Compute (1 + i)⁸:
A-16i
B16
C16i
D-16
Answer & Solution
Correct answer: B. 16
(1+i) = √2 cis(π/4). So (1+i)⁸ = (√2)⁸ × cis(8π/4) = 16 × cis(2π) = 16 × 1 = 16.
Related questions
The number of distinct $n$th roots of a non-zero complex number is:Using De Moivre's theorem, $(\cos\theta + i in\theta)^4$ equals:The modulus of $z = 5 + 12i$ is:The conjugate of $z = 3 - 4i$ is:Find (3 + 4i)/(1 + 2i):If z1, z2, z3 are vertices of an equilateral triangle, then:Solve: z² + z + 1 = 0:For roots of unity ω^n = 1: which is the SMALLEST n such that 1 + ω + ω² + ... + ω^(n-1) =