Chap 1 The Number System

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online