Solve for n, (-1/2+ iota sqrt(3)/2)^3 + (-1/2-iota sqrt(3)/2)^(2n)=2 ?

2 Answers
Shwetank Mauria
Mar 12, 2017

n=(3m)/4, where m is an integer

Explanation:

We have to evaluate (-1/2+isqrt3/2)^3+(-1/2-isqrt3/2)^(2n)

As (-1/2+isqrt3/2)=cos((2pi)/3)+isin((2pi)/3) and

(-1/2-isqrt3/2)=cos((4pi)/3)+isin((4pi)/3)

using De Moivre's Theorem

(-1/2+isqrt3/2)^3=cos((2pi)/3xx3)+isin((2pi)/3xx3)

= cos(2pi)+isin(2pi)=1

and (-1/2-isqrt3/2)^(2n)=cos((4pi)/3xx2n)+isin((4pi)/3xx2n)

= cos((8npi)/3)+isin((8npi)/3)

Therefore (-1/2+isqrt3/2)^3+(-1/2-isqrt3/2)^(2n)=2

cos((8npi)/3)+isin((8npi)/3)=1=cos(2mpi)+isin(2mpi),

where m is an integer

i.e. (8npi)/3=2mpi or m=(4n)/3

and n=(3m)/4, where m is an integer

Cesareo R.
Mar 12, 2017

n=3/2 k with k = 1,2,3,cdots

Explanation:

(-1/2+ iota sqrt(3)/2)^3 + (-1/2-iota sqrt(3)/2)^(2n)=2

We will using the complex number exponential representation

x+iy=rho e^(i phi) where rho=sqrt(x^2+y^2) and phi=arctan(y/x)

So rho = sqrt((1/2)^2+(sqrt3 /2)^2) = 1 and phi=arctan(sqrt(3))=pi/3

then we have

e^(i 3 phi)+e^(-i 2n phi)=2

or

e^(i 3 phi)+e^(-i3phi((2n)/3))=2

but e^(i3phi)= 1 and e^(-i3phi)= 1 so with (2n)/3=k, k=1,2,3,cdots

1+1^k=2

then

n=3/2 k with k = 1,2,3,cdots

Related questions
See all questions in Roots of Complex Numbers
Impact of this question
3602 views around the world
You can reuse this answer
Creative Commons License