Question

Complex analysis

(i) If \(f\) is differentiable at \(z_{0}\) then \(f\) is continuous at \(z_{0}\).

(ii) If \(f\) and \(g\) are differentiable at \(z_{0}\), then \(f+g\) and \(f g\) also are, and \((f+g)^{\prime}\left(z_{0}\right)=f^{\prime}\left(z_{0}\right)+g^{\prime}\left(z_{0}\right) \quad\) (sum rule); \((f g)^{\prime}\left(z_{0}\right)=f^{\prime}\left(z_{0}\right) g\left(z_{0}\right)+f\left(z_{0}\right) g^{\prime}\left(z_{0}\right) \quad\) (product rule). If in addition \(g\left(z_{0}\right) \neq 0\), then \(f / g\) is differentiable at \(z_{0}\), and \(\left(\frac{f}{g}\right)^{\prime}\left(z_{0}\right)=\frac{f^{\prime}\left(z_{0}\right) g\left(z_{0}\right)-f\left(z_{0}\right) g^{\prime}\left(z_{0}\right)}{g\left(z_{0}\right)^{2}} \quad\) (quotient rule).

(iii) If \(f\) is differentiable at \(z_{0}\) and \(g\) is differentiable at \(f\left(z_{0}\right)\), then the composite function \(g \circ f\) is differentiable at \(z_{0}\) and

$$ (g \circ f)^{\prime}\left(z_{0}\right)=g^{\prime}\left(f\left(z_{0}\right)\right) f^{\prime}\left(z_{0}\right) \quad \text { (chain rule). } $$

The proofs are left to the reader.

Exercise II.3.1. Prove statements (i)-(iii) in detail.

Answer 1