1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

76 CHAPTER 2 • COMPLEX FUNCTIONS

• EXAMPLE 2.18 Show that the polynomial function given by

is continuous at each point ZQ in the complex plane.

Solution If ao is the constant function, then Jim ao = ao; and if ai ¥- 0,

z-zo

then we can use Definition 2.3 with f (z) = a 1 z and the choice o = rf.J to prove

that lim (a 1 z) = a 1 zo. Using Property (2-19) and mathematical induction, we

.z-zo

obtain

for k = 0, 1, 2, ... , n. (2-27)

We can extend Property (2-18) to a finite sum of terms and use the result of

Equation (2-27) to get

Jim P(z) = lim (tavk) = tak~ = P(zo).

z-zo z -zo k=O k=O

Conditions (2-24), (2-25), and (2-26) are satisfied, so we conclude that P is

continuous at zo.