Complex Variable

Complex Variable — Differentiation: Practice Questions & Solutions

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

Module 1: Functions of a Complex Variable — Limit, Continuity & Differentiability

Q1. If $f(z) = z^2$, find $f(1+i)$ and determine $\lim_{z \to i} f(z)$.

Solution:

Given $f(z) = z^2 = (x+iy)^2 = x^2 - y^2 + 2ixy$

Value at $z = 1+i$:

$f(1+i) = (1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i$

$\boxed{f(1+i) = 2i}$

Limit as $z \to i$:

$\lim_{z \to i} z^2 = i^2 = -1$

Since $f(z)$ is a polynomial, it is continuous everywhere, so the limit equals the function value.

$\boxed{\lim_{z \to i} z^2 = -1}$

Q2. Show that $f(z) = \overline{z}$ is continuous everywhere but not differentiable anywhere.

Solution:

Let $z = x+iy$, so $f(z) = \overline{z} = x - iy = u(x,y) + iv(x,y)$

where $u = x$ and $v = -y$.

Continuity: Both $u = x$ and $v = -y$ are continuous functions of $x$ and $y$. Therefore $f(z)=\bar{z}$ is continuous everywhere.

Differentiability: Check Cauchy-Riemann equations:

$\frac{\partial u}{\partial x} = 1, \quad \frac{\partial v}{\partial y} = -1$

$\frac{\partial u}{\partial y} = 0, \quad \frac{\partial v}{\partial x} = 0$

C-R equations require:

• $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \Rightarrow 1 = -1$ ❌ (False)

Since C-R equations are not satisfied at any point, $f(z) = \overline{z}$ is not differentiable anywhere.

$\boxed{f(z) = \overline{z} \text{ is continuous but nowhere differentiable}}$

Q3. Examine the differentiability of $f(z) = |z|^2$ at $z = 0$ and at $z \neq 0$.

Solution:

$f(z) = |z|^2 = x^2 + y^2$, so , .

At $z = 0$:

Use the definition of derivative:

$f'(0) = \lim_{z \to 0} \frac{f(z) - f(0)}{z - 0} = \lim_{z \to 0} \frac{|z|^2}{z} = \lim_{z \to 0} \frac{z\overline{z}}{z} = \lim_{z \to 0} \overline{z} = 0$

So $f(z)$ is differentiable at $z = 0$ with $f'(0) = 0$.

At $z \neq 0$:

Check C-R equations:

$\frac{\partial u}{\partial x} = 2x, \quad \frac{\partial v}{\partial y} = 0$

$\frac{\partial u}{\partial y} = 2y, \quad \frac{\partial v}{\partial x} = 0$

For C-R: $2x = 0$ and $2y = 0$, which gives $x = 0, y = 0$.

Thus C-R equations are satisfied only at $z = 0$.

$\boxed{f(z) = |z|^2 \text{ is differentiable only at } z = 0}$

Module 2: Analytic Functions & Cauchy-Riemann Equations

Q4. State the Cauchy-Riemann equations in Cartesian form. Verify them for $f(z) = e^z$.

Solution:

Cauchy-Riemann Equations (Cartesian):

If $f(z) = u(x,y) + iv(x,y)$ is differentiable at a point, then:

$\boxed{\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \quad \text{and} \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}}$

Verification for $f(z) = e^z$:

$e^z = e^{x+iy} = e^x(\cos y + i\sin y)$

So:

• $u = e^x \cos y$
• $v = e^x \sin y$

Compute partial derivatives:

$\frac{\partial u}{\partial x} = e^x \cos y, \quad \frac{\partial v}{\partial y} = e^x \cos y \quad \checkmark$

$\frac{\partial u}{\partial y} = -e^x \sin y, \quad -\frac{\partial v}{\partial x} = -e^x \sin y \quad \checkmark$

Both C-R equations are satisfied for all $(x,y)$.

$\boxed{e^z \text{ is analytic everywhere (entire function)}}$

Q5. Show that $f(z) = z^3$ is analytic and find its derivative.

Solution:

$f(z) = z^3 = (x+iy)^3 = x^3 + 3x^2(iy) + 3x(iy)^2 + (iy)^3$

$= x^3 - 3xy^2 + i(3x^2y - y^3)$

So:

• $u = x^3 - 3xy^2$
• $v = 3x^2y - y^3$

Check C-R:

$\frac{\partial u}{\partial x} = 3x^2 - 3y^2, \quad \frac{\partial v}{\partial y} = 3x^2 - 3y^2 \quad \checkmark$

$\frac{\partial u}{\partial y} = -6xy, \quad -\frac{\partial v}{\partial x} = -6xy \quad \checkmark$

C-R equations are satisfied everywhere and partial derivatives are continuous.

Therefore $f(z) = z^3$ is analytic everywhere.

Derivative:

$f'(z) = \frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x} = (3x^2 - 3y^2) + i(6xy)$

$= 3(x^2 - y^2 + 2ixy) = 3(x+iy)^2 = 3z^2$

$\boxed{f'(z) = 3z^2}$

Q6. Determine where $f(z) = x^2 + iy^2$ is analytic.

Solution:

$u = x^2$, $v = y^2$.

C-R equations:

$\frac{\partial u}{\partial x} = 2x, \quad \frac{\partial v}{\partial y} = 2y \Rightarrow 2x = 2y \Rightarrow x = y$

$\frac{\partial u}{\partial y} = 0, \quad -\frac{\partial v}{\partial x} = 0 \quad \checkmark \text{ (always)}$

C-R equations are satisfied only when $x = y$, i.e., along the line $y = x$.

Since analyticity requires C-R to hold in a neighborhood (open set), not just on a line:

$\boxed{f(z) = x^2 + iy^2 \text{ is nowhere analytic}}$

Q7. If $f(z) = u + iv$ is analytic, show that $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$.

Solution:

Since $f(z)$ is analytic, C-R equations hold:

$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \quad \text{...(1)}$

$\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \quad \text{...(2)}$

Differentiate (1) w.r.t. $x$:

$\frac{\partial^2 u}{\partial x^2} = \frac{\partial^2 v}{\partial x \partial y} \quad \text{...(3)}$

Differentiate (2) w.r.t. $y$:

$\frac{\partial^2 u}{\partial y^2} = -\frac{\partial^2 v}{\partial y \partial x} \quad \text{...(4)}$

Since mixed partial derivatives are equal (assuming continuity):

$\frac{\partial^2 v}{\partial x \partial y} = \frac{\partial^2 v}{\partial y \partial x}$

Adding (3) and (4):

$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{\partial^2 v}{\partial x \partial y} - \frac{\partial^2 v}{\partial y \partial x} = 0$

$\boxed{\nabla^2 u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0}$

Module 3: Cauchy-Riemann Equations in Polar Form

Q8. State and derive the Cauchy-Riemann equations in polar form.

Solution:

Let $z = re^{i\theta}$, so $f(z) = u(r,\theta) + iv(r,\theta)$.

Polar C-R Equations:

$\boxed{\frac{\partial u}{\partial r} = \frac{1}{r}\frac{\partial v}{\partial \theta} \quad \text{and} \quad \frac{\partial v}{\partial r} = -\frac{1}{r}\frac{\partial u}{\partial \theta}}$

Or equivalently:

$r\frac{\partial u}{\partial r} = \frac{\partial v}{\partial \theta}, \quad r\frac{\partial v}{\partial r} = -\frac{\partial u}{\partial \theta}$

Derivation:

We have $x = r\cos\theta$, $y = r\sin\theta$.

By chain rule:

$\frac{\partial u}{\partial r} = \frac{\partial u}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial r} = \frac{\partial u}{\partial x}\cos\theta + \frac{\partial u}{\partial y}\sin\theta$

$\frac{\partial v}{\partial \theta} = \frac{\partial v}{\partial x}\frac{\partial x}{\partial \theta} + \frac{\partial v}{\partial y}\frac{\partial y}{\partial \theta} = \frac{\partial v}{\partial x}(-r\sin\theta) + \frac{\partial v}{\partial y}(r\cos\theta)$

Using Cartesian C-R: $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ and :

$\frac{\partial v}{\partial \theta} = r\sin\theta\frac{\partial u}{\partial y} + r\cos\theta\frac{\partial u}{\partial x} = r\left(\frac{\partial u}{\partial x}\cos\theta + \frac{\partial u}{\partial y}\sin\theta\right) = r\frac{\partial u}{\partial r}$

Similarly for the second equation.

Q9. Verify the polar C-R equations for $f(z) = z^n$ where $n$ is an integer.

Solution:

$z^n = (re^{i\theta})^n = r^n e^{in\theta} = r^n(\cos n\theta + i\sin n\theta)$

So:

• $u = r^n \cos n\theta$
• $v = r^n \sin n\theta$

Compute derivatives:

$\frac{\partial u}{\partial r} = nr^{n-1}\cos n\theta$

$\frac{1}{r}\frac{\partial v}{\partial \theta} = \frac{1}{r} \cdot r^n \cdot n\cos n\theta = nr^{n-1}\cos n\theta \quad \checkmark$

$\frac{\partial v}{\partial r} = nr^{n-1}\sin n\theta$

$-\frac{1}{r}\frac{\partial u}{\partial \theta} = -\frac{1}{r}(-nr^n \sin n\theta) = nr^{n-1}\sin n\theta \quad \checkmark$

\boxed{\text{Polar C-R equations are satisfied for all } r > 0}

Module 4: Harmonic Functions

Q10. Show that $u = x^3 - 3xy^2$ is harmonic and find its harmonic conjugate $v$.

Solution:

Step 1: Verify harmonic

$\frac{\partial u}{\partial x} = 3x^2 - 3y^2, \quad \frac{\partial^2 u}{\partial x^2} = 6x$

$\frac{\partial u}{\partial y} = -6xy, \quad \frac{\partial^2 u}{\partial y^2} = -6x$

$\nabla^2 u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 6x - 6x = 0$

$\boxed{u \text{ is harmonic}}$

Step 2: Find harmonic conjugate $v$

Using C-R equations:

$\frac{\partial v}{\partial y} = \frac{\partial u}{\partial x} = 3x^2 - 3y^2$

Integrate w.r.t. $y$:

$v = 3x^2y - y^3 + \phi(x)$

Also:

$\frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y} = 6xy$

From our expression:

$\frac{\partial v}{\partial x} = 6xy + \phi'(x) = 6xy$

So $\phi'(x) = 0 \Rightarrow \phi(x) = C$.

Taking $C = 0$:

$\boxed{v = 3x^2y - y^3}$

The corresponding analytic function is $f(z) = u + iv = z^3$.

Q11. Show that $u = e^x \cos y$ is harmonic and find the analytic function $f(z) = u + iv$.

Solution:

Verify harmonic:

$\frac{\partial u}{\partial x} = e^x \cos y, \quad \frac{\partial^2 u}{\partial x^2} = e^x \cos y$

$\frac{\partial u}{\partial y} = -e^x \sin y, \quad \frac{\partial^2 u}{\partial y^2} = -e^x \cos y$

$\nabla^2 u = e^x \cos y - e^x \cos y = 0 \quad \checkmark$

Find $v$:

$\frac{\partial v}{\partial y} = \frac{\partial u}{\partial x} = e^x \cos y$

$v = e^x \sin y + \phi(x)$

$\frac{\partial v}{\partial x} = e^x \sin y + \phi'(x) = -\frac{\partial u}{\partial y} = e^x \sin y$

So $\phi'(x) = 0 \Rightarrow \phi(x) = C$.

$\boxed{v = e^x \sin y, \quad f(z) = e^x(\cos y + i\sin y) = e^z}$

Q12. If $u = \frac{x}{x^2+y^2}$ is harmonic, find $$ and the analytic function .

Solution:

Verify harmonic (optional, but good practice):

$\frac{\partial u}{\partial x} = \frac{y^2-x^2}{(x^2+y^2)^2}, \quad \frac{\partial^2 u}{\partial x^2} = \frac{2x(x^2-3y^2)}{(x^2+y^2)^3}$

$\frac{\partial u}{\partial y} = \frac{-2xy}{(x^2+y^2)^2}, \quad \frac{\partial^2 u}{\partial y^2} = \frac{2x(3y^2-x^2)}{(x^2+y^2)^3}$

$\nabla^2 u = 0 \quad \checkmark$

Find $v$ using Milne-Thomson method:

$f'(z) = \frac{\partial u}{\partial x} - i\frac{\partial u}{\partial y}$

Replacing $x$ by $z$ and $y$ by $0$:

$\frac{\partial u}{\partial x}\Big|_{(z,0)} = \frac{0-z^2}{(z^2)^2} = -\frac{1}{z^2}$

$\frac{\partial u}{\partial y}\Big|_{(z,0)} = 0$

So:

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

$f(z) = \frac{1}{z} + C = \frac{x-iy}{x^2+y^2} + C$

Therefore:

$\boxed{v = -\frac{y}{x^2+y^2}, \quad f(z) = \frac{1}{z}}$

Module 5: Milne-Thomson Method

Q13. State Milne-Thomson method and use it to find $f(z)$ given $u = x^2 - y^2$.

Solution:

Milne-Thomson Method:

If $f(z) = u + iv$ is analytic and either $u$ or $v$ is given, then:

$f'(z) = \frac{\partial u}{\partial x} - i\frac{\partial u}{\partial y}$

Replace $x$ by $z$ and $y$ by $0$ to get $f'(z)$, then integrate.

Application:

Given $u = x^2 - y^2$:

$\frac{\partial u}{\partial x} = 2x, \quad \frac{\partial u}{\partial y} = -2y$

$f'(z) = 2x - i(-2y) = 2x + 2iy = 2(x+iy) = 2z$

Replacing $x \to z, y \to 0$:

$f'(z) = 2z$

$f(z) = z^2 + C$

$\boxed{f(z) = z^2}$

Q14. Using Milne-Thomson method, find the analytic function $f(z)$ whose real part is $u = e^{2x}\cos 2y$.

Solution:

$\frac{\partial u}{\partial x} = 2e^{2x}\cos 2y, \quad \frac{\partial u}{\partial y} = -2e^{2x}\sin 2y$

By Milne-Thomson:

$f'(z) = \frac{\partial u}{\partial x} - i\frac{\partial u}{\partial y} = 2e^{2x}\cos 2y + 2ie^{2x}\sin 2y$

Replace $x \to z, y \to 0$:

$f'(z) = 2e^{2z}\cos 0 + 2ie^{2z}\sin 0 = 2e^{2z}$

$f(z) = e^{2z} + C$

$\boxed{f(z) = e^{2z}}$

Q15. Find the analytic function $f(z)$ if $v = \frac{x}{x^2+y^2}$.

Solution:

Given $v$, we use:

$f'(z) = \frac{\partial v}{\partial y} + i\frac{\partial v}{\partial x}$

Compute:

$\frac{\partial v}{\partial y} = -\frac{2xy}{(x^2+y^2)^2}$

$\frac{\partial v}{\partial x} = \frac{(x^2+y^2) - x(2x)}{(x^2+y^2)^2} = \frac{y^2-x^2}{(x^2+y^2)^2}$

So:

$f'(z) = -\frac{2xy}{(x^2+y^2)^2} + i\frac{y^2-x^2}{(x^2+y^2)^2}$

Replace $x \to z, y \to 0$:

$f'(z) = 0 + i\frac{-z^2}{z^4} = -\frac{i}{z^2}$

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

$\boxed{f(z) = \frac{i}{z}}$

Module 6: Conformal Mapping

Q16. Define conformal mapping. Show that $w = z^2$ is conformal except at $z = 0$.

Solution:

Definition: A mapping $w = f(z)$ is conformal at a point $z_0$ if it preserves angles between curves passing through $z_0$, both in magnitude and sense (direction).

A mapping is conformal at $z_0$ if $f(z)$ is analytic at $z_0$ and $$.

For $w = z^2$:

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

• $f(z)$ is analytic everywhere.
• $f'(z) = 0$ when $z = 0$.

Therefore, $w = z^2$ is conformal at all points except $z = 0$.

At $z = 0$, angles are doubled (not preserved). For example, the lines $\arg z = 0$ and $\arg z = \frac{\pi}{2}$ map to and , so the right angle becomes a straight angle.

$\boxed{w = z^2 \text{ is conformal for all } z \neq 0}$

Q17. Find the image of the line $x = 1$ under the mapping $w = \frac{1}{z}$.

Solution:

Let $z = x + iy = 1 + iy$.

$w = \frac{1}{z} = \frac{1}{1+iy} = \frac{1-iy}{1+y^2} = \frac{1}{1+y^2} - i\frac{y}{1+y^2}$

Let $w = u + iv$:

$u = \frac{1}{1+y^2}, \quad v = -\frac{y}{1+y^2}$

Note that:

$u^2 + v^2 = \frac{1+y^2}{(1+y^2)^2} = \frac{1}{1+y^2} = u$

So:

$u^2 + v^2 = u \Rightarrow u^2 - u + v^2 = 0 \Rightarrow \left(u-\frac{1}{2}\right)^2 + v^2 = \frac{1}{4}$

$\boxed{\text{The image is a circle with center } \left(\frac{1}{2}, 0\right) \text{ and radius } \frac{1}{2}}$

Q18. Find the image of the infinite strip 0 < y < \frac{1}{2} under $w = \frac{1}{z}$.

Solution:

Using $w = u + iv = \frac{1}{z} = \frac{x-iy}{x^2+y^2}$:

$u = \frac{x}{x^2+y^2}, \quad v = -\frac{y}{x^2+y^2}$

Note that $\frac{v}{u} = -\frac{y}{x}$, so .

Also: $u^2 + v^2 = \frac{1}{x^2+y^2}$

From $v = -\frac{y}{x^2+y^2} = -y(u^2+v^2)$:

$y = -\frac{v}{u^2+v^2}$

For 0 < y < \frac{1}{2}:

0 < -\frac{v}{u^2+v^2} < \frac{1}{2}

Since y > 0, we need v < 0.

The upper bound:

-\frac{v}{u^2+v^2} < \frac{1}{2} \Rightarrow -2v < u^2+v^2 \Rightarrow u^2 + v^2 + 2v > 0

u^2 + (v+1)^2 > 1

So the image is the region outside the circle $u^2 + (v+1)^2 = 1$ with v < 0.

\boxed{\text{Image: } u^2 + (v+1)^2 > 1, \quad v < 0}

Module 7: Möbius Transformation (Bilinear Transformation)

Q19. Define Möbius transformation and state its properties.

Solution:

Definition: A Möbius transformation (or bilinear transformation) is a mapping of the form:

$\boxed{w = \frac{az+b}{cz+d}, \quad ad-bc \neq 0}$

where $a, b, c, d$ are complex constants.

Properties:

1. Conformality: Möbius transformations are conformal (angle-preserving) everywhere except at the pole $z = -\frac{d}{c}$.

2. Circle-preserving: Maps circles and straight lines in the $z$-plane to circles or straight lines in the $w$-plane.

Q20. Find the Möbius transformation that maps $z = 0, 1, \infty$ to $w = i, 1, -i$ respectively.

Solution:

Use the cross-ratio formula. If $z_1, z_2, z_3$ map to $w_1,{}_{}$:

$\frac{(w-w_1)(w_2-w_3)}{(w-w_3)(w_2-w_1)} = \frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}$

Given: $z_1 = 0, z_2 = 1, z_3 = \infty$ and .

For $z_3 = \infty$, the term $(z-z_3)$ simplifies. The formula becomes:

$\frac{(w-i)(1-(-i))}{(w-(-i))(1-i)} = \frac{(z-0)(1-\infty)}{(z-\infty)(1-0)} \Rightarrow \frac{(w-i)(1+i)}{(w+i)(1-i)} = \frac{z}{1}$

Simplify:

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

Note that $\frac{1+i}{1-i} = \frac{(1+i)^2}{2} = \frac{2i}{2} = i$.

So:

$\frac{i(w-i)}{w+i} = z \Rightarrow i(w-i) = z(w+i)$

$iw + 1 = zw + zi$

$w(i - z) = zi - 1$

$w = \frac{zi - 1}{i - z} = \frac{1 - iz}{z - i}$

$\boxed{w = \frac{1 - iz}{z - i}}$

Q21. Find the Möbius transformation that maps the unit circle $|z| = 1$ onto the real axis $\text{Im}(w) = 0$ with $z = 1 \to w = 0$, , .

Solution:

Given mappings:

• $z = 1 \to w = 0$
• $z = i \to w = 1$
• $$

Since $z = -1$ maps to $w = \infty$, the denominator must vanish at $z = -1$.

So the transformation has the form:

$w = k\frac{z-1}{z+1}$

Use $z = i \to w = 1$:

$1 = k\frac{i-1}{i+1} = k\frac{(i-1)^2}{(i+1)(i-1)} = k\frac{-1-2i+1}{-1-1} = k\frac{-2i}{-2} = ki$

So $k = -i$.

$\boxed{w = -i\frac{z-1}{z+1} = i\frac{1-z}{z+1}}$

Q22. Find the fixed points of $w = \frac{2z+3}{z+4}$.

Solution:

Fixed points satisfy $w = z$:

$z = \frac{2z+3}{z+4}$

$z(z+4) = 2z+3$

$z^2 + 4z = 2z + 3$

$z^2 + 2z - 3 = 0$

$(z+3)(z-1) = 0$

$\boxed{\text{Fixed points: } z = 1 \text{ and } z = -3}$

Module 8: Mixed Exam-Level Problems

Q23. If $f(z) = u + iv$ is analytic and $u - v = e^x(\cos y - \sin y)$, find .

Solution:

Given $u - v = e^x(\cos y - \sin y)$.

Differentiate w.r.t. $x$:

$\frac{\partial u}{\partial x} - \frac{\partial v}{\partial x} = e^x(\cos y - \sin y) \quad \text{...(1)}$

Differentiate w.r.t. $y$:

$\frac{\partial u}{\partial y} - \frac{\partial v}{\partial y} = e^x(-\sin y - \cos y) \quad \text{...(2)}$

Using C-R equations: $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ and .

From (1):

$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = e^x(\cos y - \sin y) \quad \text{...(3)}$

From (2):

$\frac{\partial u}{\partial y} - \frac{\partial u}{\partial x} = e^x(-\sin y - \cos y) \quad \text{...(4)}$

Add (3) and (4):

$2\frac{\partial u}{\partial y} = -2e^x \sin y \Rightarrow \frac{\partial u}{\partial y} = -e^x \sin y$

Subtract (4) from (3):

$2\frac{\partial u}{\partial x} = 2e^x \cos y \Rightarrow \frac{\partial u}{\partial x} = e^x \cos y$

Integrate $\frac{\partial u}{\partial x} = e^x \cos y$ w.r.t. $x$:

$u = e^x \cos y + \phi(y)$

Differentiate w.r.t. $y$:

$\frac{\partial u}{\partial y} = -e^x \sin y + \phi'(y) = -e^x \sin y$

So $\phi'(y) = 0 \Rightarrow \phi(y) = C$.

$u = e^x \cos y$

Then:

$v = u - e^x(\cos y - \sin y) = e^x \cos y - e^x \cos y + e^x \sin y = e^x \sin y$

$\boxed{f(z) = e^x(\cos y + i\sin y) = e^z}$

Q24. Show that an analytic function with constant modulus is constant.

Solution:

Let $f(z) = u + iv$ be analytic and $|f(z)| = c$ (constant).

Then $u^2 + v^2 = c^2$.

Differentiate w.r.t. $x$:

$2u\frac{\partial u}{\partial x} + 2v\frac{\partial v}{\partial x} = 0 \Rightarrow u\frac{\partial u}{\partial x} + v\frac{\partial v}{\partial x} = 0 \quad \text{...(1)}$

Differentiate w.r.t. $y$:

$2u\frac{\partial u}{\partial y} + 2v\frac{\partial v}{\partial y} = 0 \Rightarrow u\frac{\partial u}{\partial y} + v\frac{\partial v}{\partial y} = 0 \quad \text{...(2)}$

Using C-R equations in (2):

$-u\frac{\partial v}{\partial x} + v\frac{\partial u}{\partial x} = 0 \quad \text{...(3)}$

From (1) and (3), we have a system:

$u\frac{\partial u}{\partial x} + v\frac{\partial v}{\partial x} = 0$

$v\frac{\partial u}{\partial x} - u\frac{\partial v}{\partial x} = 0$

In matrix form:

$\begin{pmatrix} u & v \\ v & -u \end{pmatrix} \begin{pmatrix} \frac{\partial u}{\partial x} \\ \frac{\partial v}{\partial x} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$

The determinant is $-u^2 - v^2 = -c^2 \neq 0$ (unless ).

By Cramer's rule, the only solution is:

$\frac{\partial u}{\partial x} = 0, \quad \frac{\partial v}{\partial x} = 0$

By C-R, all partial derivatives are zero. Therefore $u$ and $v$ are constants.

$\boxed{f(z) \text{ is constant}}$

Q25. Find the bilinear transformation that maps the points $1, i, -1$ onto $0, 1, \infty$.

Solution:

Using cross-ratio:

$\frac{(w-0)(1-\infty)}{(w-\infty)(1-0)} = \frac{(z-1)(i+1)}{(z+1)(i-1)}$

Simplifying (since terms with $\infty$ cancel appropriately):

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

Compute $\frac{i+1}{i-1}$:

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

So:

$w = -i\frac{z-1}{z+1} = i\frac{1-z}{z+1}$

$\boxed{w = i\frac{1-z}{1+z}}$

Q26. If $f(z)$ is analytic, show that $\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\right)|f(z)|^2 = 4|f'(z)|^2$.

Solution:

Let $f(z) = u + iv$, so $|f(z)|^2 = u^2 + v^2$.

$\frac{\partial}{\partial x}(u^2+v^2) = 2u\frac{\partial u}{\partial x} + 2v\frac{\partial v}{\partial x}$

$\frac{\partial^2}{\partial x^2}(u^2+v^2) = 2\left(\frac{\partial u}{\partial x}\right)^2 + 2u\frac{\partial^2 u}{\partial x^2} + 2\left(\frac{\partial v}{\partial x}\right)^2 + 2v\frac{\partial^2 v}{\partial x^2}$

Similarly for $y$:

$\frac{\partial^2}{\partial y^2}(u^2+v^2) = 2\left(\frac{\partial u}{\partial y}\right)^2 + 2u\frac{\partial^2 u}{\partial y^2} + 2\left(\frac{\partial v}{\partial y}\right)^2 + 2v\frac{\partial^2 v}{\partial y^2}$

Adding and using $\nabla^2 u = 0$, $\nabla^2 v = 0$:

$= 2\left[\left(\frac{\partial u}{\partial x}\right)^2 + \left(\frac{\partial u}{\partial y}\right)^2 + \left(\frac{\partial v}{\partial x}\right)^2 + \left(\frac{\partial v}{\partial y}\right)^2\right]$

Using C-R: $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ and :

$= 2\left[\left(\frac{\partial u}{\partial x}\right)^2 + \left(-\frac{\partial v}{\partial x}\right)^2 + \left(\frac{\partial v}{\partial x}\right)^2 + \left(\frac{\partial u}{\partial x}\right)^2\right]$

$= 4\left[\left(\frac{\partial u}{\partial x}\right)^2 + \left(\frac{\partial v}{\partial x}\right)^2\right] = 4|f'(z)|^2$

$\boxed{\nabla^2|f(z)|^2 = 4|f'(z)|^2}$

Q27. Show that $w = \frac{z-i}{z+i}$ maps the real axis onto the unit circle $|w| = 1$.