Question

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

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

answered by: Jim G.
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