How do you use DeMoivre's theorem to simplify #(sqrt3+i)^7#?

2 Answers
Narad T.
Dec 21, 2016

The answer is #=64(-sqrt3-i)#

Explanation:

De Moivre's theorem states

#(costheta+isintheta)^n=cosntheta+isinntheta#

Let #z=sqrt3+i#

#∥z∥=sqrt(3+1)=2#

#z=2*(sqrt3/2+i1/2)#

#z=r(costheta+isintheta)#

#costheta=sqrt3/2#, #=>#, #theta=pi/6#

#sintheta=1/2#, #=>#, #theta=pi/6#

#z=2(cos(pi/6)+isin(pi/6))#

Therefore,

#z^7=2^7(cos(pi/6)+isin(pi/6))^7#

#=128(cos(7/6pi)+isin(7/6pi))#

#=128(-sqrt3/2-i/2)#

#=64(-sqrt3-i)#

Daniel L.
Dec 21, 2016

See explanation.

Explanation:

De Moivre's Theorem says that:

If a complex number #z# is given in trigonometric form:

#z=r(cosvarphi+isinvarphi)#

Then #n-th# power of #z# is given as:

#z^n=|z|^n*(cosnvarphi+isinnvarphi)#

So first thing to do is to change #z=sqrt(3)+i# into trigonometric form:

#|z|=sqrt(sqrt(3)^2+1^2)=sqrt(3+1)=sqrt(4)=2#

#cosvarphi=(re(z))/r=sqrt(3)/2 => varphi=30^o#

So the trigonometric form of #z# is:

#z=2(cos30+isin30)#

Now we can calculate #z^7# according to the de Moivre's theorem:

#z^7=2^7*(cos 7*30+isin7*30)#

#z^7=128*(cos210+isin210)#

#z^7=128*(cos(180+30)+isin(180+30))#

#z^7=128*(-cos30-sin30i)#

#z^7=128*(-sqrt(3)/2-1/2i)#

Answer: #z^7=-64sqrt(3)-64i#

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