Discussion
Is \( f(z) = \bar{z} \) analytic anywhere in the plane?
Answer: C
Here \( u = x \) and \( v = -y \), so \( u_{x} = 1 \) while \( v_{y} = -1 \):
\[ u_{x} \;=\; 1 \;\neq\; -1 \;=\; v_{y}. \]
The first Cauchy-Riemann equation fails at every point, so the conjugate is nowhere analytic despite being continuous everywhere. Continuity is no guarantee of complex differentiability
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