By Hathaway A.S.

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**Example text**

Hence t∗ = arctan tan and z = −2 cos a a a + 2π = − π + 2π = + π 2 2 2 a a a + π + i sin +π cos 2 2 2 . (c) If a = π, then z = 0. Problem 2. Find all complex numbers z such that |z| = 1 and z z = 1. + z z Solution. Let z = cos x + i sin x, x ∈ [0, 2π). Then 1= z z |z 2 + z 2 | = | cos 2x + i sin 2x + cos 2x − i sin 2x| = 2| cos 2x|, + = z z |z|2 whence cos 2x = 1 1 or cos 2x = − . 2 2 If cos 2x = 12 , then x1 = π , 6 x2 = 5π , 6 x3 = 7π , 6 x4 = 11π . 6 If cos 2x = − 12 , then x5 = π , 3 x6 = 2π , 3 x7 = 4π 5π , x8 = .

The polar radius r of the geometric image of z is equal to the modulus of z. For z = 0, the modulus and argument of z are uniquely determined. Consider z = r(cos t∗ + i sin t∗ ) and let t = t∗ + 2kπ for an integer k. , every complex number z can be represented as z = r(cos t+i sin t), where r ≥ 0 and t ∈ R. The set Arg z = {t : t∗ + 2kπ, k ∈ Z} is called the extended argument of the complex number z. Therefore, two complex numbers z1 , z2 = 0 represented as z1 = r1 (cos t1 + i sin t1 ) and z2 = r2 (cos t2 + i sin t2 ) are equal if and only if r1 = r2 and t1 − t2 = 2kπ, for some integer k.

We obtain ρn = r and nϕ = t∗ + 2kπ for k ∈ Z; hence ρ = t∗ 2π +k· for k ∈ Z. n n So far, the roots of equation (1) are √ Zk = n r(cos ϕk + i sin ϕk ) for k ∈ Z. √ n r and ϕk = Now observe that 0 ≤ ϕ0 < ϕ1 < · · · < ϕn−1 < 2π, so the numbers ϕk , k ∈ {0, 1, . . , ϕ∗k = ϕk . Until now, we had n distinct roots of z0 : Z0 , Z1 , . . , Zn−1 . Consider some integer k and let r ∈ {0, 1, . . , n − 1} be the residue of k modulo n. Then k = nq + r for q ∈ Z, and ϕk = 2π t∗ 2π t∗ + (nq + r) = +r + 2qπ = ϕr + 2qπ.