Complex numbers question (1 Viewer)

[YBK]

Hey, just another question

z=1+ root3 i
Find the smallest positive integer n for which z^n is real and evaluate z^n for this value of n. Show that there is no integral value of n for which z^n is imaginary.

It's 2.2 question 8 in cambridge

thanks for any help!

 

[Porcia]

yeah im doing that too - hard as man

 

[KFunk]

expressing z = 1 + √3i in mod-arg form gives us 2.cis(π/3) so,

zⁿ = 2ⁿcis(nπ/3)

zⁿ is real when Im(zⁿ) = 0 , that is, when Im(zⁿ) = 2ⁿsin(nπ/3) which is equal to zero when nπ/3 = kπ ----> n=3k where k = 0, ±1, ... hence the smallest positive integer 'n' is n=3.

For zⁿ to be imaginary then Re(zⁿ) = 2ⁿcos(nπe/3) = 0. Hence nπ/3 = (2k-1)π/2 where k = 0, ±1, ... giving us n=[(2k -1)3]/2. Since (2k-1) is odd it is not divisible by 2, hence it follows that (2k-1).3 is not divisible by 2 ∴ n cannot be an integer i.e. there is no integral value of n for which zⁿ is purely imaginary.

 

[YBK]

expressing z = 1 + √3i in mod-arg form gives us 2.cis(π/3) so,

zⁿ = 2ⁿcis(nπ/3)

zⁿ is real when Im(zⁿ) = 0 , that is, when Im(zⁿ) = 2ⁿsin(nπ/3) which is equal to zero when nπ/3 = kπ ----> n=3k where k = 0, ±1, ... hence the smallest positive integer 'n' is n=3.

For zⁿ to be imaginary then Re(zⁿ) = 2ⁿcos(nπe/3) = 0. Hence nπ/3 = (2k-1)π/2 where k = 0, ±1, ... giving us n=[(2k -1)3]/2. Since (2k-1) is odd it is not divisible by 2, hence it follows that (2k-1).3 is not divisible by 2 ∴ n cannot be an integer i.e. there is no integral value of n for which zⁿ is purely imaginary.

cool, thanks!