z^3\left(z - \frac{z^3}{3!} + \frac{z^5}{5!} - \dots\right) = z^4 - \frac{z^6}{6} + \dots$

The lowest power is $z^4$, so $z=0$ is a zero of order 4, not 3.

Wait, let me recheck. $z^3 \cdot z = z^4$. Yes, order 4.

Actually, let me correct the question to make it order 3: $f(z) = z^2 \sin z$ would give order 3.

But let's keep it as is and solve correctly:

$f(z) = z^3 \sin z$

Taylor expansion: $\sin z = z - \frac{z^3}{6} + O(z^5)$

$f(z) = z^3\left(z - \frac{z^3}{6} + \dots\right) = z^4 - \frac{z^6}{6} + \dots$

$f(0) = 0, \quad f'(0) = 0, \quad f''(0) = 0, \quad f'''(0) = 0, \quad f^{(4)}(0) = 4! = 24 \neq 0$

$\boxed{z = 0 \text{ is a zero of order } 4}$

Module 6: Residues

Q22. Find the residue of $f(z) = \frac{1}{z^2+1}$ at $z = i$.

Solution:

$f(z) = \frac{1}{(z-i)(z+i)}$

$z = i$ is a simple pole.

$\text{Res}(f, i) = \lim_{z \to i} (z-i)f(z) = \lim_{z \to i} \frac{1}{z+i} = \frac{1}{2i} = -\frac{i}{2}$

$\boxed{\text{Res}(f, i) = \frac{1}{2i} = -\frac{i}{2}}$

Q23. Find the residue of $f(z) = \frac{e^z}{(z-1)^2}$ at .

Solution:

$z = 1$ is a pole of order 2.

For a pole of order $m$:

$\text{Res}(f, a) = \frac{1}{(m-1)!} \lim_{z \to a} \frac{d^{m-1}}{dz^{m-1}}[(z-a)^m f(z)]$

Here $m = 2$, $a = 1$:

$\text{Res}(f, 1) = \frac{1}{1!} \lim_{z \to 1} \frac{d}{dz}[e^z] = \lim_{z \to 1} e^z = e$

$\boxed{\text{Res}(f, 1) = e}$

Q24. Find all residues of $f(z) = \frac{z^2+1}{z^3-z^2}$.

Solution:

$f(z) = \frac{z^2+1}{z^2(z-1)}$

Singularities:

1. $z = 0$ (pole of order 2)
2. $z = 1$ (simple pole)

At $z = 1$:

$\text{Res}(f, 1) = \lim_{z \to 1} (z-1)f(z) = \lim_{z \to 1} \frac{z^2+1}{z^2} = \frac{2}{1} = 2$

At $z = 0$:

$\text{Res}(f, 0) = \frac{1}{1!} \lim_{z \to 0} \frac{d}{dz}\left[z^2 \cdot \frac{z^2+1}{z^2(z-1)}\right] = \lim_{z \to 0} \frac{d}{dz}\left[\frac{z^2+1}{z-1}\right]$

$= \lim_{z \to 0} \frac{2z(z-1) - (z^2+1)}{(z-1)^2} = \frac{0 - 1}{1} = -1$

At $z = \infty$:

Sum of all residues = 0 (including infinity for rational functions):

$\text{Res}(f, \infty) = -(\text{Res}(f,0) + \text{Res}(f,1)) = -(-1+2) = -1$

$\boxed{\text{Res}(f, 0) = -1, \quad \text{Res}(f, 1) = 2, \quad \text{Res}(f, \infty) = -1}$

Module 7: Cauchy's Residue Theorem & Applications

Q25. State Cauchy's Residue Theorem.

Solution:

Cauchy's Residue Theorem: If $f(z)$ is analytic inside and on a simple closed contour $C$, except for a finite number of isolated singularities $z_1, z_2, \dots, z_n$ inside , then:

$\boxed{\oint_C f(z) \, dz = 2\pi i \sum_{k=1}^{n} \text{Res}(f, z_k)}$

Q26. Evaluate $\oint_C \frac{1}{z^2+1} \, dz$ where is using the residue theorem.

Solution:

$f(z) = \frac{1}{z^2+1} = \frac{1}{(z-i)(z+i)}$

Poles at $z = i$ and $z = -i$, both inside $|z|=2$.

From Q22: $\text{Res}(f, i) = \frac{1}{2i}$

By symmetry: $\text{Res}(f, -i) = \frac{1}{-2i} = -\frac{1}{2i}$

Sum of residues:

$\text{Res}(f, i) + \text{Res}(f, -i) = \frac{1}{2i} - \frac{1}{2i} = 0$

By Residue Theorem:

$\oint_C \frac{1}{z^2+1} \, dz = 2\pi i \cdot 0 = 0$

$\boxed{\oint_{|z|=2} \frac{1}{z^2+1} \, dz = 0}$

Q27. Evaluate $\oint_C \frac{z+1}{z^2+2z+2} \, dz$ where is .

Solution:

$z^2+2z+2 = 0 \Rightarrow z = \frac{-2 \pm \sqrt{4-8}}{2} = -1 \pm i$

Both $z = -1+i$ and $z = -1-i$ have modulus:

|-1 \pm i| = \sqrt{1+1} = \sqrt{2} < 2

So both poles are inside $C$.

$f(z) = \frac{z+1}{(z+1-i)(z+1+i)}$

At $z = -1+i$:

$\text{Res}(f, -1+i) = \lim_{z \to -1+i} \frac{z+1}{z+1+i} = \frac{i}{2i} = \frac{1}{2}$

At $z = -1-i$:

$\text{Res}(f, -1-i) = \lim_{z \to -1-i} \frac{z+1}{z+1-i} = \frac{-i}{-2i} = \frac{1}{2}$

Sum of residues = $\frac{1}{2} + \frac{1}{2} = 1$

$\oint_C f(z) \, dz = 2\pi i \cdot 1 = 2\pi i$

$\boxed{\oint_{|z|=2} \frac{z+1}{z^2+2z+2} \, dz = 2\pi i}$

Q28. Evaluate $\oint_C \frac{e^{1/z}}{z} \, dz$ where is .

Solution:

Inside $C$, there is an essential singularity at $z = 0$.

Find the Laurent series of $\frac{e^{1/z}}{z}$:

$\frac{e^{1/z}}{z} = \frac{1}{z}\left(1 + \frac{1}{z} + \frac{1}{2!z^2} + \frac{1}{3!z^3} + \dots\right)$

$= \frac{1}{z} + \frac{1}{z^2} + \frac{1}{2!z^3} + \dots$

The residue is the coefficient of $\frac{1}{z}$, which is $1$.

$\oint_C \frac{e^{1/z}}{z} \, dz = 2\pi i \cdot \text{Res}(f, 0) = 2\pi i \cdot 1 = 2\pi i$

$\boxed{\oint_{|z|=1} \frac{e^{1/z}}{z} \, dz = 2\pi i}$

Q29. Evaluate $\oint_C \frac{z^2}{(z-1)(z-2)} \, dz$ where is .

Solution:

Singularities at $z = 1$ and $z = 2$, both inside $|z|=3$.

At $z = 1$:

$\text{Res}(f, 1) = \lim_{z \to 1} (z-1)\frac{z^2}{(z-1)(z-2)} = \frac{1}{1-2} = -1$

At $z = 2$:

$\text{Res}(f, 2) = \lim_{z \to 2} (z-2)\frac{z^2}{(z-1)(z-2)} = \frac{4}{2-1} = 4$

Sum of residues = $-1 + 4 = 3$

$\oint_C f(z) \, dz = 2\pi i \cdot 3 = 6\pi i$

$\boxed{\oint_{|z|=3} \frac{z^2}{(z-1)(z-2)} \, dz = 6\pi i}$

Q30. Evaluate $\int_0^{2\pi} \frac{d\theta}{2+\cos\theta}$ using contour integration.

Solution:

Substitute $z = e^{i\theta}$, so $d\theta = \frac{dz}{iz}$ and .

$\int_0^{2\pi} \frac{d\theta}{2+\cos\theta} = \oint_{|z|=1} \frac{1}{2+\frac{z+z^{-1}}{2}} \cdot \frac{dz}{iz}$

$= \oint_{|z|=1} \frac{2z}{z^2+4z+1} \cdot \frac{dz}{iz} = \frac{2}{i}\oint_{|z|=1} \frac{dz}{z^2+4z+1}$

Find roots of $z^2+4z+1=0$:

$z = \frac{-4 \pm \sqrt{16-4}}{2} = -2 \pm \sqrt{3}$

Only $z = -2+\sqrt{3} \approx -0.268$ is inside $\mathrm{\mid }z$.

$z_1 = -2+\sqrt{3}, \quad z_2 = -2-\sqrt{3}$

$\frac{1}{z^2+4z+1} = \frac{1}{(z-z_1)(z-z_2)}$

$\text{Res}\left(\frac{1}{(z-z_1)(z-z_2)}, z_1\right) = \frac{1}{z_1-z_2} = \frac{1}{2\sqrt{3}}$

Integral:

$= \frac{2}{i} \cdot 2\pi i \cdot \frac{1}{2\sqrt{3}} = \frac{2\pi}{\sqrt{3}}$

$\boxed{\int_0^{2\pi} \frac{d\theta}{2+\cos\theta} = \frac{2\pi}{\sqrt{3}}}$

Quick Reference: Residue Formulas

Singularity Classification Summary

Exam Tips & Common Mistakes

1. Cauchy's Theorem requires analyticity inside AND on the contour. If there's a singularity inside, use the Residue Theorem instead.

2. For simple poles of the form $f(z) = \frac{g(z)}{h(z)}$ where $$ and : This is often faster than the limit formula.

Summary

This unit provides a complete problem-solving toolkit for Complex Integration:

• Line Integrals: Direct evaluation via parametrization
• Cauchy's Integral Theorem: Zero integral for analytic functions
• Cauchy's Integral Formula: Evaluating integrals using function values
• Generalized CIF: Connecting integrals to derivatives
• Taylor Series: Power series expansions in disks
• Laurent Series: Expansions in annuli, revealing singular behavior
• Singularities: Removable, poles, and essential classification
• Zeros: Orders and relationship to poles
• Residues: Computing coefficients of $(z-a)^{-1}$
• Residue Theorem: Powerful tool for closed contour integrals
• Applications: Real definite integrals via contour integration

Master these 30 solved problems and you'll be fully prepared for any Complex Integration examination!