Discussion
For \(z = x + iy\), let \(f(z) = x^{2} + y^{2} + 2xyi\). Then :
Answer: B
\(f\) is nowhere analytic.
Here \(u = x^{2} + y^{2}\) and \(v = 2xy\). The Cauchy-Riemann equations need \(u_{x} = v_{y}\) and \(u_{y} = -v_{x}\):
\[ \begin{aligned} u_{x} &= 2x = v_{y} \\ u_{y} &= 2y,\quad -v_{x} = -2y \end{aligned} \]
The second holds only when \(y = 0\). So \(f\) is complex differentiable exactly on the real axis. Analyticity at a point needs differentiability on a whole neighbourhood, and every disc around a real point contains points with \(y \ne 0\), so \(f\) is analytic nowhere.
That is the trap in option D: differentiable on the real axis is not analytic there. 'Everywhere' and 'only at 0' are both ruled out by the same computation.
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