Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.25)**
**1. Complex Analysis Principles (Cauchy-Riemann Equations)**:
- Given function: $f(z) = z + |z - a|^2$, where $a \in \mathbb{R}$.
- Let $z = x + iy$. Then $|z - a|^2 = (x - a)^2 + y^2$.
- $f(z) = x + iy + (x - a)^2 + y^2 = u(x, y) + i v(x, y)$ with:
$u(x, y) = x + (x - a)^2 + y^2, \quad v(x, y) = y$
**2. Step-by-Step Evaluation**:
- **Continuity**: Both $u(x,y)$ and $v(x,y)$ are elementary polynomials in $x$ and $y$. Thus, $f(z)$ is continuous everywhere on $\mathbb{C}$, including at $(a, a)$. **(A) is TRUE**.
- **Complex Differentiability**: Requires Cauchy-Riemann (CR) equations:
$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$
- Partial derivatives:
$\frac{\partial u}{\partial x} = 1 + 2(x - a), \quad \frac{\partial u}{\partial y} = 2y$
$\frac{\partial v}{\partial x} = 0, \quad \frac{\partial v}{\partial y} = 1$
- Evaluating CR equations:
$1 + 2(x - a) = 1 \implies 2(x - a) = 0 \implies x = a$
$2y = -0 \implies y = 0$
- The CR equations are satisfied **only at the single point** $(x, y) = (a, 0)$.
- At $(a, a)$: $y = a \ne 0$, so CR fails $\implies$ Not differentiable at $(a, a)$. (B) is FALSE.
- At $(a, 0)$: CR equations and partial derivatives are continuous $\implies f(z)$ is complex-differentiable at $(a, 0)$. **(C) is TRUE**.
- Analyticity requires differentiability in an open neighborhood. Since $f(z)$ is differentiable only at a single isolated point, it is **not analytic** anywhere. (D) is FALSE.
**3. Official Verification**:
The verified official IIT Master Answer is **Options (A, C)**.