Question

How do you use DeMoivre's theorem to simplify `(2+2i)^6`?

Answer 1

`-512i`

Explanation:

Before applying `color(blue)"De Moivre's theorem"` we requireto convert the complex number into trigonometric form.

To convert from `color(blue)"complex to trigonometric form"`

`color(orange)"Reminder"`

`color(red)(|bar(ul(color(white)(a/a)color(black)(z=x+yi=r(costheta+isintheta))color(white)(a/a)|)))" where"`

`color(red)(|bar(ul(color(white)(a/a)color(black)(r=sqrt(x^2+y^2))color(white)(a/a)|)))" and " color(red)(|bar(ul(color(white)(a/a)color(black)(theta=tan^-1(y/x))color(white)(a/a)|)))`

For 2 + 2i , we have x = 2 and y = 2

`rArrr=sqrt(2^2+2^2)=sqrt8=2sqrt2`

Now 2 + 2i is in the 1st quadrant and so we must ensure that `theta` is in the 1st quadrant.

`rArrtheta=tan^-1(2/2)-tan^-1(1)=pi/4" in 1st quadrant"`

`rArr2+2i=2sqrt2(cos(pi/4)+isin(pi/4))color(blue)" in trig form"`

`color(blue)"DeMoivre's theorem"` states that.

`color(red)(|bar(ul(color(white)(a/a)color(black)((r(costheta+isintheta))^n=r^n(cos(ntheta)+isin(ntheta)))|)))ninQQ`

`rArr(2sqrt2(cos(pi/4)+isin(pi/4)))^6=`

`(2sqrt2)^6(cos(6xxpi/4)+isin(6xxpi/4))`

`=512(cos((3pi)/2)+isin((3pi)/2))`

`color(orange)"Reminder " color(red)(|bar(ul(color(white)(a/a)color(black)(cos((3pi)/2)=0" and " sin((3pi)/2)=-1)color(white)(a/a)|)))`

`rArr512(cos((3pi)/2)+isin((3pi)/2))=512(0-i)`

`rArr(2+2i)^6=color(red)(|bar(ul(color(white)(a/a)color(black)(-512i)color(white)(a/a)|)))`