0. Differentiability and the Cauchy-Riemann Equations 7

0.6. Suppose that / is differentiate at z and that g is differentiate at f(z).

Show that the composition g o f is differentiable at z1 with

(gof)'(z)=g'(f(z))f'(z).

Note. An obvious approach to Exercise 6 is to write

9Qf(z)-gof(w)

=

(g(f(z))-g(f(w))\ /f(z)-f(w)\

z-w \ f(z)-f(w) J\ z-w J

and use Exercise 2, which shows that f(w) tends to f(z) as w approaches z.

There is a slight problem with this approach: It may happen that f(w) =

f(z) for w near z, in which case we are dividing by 0.

If you don't see how tofixthis problem, you can find the solution in any

calculus book. Better is to simply avoid the problem by using Exercise 0.1

instead of the definition of the derivative via quotients!

A slightly informal version of a solution using Exercise 0.1 would use the

exercise twice to show that

go f(z + h) = g(f(z) + f'(z)h + o(h))

= /(/(*)) + g'(f(z))(f(z)h + o(h)) + o(f'(z)h + o(h))

= 9(f(z)) + g'(f(z))f(z)h + g'(f(z))o(h) + o(f'(z)h + o(h))

= g(f(z)) + g'(f(z))f'(z)h + o(h),

and then invoke the exercise once more to conclude that (g o

f)f(z)

—

g'{f(z))f'{z). The whole point of this "o" notation is to allow one to per-

form this sort of manipulation with error terms. However, if we're new to

all this we probably want to give a more careful argument; in this regard

you might note that f(z + h) = f(z) +

ff(z)h

+ o(h) holds if and only if

f(z + h) — f(z) +

ff{z)h

+ E(h), where E satisfies this condition: For every

e 0 there exists S 0 such that \E(h)\ e\h\ for all h with \h\ S. (The

point to the rephrasing is to avoid the division in the definition of "o(/i)".)

The next two exercises illustrate the problems that arise in attempting to

give a version of Proposition 0.1 involving only partial derivatives (without

extra hypotheses like continuity of the partials).

0.7. Define / : C - C by

ro (x = o),

f(x + iy)=l 0 fo = 0),

v 1 (otherwise).

Show that / satisfies the Cauchy-Riemann equations at the origin although

/ is not complex-differentiable at the origin.