Lösung 3.3:2c

Aus Online Mathematik Brückenkurs 2

We write z and the right-hand side −1−i in polar form

z−1−i=r(cos+isin)=2cos45+isin45.

Using de Moivre's formula, the equation can now be written as

r5(cos5+isin5)=2cos45+isin45.

If we identify the magnitude and argument on both sides, we get

r55=2=45+2n(n is an arbitrary integer).

(The arguments 5 and 54 can differ by a multiple of 2 and still correspond to the same complex number.)

This gives that

r=52=21215=2110=5145+2n=4+52n(n is an arbitrary integer).

If we investigate the argument more closely, we see that it assumes essentially only five different values,

4, 4+52, 4+54, 4+56 and 4+58

since these angle values then repeat to within a multiple of 2.

In summary, the solutions are

z=2110cos4+52n+isin4+52n

for n=0, 1, 2, 3 and 4.