Discussion
If \(f(z) = u + iv\) is analytic on a domain and \(v = y + 2xy\), then \(u =\)
Answer: B
The conjugate is \(u = x + x^{2} - y^{2} + c\).
For \(f = u + iv\) to be analytic, the Cauchy-Riemann equations \(u_{x} = v_{y}\) and \(u_{y} = -v_{x}\) must hold. With \(v = y + 2xy\):
\[ \begin{aligned} u_{x} &= v_{y} = 1 + 2x \\ u &= x + x^{2} + \phi(y) \\ u_{y} &= \phi'(y) = -v_{x} = -2y \\ \phi(y) &= -y^{2} + c \end{aligned} \]
So \(u = x + x^{2} - y^{2} + c\) and \(f(z) = z + z^{2} + c\). The option \(x^{2} - y^{2} + c\) drops the \(x\) that comes from the \(y\) in \(v\); \(x + 2xy + c\) just copies the form of \(v\); and \(y^{2} - x^{2} + y + c\) has the signs reversed and fails \(u_{x} = 1 + 2x\).
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