Chap 1 The Number System

# Furthermore pz z 1z 2 2 so that two other solutions

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Unformatted text preview: n = rn (cos nθ + i sin nθ) = R(cos α + i sin α). It follows that rn = R, and we can take nθ = α + 2kπ, so that (14) θ= α + 2kπ , n where k = 0, 1, . . . , n − 1. where k = 0, 1, . . . , n − 1, Note that no two values of θ in (14) diﬀer by an integer multiple of 2π . It follows that (15) z= √ n R cos α + 2kπ α + 2kπ + i sin n n , where k = 0, 1, . . . , n − 1, give n distinct complex numbers. On the other hand, it follows from (15) and de Moivre’s theorem that each of the n numbers in (15) satisﬁes z n = c. We have proved the following result. PROPOSITION 1F. Suppose that c = R(cos α + i sin α), where c, α ∈ R and c > 0. Then the solutions of the equation z n = c are given by (15). Example 1.7.1. The 7-th roots of 1 − i can be calculated as follows. Note here that c=1−i= √ 2 cos 7π 7π + i sin 4 4 (observe that it is not necessary to use the principal argument). It follows from Proposition 1F that the 7-th roots of 1 − i are given by z= 14 √ 2 cos π...
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## This note was uploaded on 02/01/2009 for the course MATH 2343124 taught by Professor Staff during the Fall '08 term at UCSD.

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