## Homework 1

Hint for complex chain rule:

Let

$ \frac{f(z)-f(a)}{z-a} - f'(a) = E_f(z). $

Then

$ f(z)= f(a)+f'(a)(z-a)+E_f(z)(z-a). $

Note that if $ f(z) $ is complex differentiable at $ a $, then

$ \lim_{z\to a} E_f(z)=0. $

Next, let

$ \frac{g(w)-g(A)}{w-A} - g'(A) = E_g(w), $

where $ A=f(a). $

Now

$ g(w)= g(A)+g'(A)(w-A)+E_g(w)(w-A). $

To prove the Chain Rule, let $ w=f(z) $ in the formula above and write out the difference quotient for $ g(f(z)) $. The limit should become obvious. You will need to use the fact that $ f $ complex differentiable at $ a $ implies that $ f $ is continuous at $ a $ in order to deduce that $ E_g(f(z)) $ tends to zero as $ z $ tends to $ a $. --Steve Bell