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Additional info for Complex Functions c-1 - Examples concerning Complex Numbers
D) It follows by inspection that z 3 + z 2 + z + 1)z 2 (z + 1) + (z + 1) = (z + 1) z 2 + 1 , and we conclude that the roots are −1, i and −i. Alternatively it is seen that z 3 +z 2 +z +1 are the ﬁrst four terms of a quotient series. Therefore, if z = 1, then z 3 + z 2 + z1 = z4 − 1 , z−1 z = 1. The numerator z 4 −1 has the four roots 1, i, −1, −i, and when z = 1, it follows that z 3 +z 2 +z+1 = 4 = 0. Therefore, we conclude that the three roots are i, −1, −i. 8 Prove that all the roots of the equation (z + 1)5 + z 5 = 0 1 lie on the line x = − .
NNE Pharmaplan offers me freedom with responsibility as well as the opportunity to plan my own time. com NNE Pharmaplan is the world’s leading engineering and consultancy company focused exclusively on the pharma and biotech industries. NNE Pharmaplan is a company in the Novo Group. com 35 Complex Funktions c-1 Polar form of complex numbers The real part is 1 + cos θ + cos 2θ + · · · + cos n θ = Re = Re = ei(n+1)θ − 1 ei θ − 1 1 = Re cos(n + 12 )θ + i sin(n + 12 )θ − cos θ2 + i sin θ2 2i sin θ2 = 1 ei(n+ 2 )θ − e−i 2 θ θ θ ei 2 − e−i 2 sin(n + 12 )θ 2 sin θ2 + sin θ2 2 sin θ2 1 sin(n + 12 )θ + .
Then by a multiplication by 2z and a rearrangement we get (4) z 2 − 2a z + 1 = 0, the solutions of which are z =a± a2 − 1. Denote the roots by z1 and z2 . The structure of (4) shows that the product of the roots is 1, hence z1 · z2 = 1. g. |z1 | ≥ 1, then we necessarily must |z2 | ≤ 1, and vice versa. This shows that we have precisely one solution in the open unit disc |z| < 1, if and only if the other solution lies in the open complementary set |z| > 1of the closed unit disc. com 55 Complex Funktions c-1 Equations of second degree The condition that this is not the case is equivalent to |z1 | = |z2 | = 1.