# Complex variables and applications by Brown J., Churchill R. By Brown J., Churchill R.

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17 Limits Involving the Point at Infinity 51 N P O z FIGURE 26 point z in the plane. By letting the point N of the sphere correspond to the point at infinity, we obtain a one to one correspondence between the points of the sphere and the points of the extended complex plane. The sphere is known as the Riemann sphere, and the correspondence is called a stereographic projection. Observe that the exterior of the unit circle centered at the origin in the complex plane corresponds to the upper hemisphere with the equator and the point N deleted.

2kπ n (k = 0, 1, 2, . . , n − 1). When n = 2, these roots are, of course, ±1. When n ≥ 3, the regular polygon at whose vertices the roots lie is inscribed in the unit circle |z| = 1, with one vertex corresponding to the principal root z = 1 (k = 0). In view of expression (3), Sec. 9, these roots are simply 1, ωn , ωn2 , . . , ωnn−1 where ωn = exp i 2π n . See Fig. 13, where the cases n = 3, 4, and 6 are illustrated. Note that ωnn = 1. y y 1x y 1x 1x FIGURE 13 √ 1/2 EXAMPLE √3. The two values ck (k = 0, 1) of ( 3 + i) , which are the square roots of 3 + i, are found by writing √ π 3 + i = 2 exp i + 2kπ (k = 0, ±1, ±2, .

Nn−1 where ωn = exp i 2π n . See Fig. 13, where the cases n = 3, 4, and 6 are illustrated. Note that ωnn = 1. y y 1x y 1x 1x FIGURE 13 √ 1/2 EXAMPLE √3. The two values ck (k = 0, 1) of ( 3 + i) , which are the square roots of 3 + i, are found by writing √ π 3 + i = 2 exp i + 2kπ (k = 0, ±1, ±2, . ) 6 and (see Fig. 14) (3) ck = √ π + kπ 2 exp i 12 (k = 0, 1). Euler’s formula tells us that √ √ π π π = 2 cos + i sin , c0 = 2 exp i 12 12 12 and the trigonometric identities 10/29/07 3:32pm 28 Brown-chap01-v3 sec.