Discussion
Let \(f(z)\) be an analytic function such that
\[ Re(f'(z)) = 3x^{2} - 4y - 3y^{2}, \]
\[ f(i) = 0, \; f'(0) = 0, \]
where \(i = \sqrt{-1}\). Then the value of \(f(1)\) is equal to
Answer: B
The value is \(f(1) = 1 + 5i\).
Look for an analytic \(f'(z)\) whose real part is \(3x^{2}-4y-3y^{2}\). Since \(\text{Re}(3z^{2}) = 3x^{2}-3y^{2}\) and \(\text{Re}(4iz) = -4y\):
\[ f'(z) = 3z^{2} + 4iz + ic \]
for a real constant c, and f'(0) = 0 gives c = 0. Integrating:
\[ \begin{aligned} f(z) &= z^{3} + 2iz^{2} + K \\ f(i) &= -i - 2i + K = 0 \Rightarrow K = 3i \\ f(1) &= 1 + 2i + 3i = 1 + 5i \end{aligned} \]
Once f'(z) is fixed, f(i) = 0 fixes K, so f(1) has only one possible value. Dropping K would give 1 + 2i, which is not offered; 4 + 2i, 4 - 2i and 1 - i do not come from any f with this derivative that vanishes at i.
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