First slide

Algebra of complex numbers

Question

If $a, b, x, y \in R, ω \neq 1,$ is a cube root of unity and ${(a + b ω)}^{7} = x + y ω,$ then ${(b + a ω)}^{7}$ equals

Moderate

A

$y + x ω$
B

$- y - x ω$
C

$x + y ω$
D

$- x - y ω$

Solution

Taking conjugate, we get
$\begin{array}{l} {(a + b ω^{2})}^{7} = x + y ω^{2} \Rightarrow {(a ω^{3} + b ω^{2})}^{7} = x ω^{3} + y ω^{2} \\ ω^{14} (a ω + b)^{7} = ω^{2} (x ω + y) \\ \Rightarrow (a ω + b)^{7} = x ω + y \end{array}$

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