PYQ on Complex Analysis

In the examination, questions from Complex Analysis are often asked in the form of evaluating complex integrals, finding residues, checking analyticity, or determining harmonic functions.

Typical methods include the use of Cauchy-Riemann equations, Cauchy’s Integral Theorem and Formula, Residue Theorem, Laurent series, and conformal mapping, unlike real-variable calculus, which deals primarily with functions of real variables. Complex analysis questions often involve functions with singularities, poles, or branch points, and require techniques like computing residues at poles, expanding functions in Taylor or Laurent series, or finding harmonic conjugates.

Short Question on Complex Analysis

Question 1: Find the residue of f(z)=\dfrac{e^z}{(z-1)^2} ​ at z = 1.

Question 2: Find the residue of f(z) = \dfrac{\cos z}{z^2+ \pi^2} at z=iπ.

Question 3: Find the residue of f(z) =\dfrac{1}{(z^2+1)^2}​ at z = i.

Question 4: Show that f(z)=z³ is analytic everywhere.

Question 5: Check whether the function f(z)=\frac{z}{z+1} is continuous at z = 0.

Question 6: Determine all points where f(z) = ∣z∣ is differentiable.

Question 7: Show that u(x, y) = e^xycos⁡y is harmonic. Also find its harmonic conjugate v(x, y).

Question 8: Evaluate\lim_{z\to 0}\frac{z^2}{|z|}.

Question 9: Expand f(z) = \dfrac{1}{z(1-z)}​ in a Laurent series valid for 0 < ∣z∣ < 1.

Question 10 : Show that f(z) = z² + 3z + 2 is analytic everywhere and find f′(z).

Long Questions on Complex Analysis

Question 11: Evaluate the integral \oint_{|z|=2} \frac{z^2}{z^2 - 1} \, dzwhere the contour is the positively oriented circle |z| = 2.

Question 12: Let f(z) = u(x,y) + iv(x,y) be defined by

f(z) = u(x, y) + iv(x, y), u(x, y) = x³ − 3xy², v(x, y) = 3x²y − y³.

(a) Show that f(z) is analytic.
(b) Verify that u(x, y) and v(x, y) are harmonic functions.

Check if you were right - full answer with solution below.

Solution 1:

Pole of order 2 at z = 1.

For a pole of order m = 2,

Res⁡(f, 1) = \frac{1}{(m-1)!} \lim_{z\to1}\frac{d}{dz}\big[(z-1)^2 f(z)\big]

=\lim_{z\to1}\frac{d}{dz}e^z

=\lim_{z\to1} e^z = e¹ = e.

Solution 2:

Simple pole at z = iπ.

Compute

Res(f,iπ) = \lim_{z \to i\pi} (z - i\pi)\,\frac{z - i\pi}{(z + i\pi)\cos z}

= \frac{i\pi + i\pi}{\cos(i\pi)}

= \frac{2 i\pi}{\cosh \pi}.

Solution 3:

Pole at z = i of order 2.

Write z² + 1 = (z − i)(z + i).

Use formula for order-2:

Res⁡(f,i) = \lim_{z\to i}\frac{d}{dz}\Big[(z-i)^2 \frac{1}{(z^2+1)^2}\Big] = \lim_{z\to i}\frac{d}{dz}\frac{1}{(z+i)^2}

Differentiate: \dfrac{d}{dz}(z+i)^{-2} = -2(z+i)^{-3}.

Evaluate at z = i : −2(i + i)⁻³ = −2(2i)⁻³
Now compute (2i)³= 8i³= 8(−i) = −8. So (2i)⁻³ = 1/(−8i) = −1/(8i).
Multiply: −2⋅(−1/8i) = 2/8i = 1/4i=−i/4 (since 1/i =−i).
Thus Res⁡(f,i) =−i/4.

Solution 4:

Write in Cartesian form: z = x + iy Expand:

z³ = (x + iy)³ = (x³ − 3xy²) + i(3x²y − y³)

Compute partials:

u_x = 3x² − 3y², v_y = 3x² − 3y²

u_y​ = − 6xy, v_x​ = − 6xy

C–R equations hold everywhere.
Derivatives are continuous.
Hence the function is analytic in entire complex plane.

Solution 5:

Evaluate limit: \lim_{z\to 0}\frac{z}{z+1}=0.

Function value:

f(0) = 0

Both match and denominator ≠ 0 near origin.
Therefore function is continuous at z = 0.

Solution 6:

Write f(z)= \sqrt{x^2+y^2}

Compute partials:

u_x ​ = x/\sqrt{x^{2} + y^{2}} , u_{y ​}= y/\sqrt{x^{2} + y^{2}}

But imaginary part = 0 → CR:

u_x = 0, u_y = 0

This occurs only at x = y = 0.
Check continuity of partials: undefined at origin.
Hence f(z) = |z| is differentiable nowhere.

Solution 7:

Check Laplace equation:

• u_x​ = e^xcosy, u_xx​ = e^xcosy
• u_y​ = −e^xsiny, u_yy​ = −e^xcosy
• u_xx​+u_yy​ = e^xcosy − e^xcosy = 0

Hence u is harmonic.

To find harmonic conjugate, use CR: v_x​ = −u_y​ = e^xsiny

Integrate wrt x: v = e^x sin y + C(y)

Now use v_y= u_x :

e^xcos⁡y + C′(y) = e^xcos⁡y ⇒ C′ =0.

v(x, y) = e^x sin y

Solution 8:

Let z=re^iθ

Then

\frac{z^2}{|z|}=\frac{r^2e^{i2\theta}}{r}=re^{i2\theta}.

As r → 0 the whole expression → 0 regardless of θ.Thus limit exists and equals 0.

Solution 9:

f(z) = 1/z(1 − z) = 1/z.

For ∣z∣<1 use the geometric series 1/1−z = \sum_{n=0}^{\infty}z^n

Hence,

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

= z^{-1} + 1 + z + z^{2} + \cdots

valid for 0 < ∣z∣<1. (The principal part is z⁻¹; the residue at z = 0 is 1.)

Solution 10:

f(z) = z² + 3z + 2 is a polynomial, which is analytic everywhere in \mathbb{C}.

Its derivative:

f′(z) = 2z + 3

Solution 11 :

Factorize the denominator:

z² - 1 = (z - 1)(z + 1)

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

Identify singularities:

z = 1, z = -1

Both singularities are inside the contour |z|=2.

Use the Residue Theorem:

\oint_{|z|=2} f(z)\, dz = 2\pi i \sum \text{Res}(f, z_k)

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

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

= \lim_{z \to -1} \frac{z^2}{z-1}

= -\frac{1}{2}

Apply residue Theorem:

\oint_{|z|=2} \frac{z^2}{z^2-1} dz = 2\pi i \left(\frac{1}{2} + \left(-\frac{1}{2}\right)\right) = 0

Solution 12:

Cauchy-Riemann (CR) equations:

u_x ​= v_y​, u_y​ = −v_x​

Compute partial derivatives:

• u_x = 3x² − 3y²,u_y = −6xy
• v_x​ = 6xy, v_y​ = 3x² − 3y²

Check CR equations:

• u_x​ = v_y​ ⇒ 3x² − 3y² = 3x² −3y²
• u_y = −v_x ⇒ −6xy = −6xy

f(z) satisfies CR equations and is analytic.

(b) A function u(x,y) is harmonic if:

u_{xx ​} + u_yy​ = 0

Compute second derivatives:

• u_xx​ = 6x, u_yy​ = −6x ⇒u_xx​ + u_yy​ = 0
• v_xx ​= 6y, v_yy​ = −6y ⇒ v_xx​ +v_yy​ = 0

Hence u(x,y) and v(x,y) are harmonic.