The value of $(1 + i) (1 + i^{2}) (1 + i^{3}) (1 + i^{4})$ is
(a) 2
(b) 0
(c) 1
(d) i

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Solution

(b) 0
(1+ i) (1 + i²) (1 + i³) (1 + i⁴)
= (1+ i) (1 $-$ 1) (1 $-$ i) (1 + 1) ( $∵$ i² = $-$ 1, i³ = $-$ i and i⁴ = 1)
= (1 + i) (0) (1 $-$ i) (2)
= 0

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