First slide

Algebra of complex numbers

Question

If $z = i (1 + \sqrt{3})$ then $z^{4} + 2 z^{3} + 4 z^{2} + 5$ is equal to-

Moderate

A

5
B

-5
C

$2 \sqrt{3} i$
D

$- 2 \sqrt{3} i$

Solution

$z = i (1 + \sqrt{3}) = - 1 + \sqrt{3} i = 2 ω$
where $ω \neq 1$ is a cube root of unity
$\begin{array}{l} ∴ z^{4} + 2 z^{3} + 4 z^{2} + 5 \\ = (2 ω)^{4} + 2 (2 ω)^{3} + 4 (2 ω)^{2} + 5 \\ = 16 ω^{4} + 16 ω^{3} + 16 ω^{2} + 5 \\ = 16 (ω + 1 + ω^{2}) + 5 = 16 (0) + 5 = 5 \end{array}$

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