First, we express `1 - 2j` in polar form:

`sqrt(1^2 + (-2)^2)=sqrt(5)`, and

`arctan((-2)/1) =296.6^"o"`

So

`1-2j=sqrt5\ /_ \ 296.6^text(o)`

Then

`(1-2j)^6=(sqrt5)^6/_ \ [6xx296.6^text(o)]`

`=125\ /_ \ [1779.3903^text(o)]`

`=125\ /_ \ [339.39^text(o)]`

(The last line is true because `360° × 4 = 1440°`, and we substract this from `1779.39°`.)

In rectangular form,

x = 125 cos 339.39° = 117

y = 125 sin 339.39° = -44

So (1 - 2j)6= 117 - 44j