First slide

Algebra of complex numbers

Question

$If Z = \frac{\sqrt{3} + i}{2}, then {[Z^{101} + i^{103}]}^{105} is equal to$

Moderate

A

$Z$
B

$Z^{2}$
C

$Z^{3}$
D

1

Solution

$\begin{aligned} Given, z = \frac{\sqrt{3} + i}{2} = \frac{i [1 - i \sqrt{3}]}{2} = - i w \\ Consider, z^{101} + i^{103} = (- i w)^{101} + i^{103} = - i w^{2} - i = i w \\ Now, {[z^{101} + i^{103}]}^{105} = (i w)^{105} = i \\ Also z = - i w \\ z^{2} = (- i w)^{2} = i^{2} w^{2} = - w^{2} \\ z^{3} = (- i w)^{3} = - i^{3} \cdot w^{3} = i \\ [z^{101} + i^{103}] = z^{3} \end{aligned}$

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