By Mejlbro L.

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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.

The trick is to multiply by w − 1 = 0. I. By the multiplication by w − 1 = 0, followed by a calculation and a reduction we get by the assumption that (w − 1)(wn−1 + wn−2 + · · · + w + 1) = (wn − wn−1 ) + (wn−1 − wn−2 ) + · · · + (w 2 − w) + (w − 1) = wn − 1 = 0. Then one of the factors must be 0. Now, w − 1 = 0, so we conclude that 1 + w + w2 + · · · + wn−1 = 0. 2 Solve the equations (1) z 3 = 1, (2) z 3 = i. Sketch ﬁgures! A. Two binomial equations. D. Solve by using modulus and argument. Alternatively it is possible to ﬁnd a root by inspection.

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.

### Calculus Analyse lc-4, Complex Numbers Examples by Mejlbro L.

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