Math miscellaneousHard
Question
If α + iβ = (√3 + i)3n1(1 - i)-2n2, then
Options
A.α = 0 if n1 and n2 are both even
B.β = 0 if n1 and n2 are both odd
C.α = 0 if only one of n1 and n2 is odd
D.|α + iβ| if an even number provided n2 ≠ 3n1
Solution
α + iβ = 23n1. e
(√2)-2n2e
= 23n1 - n2e3n1 - n2ei(n1 + n2)
α = 23n1 - n2 cos(n1 + n2)
= 0
when (n1 + n2) is odd
β = 23n1 - n2 sin(n1 + n2)
= 0
when n1 + n2 is even
⇒ |α + iβ| = 23n1 + n2 is even for all n1 and n2 3n1 ≠ n2 except
(√2)-2n2e = 23n1 - n2e3n1 - n2ei(n1 + n2)α = 23n1 - n2 cos(n1 + n2)
= 0when (n1 + n2) is odd
β = 23n1 - n2 sin(n1 + n2)
= 0when n1 + n2 is even
⇒ |α + iβ| = 23n1 + n2 is even for all n1 and n2 3n1 ≠ n2 except
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