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1) If z = ±1, then P (±1) = 1 − 4i − 1 + 4i = 0. D. 2) If z=± √ 2 17 − 2 + i 1 2 √ 2 17 + 2 , then √ 1 2 · (−2) + 2i 4 · 17 − 4 4 1 {−4 + 2i · 4 · 2} = −1 + 4i, = 4 hence by insertions, z2 = P (z) = (−1 + 4i)2 − 4i(−1 + 4i) − 1 + 4i = (−1 + 4i) · {−1 + 4i + 1 − 4i} = 0. D. 8 Find the solutions z of the equation z 2 − z + 1 − i = 0. Then find all complex solutions y of the equation e2y − ey + 1 − i = 0. A. Complex roots. D. One may either use the solution formula, or find a root by inspection. I. 1) By a small inspection we see that 0 = z 2 − z + 1 − i = (z 2 + 1) − (z + i) = (z + i)(z − 1 − i), thus the roots are 1 + i and −i.

Alternatively, P (x) = x4 + 16 = x2 = = 2 + 42 + 2 · 4 · x2 − 8x2 2 √ 2 x2 + 4 − 2 2 x √ √ x2 − 2 2 x + 4 x2 + 2 2 x + 4 by the rule a2 − b2 = (a + b)(a − b). 5 1) Solve the equation x8 + 1 = 0. 2) Write the polynomial P (x) = x8 +1 as a product of polynomials of second degree of real coefficients. A. A binomial equation and a factorization of a real polynomial into real factors of second degree. The second item is very difficult. D. 1) We solve this as a binomial equation. 2) a) Complex conjugated roots are paired.

E. com 42 Calculus Analyse 1c-4 The equation of second degree where we have solve the equation as a binomial equation. Here, π π (1) π8 = cos + i sin 8 8 π 1 + cos 4 1 − cos π4 +i = 2 2 √ √ i 1 2+ 2+ 2 − 2, = 2 2 √ √ 1 i (1) +π π2 = (1) π8 · i = − 2− 2+ 2 + 2, 2 2 √ √ 1 i (1) π8 +π = −(1) π8 = − 1+ 2− 2 − 2, 2 2 √ √ 1 i (1) π8 + 3π = (1) π8 · (−i) = 2− 2− 2 + 2. 2 2 2 √ All things put together we see that for ± ± i we get ± 1 2 2+ √ 2+i 2− √ 2 , ± 1 2 2− √ 2−i 2+ √ 2 . 7 Solve the equation of second degree, z 2 − 4iz − 1 + 4i = 0.

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