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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