## 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