Analytic Functions: Complex Differentiability

At a Glance

• Frequency: 1 sub-part across 1 of 13 years (2024)
• Priority tier: T4
• Marks (count): 20 (1)
• Average solve time: ~25 min
• Difficulty mix: medium 1
• Section: B | Dominant type: computation

Why This Chapter Matters

Analytic functions are the central objects of complex analysis, and UPSC has tested complex differentiability checks using the Cauchy-Riemann equations. The 2024 question required checking whether a given function is analytic and computing its derivative — a direct application of the C-R criterion with continuous partials. Mastering this atom is essential groundwork for contour integration and Laurent series topics that appear far more frequently.

Minimum Theory

Complex Differentiability

A function $f : \mathbb{C} \to \mathbb{C}$ is complex-differentiable (has a complex derivative) at $z_0$ if the limit

$f'(z_0) = \lim_{h \to 0} \frac{f(z_0 + h) - f(z_0)}{h}$

exists and is the same for every direction in which $h \to 0$ in $\mathbb{C}$. This is strictly stronger than real differentiability: the limit must agree whether $h \to 0$ along the real axis, the imaginary axis, or any other path.

Analytic (Holomorphic) Functions

$f$ is analytic (or holomorphic) at $z_0$ if it is complex-differentiable throughout some open neighbourhood of $z_0$. Being analytic at a single isolated point is not enough; differentiability must hold in a disk around $z_0$.

The Cauchy-Riemann Equations

Write $z = x + iy$, $f(z) = u(x,y) + iv(x,y)$ where $u, v$ are real-valued.

Necessary condition. If $f'(z_0)$ exists, then at $(x_0, y_0)$:

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

Sufficient condition. If $u$ and $v$ have continuous partial derivatives in a neighbourhood of $(x_0, y_0)$ and satisfy the C-R equations there, then $f$ is complex-differentiable at $z_0$, with

$f'(z_0) = \frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x} = \frac{\partial v}{\partial y} - i\frac{\partial u}{\partial y}.$

Key Consequences

• If $f$ is analytic in a domain $D$, both $u$ and $v$ are harmonic in $D$: $\nabla^2 u = 0$ and $\nabla^2 v = 0$.
• $u$ and $v$ are called conjugate harmonic functions.
• The modulus $|f'(z)|$ gives the local scale factor of the mapping; $\arg f'(z)$ gives the local rotation.

Standard Analytic Functions

Non-Analytic Functions

• $\overline{z} = x - iy$: C-R gives $1 = -1$, contradiction — nowhere analytic.
• $|z|^2 = x^2 + y^2$: differentiable only at $z = 0$, not analytic anywhere.
• Any function involving $\overline{z}$ explicitly is not analytic.

Question Archetypes

check-analyticity (1 question; 2024)

Recognition Cues

• Question asks to “check”, “test”, or “determine” analyticity of an explicit $f(z)$.
• Function is given in terms of $x, y$ or in terms of $z$ and $\bar{z}$.
• May ask you to find $f'(z)$ as a follow-up.

Solution Template

1. Write $f(z) = u(x,y) + iv(x,y)$ by separating real and imaginary parts.
2. Compute all four partial derivatives: $u_x, u_y, v_x, v_y$.
3. Check C-R equations: $u_x = v_y$ and $u_y = -v_x$.
4. Check continuity of partial derivatives.
5. If C-R holds and partials are continuous: state $f$ is analytic and compute $f'(z) = u_x + iv_x$.
6. If C-R fails anywhere: state $f$ is not analytic at those points (or nowhere analytic).

Worked Example

2024 Paper 2, 2024-P2-Q1a (20 marks)

Show that $f(z) = e^z$ is analytic everywhere and find $f'(z)$.

Write $z = x + iy$. Then

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

So $u(x,y) = e^x\cos y$ and $v(x,y) = e^x\sin y$.

Compute partial derivatives:

$u_x = e^x\cos y, \quad u_y = -e^x\sin y,$ $v_x = e^x\sin y, \quad v_y = e^x\cos y.$

Check C-R equations:

$u_x = e^x\cos y = v_y \quad \checkmark$ $u_y = -e^x\sin y = -v_x \quad \checkmark$

All four partials are continuous everywhere on $\mathbb{C}$. By the sufficiency theorem, $f(z) = e^z$ is analytic (entire) on $\mathbb{C}$.

The derivative is:

$f'(z) = u_x + iv_x = e^x\cos y + i\,e^x\sin y = e^{x+iy} = e^z.$

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

$\blacksquare$

Common Traps

• Confusing the necessary condition with sufficient: C-R alone (without continuous partials) only proves differentiability is possible, not guaranteed — state the sufficiency criterion explicitly.
• Computing partials carelessly: $u_y$ and $v_x$ are the ones that produce the minus sign; students often mix them up.
• Claiming $|z|^2 = x^2+y^2$ is analytic because it is real and smooth — it is not; check C-R.
• Forgetting that $\overline{z}$, $\text{Re}(z)$, $\text{Im}(z)$ are not analytic (except at isolated points at best).

Marks-Aware Writing

This is a 20-mark Section B question. UPSC expects full working:

1. State the definition of analyticity (or reference C-R as the test).
2. Separate $u$ and $v$ explicitly — do not skip this.
3. Display all four partial derivatives.
4. Verify both C-R equations with explicit equality.
5. Cite the sufficiency theorem (continuous partials + C-R $\Rightarrow$ analytic).
6. Write out $f'(z)$ using the C-R formula, simplify to closed form in $z$.

Partial credit is awarded at each of these steps, so structured presentation pays.

Practice Set

Only one historical question on this atom (shown above).