Find the value of $(1 + i) (1 + i^{2}) (1 + i^{3}) (1 + i^{4})$ .

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Solution

$(1 + i) (1 + i^{2}) = 1 + i + i^{2} + i^{2} . i = 1 + i - 1 - i = 0$
$(1 + i^{3}) (1 + i^{4}) = 1 + i^{2} . i + i^{2} . i^{2} + i^{2} . i^{2} . i^{2} . i$
$= 1 - i + 1 - i = 2 - 2 i$
$= 0 \times (2 - 2 i)$
$= 0$

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Similar questions

(1 + i) (1 + i^{2}) (1 + i^{3}) (1 + i^{4})

(1 + i) (1 + i^{2}) (1 + i^{3}) (1 + i^{4})

$The value of (1 + i) (1 + i^{2}) (1 + i^{3}) (1 + i^{4}) is$

i^{18} - {3.1}^{7} + i^{2} (1 + i^{4}) (- i)^{26}

A B C D

O

1, 2, 3

4

I_{1}, I_{2}, I_{3} & I_{4}

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