7.4 • $INGULARITIES, ZEROS, AND POLES 279- EXAMPLE 7.10 From Theorem 7.10 we see that the function
z1 z11 z15
f (z) = zsin z^2 = z^3 - - + - - - + · · ·
3! 5! 7!
has a zero of order 3 at z = 0. Definition 7.6 confirms this faet because
f ' (z) = 2z^2 cosz^2 + sinz^2 ,
f" (z) = 6z cosz^2 - 4z^3 sin z^2 ,
f '" (z) = 6 cos z^2 - 8z^4 cos z^2 - 24z^2 sin z^2.
Then f (O) = f' (O) = !" (O) = 0, but!"' (O) = 6 =F 0.