Correct answer: (B) If g and h are differentiable at z0, f is differentiable there.
Explanation
Differentiation is linear: if g and h have complex derivatives at z0, their sum has derivative g′(z0)+h′(z0). Thus **B** is always true.
The converse fails because nondifferentiable terms can cancel, and continuity alone is weaker than complex differentiability. Real and imaginary parts viewed as separate complex-valued functions need not be holomorphic; for example Re(z) is not, although f(z)=z is.
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