Discussion
Let \(f : \mathbf{C} \to \mathbf{C}\) be a bijective analytic function. Then :
Answer: A
\(f\) must be a linear polynomial, \(f(z) = az + b\) with \(a \ne 0\).
An entire injective function cannot have an essential singularity at \(\infty\): by Casorati-Weierstrass, \(f(\{|z| \gt 1\})\) would be dense and would meet the open set \(f(\{|z| \lt 1\})\), contradicting injectivity. So \(f\) is a polynomial, and an injective polynomial must have degree 1, since a polynomial of degree \(n \ge 2\) takes most values \(n\) times.
So \(f'(z) = a \ne 0\) everywhere, ruling out (B), and \(f^{-1}(w) = (w - b)/a\) is continuous and analytic, ruling out (C) and (D).
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