Establish the identity (ez)n = enz (n = 0, 1, 2, . . .) in the following
Question:
(ez)n = enz (n = 0, ±1, ±2, . . .)
in the following way.
(a) Use mathematical induction to show that it is valid when n = 0, 1, 2, . . . .
(b) Verify it for negative integers n by first recalling from Sec. 7 that
zn = (z−1)m (m = −n = 1, 2, . . .)
when z ≠ 0 and writing (ez)n = (1/ez)m. Then use the result in part (a), together with the property 1/ez = e−z (Sec. 29) of the exponential function.
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Related Book For
Complex Variables and Applications
ISBN: 978-0073051949
8th edition
Authors: James Brown, Ruel Churchill
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