# New PDF release: Cours de calcul differentiel By Serret J.-A.

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Dieses Lehrbuch ist der erste Band einer dreiteiligen Einf? hrung in die research. Es ist durch einen modernen und klaren Aufbau gepr? gt, der versucht den Blick auf das Wesentliche zu richten. Anders als in den ? blichen Lehrb? chern wird keine okay? nstliche Trennung zwischen der Theorie einer Variablen und derjenigen mehrerer Ver?

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Example text

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.