1+√3 i = 2cis π/3
1-√3 i = 2cis -π/3
so,
(1+√3 i)^9 = 2^9 cis (π/3 * 9) = 512 cis 3π = -512
now finish it off
Write 1+iroot 3 and 1-iroot3 in trig form and simplify (1+iroot3)^9 +(1-iroot3)^9 in x+iy form.
2 answers
1+iroot 3
= 1 + √3
tanθ = √3/1 -----> θ = 60 degrees or θ = π/3 radians
r = √(1 + 3) = 2
1 + √3 = 2(cosπ/3 + i sinπ/3)
so (1 + √3)^9 = 2^9(cos 9*π/3 + i sin 9*π/3)
= 512(cos 3π + i sin 3π)
= 512(-1 + 0i) = -512
do the same for 1 - √3 and then (1 - √3)^9
= 1 + √3
tanθ = √3/1 -----> θ = 60 degrees or θ = π/3 radians
r = √(1 + 3) = 2
1 + √3 = 2(cosπ/3 + i sinπ/3)
so (1 + √3)^9 = 2^9(cos 9*π/3 + i sin 9*π/3)
= 512(cos 3π + i sin 3π)
= 512(-1 + 0i) = -512
do the same for 1 - √3 and then (1 - √3)^9