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2) From z 3 = 1 + i = ( 2) π4 +2pπ we get √ 6 π z = ( 2) 12 , p = 0, 1, 2. + 2ππ 3 Here, √ 6 π ( 2) 12 = = = = √ 6 π π + i sin 2 cos 12 12 ⎫ ⎧ π π⎪ ⎪ ⎪ ⎪ ⎬ ⎨ 1 + cos 1 − cos √ 6 6 +i 6 2 ⎪ ⎪ 2 2 ⎪ ⎪ ⎭ ⎩ ⎫ ⎧ √ √ ⎬ ⎨ 1 √ 3 1 3 6 1+ +i 1− 2 ⎩ 2 2 2 2 ⎭ √ 6 2 √ 1 (4 + 2 3) + i 8 √ 1 (4 − 2 3) 8 √ 6 √ √ 3 √ ( 3 + 1)2 + i ( 3 − 1)2 = 2 2 √ √ 1 √ = 3 + 1 + i( 3 − 1) . 3 2 2 2π 9π 3π π + = = , thus Furthermore, 12 3 12 4 √ √ 6 6 ( 2) 3π = 2 4 1 √ (−1 + i) 2 1 = √ (−1 + i). com 50 Calculus Analyse 1c-4 The binomial equation Finally, √ 6 π ( 2) 12 + 4π 3 √ 1 √ 1 6 (−1 + i) · (− 3 + i) = ( 2) 3π · (1) 2π = √ 3 3 4 2 2 √ 1 √ √ { 3 − 1 − i( 3 + 1)}.

D. First solve (1). Apply (1) in the solution of (2), and then use both to solve (3). 5 –1 Figure 12: Graphical solution of the binomial equation z 3 = 1. I. 1) From z 3 = (1)2pπ we get z = (1) 2pπ , p = 0, 1, 2, thus the solutions are 3 √ √ 1 1 3 3 1, − +i , − −i . 2 2 2 2 2 1 –8 –6 –4 –2 0 2 –1 –2 Figure 13: Graphical solution of the binomial equation z 3 = −8. 2) From z 3 = (−2)3 · (1)2pπ we get z = −2 · (1) 2pπ , 3 p = 0, 1, 2. Thus, by (1) the solutions are √ √ −2, 1 − i 3, 1 + i 3. com 56 Calculus Analyse 1c-4 The binomial equation 3) By putting w = z 3 the equation is reduced to 0 = w2 + 7w − 8 = (w − 1)(w + 8) = (z 3 − 1)(z 3 + 8).

This can be done in more than one way, although the methods in principle rely on the same idea. I. First variant. From (z + 1)2 = z 2 + 2z + 1, we get the inspiration of performing the following rearrangement, 0 = z 2 + 2z − 2 − 4i = (z 2 + 2z + 1) − (3 + 4i) = (z + 1)2 − (4 + i2 + 4i) = (z + 1)2 − (2 + i)2 = (z + 3 + i)(z − 1 − i). It is immediately seen that the roots are −3 − i and 1 + i. com 34 Calculus Analyse 1c-4 The equation of second degree Second variant. An unconscious application of the solution formula from high school gives the awkward solutions √ √ z = −1 ± 1 + 2 + 4i = −1 ± 3 + 4i.