How do you use De Moivre's theorem to express (1 + i)^8?

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Answer 1 · 2018-03-03T09:08:31

#z=16#

De Moivre's theorem states if #color(orange)(z=r(cos(theta)+isin(theta)))#,

then #color(orange)(z^n=r^n(cos(ntheta)+isin(ntheta)))#

First step convert from complex form to trig form

#a+birarrr(cos(theta)+isin(theta))#

By using

#r=sqrt(a^2+b^2)# and #theta=arctan(b/a)#

We have the number #z=1+i#, thus

#r=sqrt(1^2+1^2)=sqrt(2)#

#theta=arctan(1/1)=pi/4#

#z=sqrt(2)(cos(pi/4)+isin(pi/4))larr"Trig form"#

Second step apply De Moivre's Theorem

#z^8=(sqrt(2)(cos(pi/4)+isin(pi/4)))^8#

#=sqrt(2)^8(cos(8pi/4)+isin(8pi/4))#

#=16(cos(2pi)+isin(2pi)#

#=16#