Let $α^{k}$ where K = 0, 1, 2, ,…….. 2023 are 2024th root of unity. If $Z_{1}$ and $Z_{2}$ be any two complex numbers such that $|Z_{1}| = |Z_{2}| = \frac{1}{\sqrt{1012}}$ then the value of $\sum_{K = 0}^{2023} {|Z_{1} + α^{K} Z_{2}|}^{2}$

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answer is 4.

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Detailed Solution

$Z^{2024}$ = 1 $\Rightarrow Z = e^{{(\frac{2 K π}{2024})}^{i}} = α^{K}$ (given)
Let $Z_{1}$ = $r e^{i θ_{1}}$ , $Z_{2} = r e^{i θ_{2}}$ , where $r = \frac{1}{\sqrt{1012}}$

$∴ {|Z_{1} + α^{K} Z_{2}|}^{2} = (Z_{1} + α^{K} Z_{2}) ({\bar{Z}}_{1} + \bar{α^{K}} {\bar{Z}}_{2})$

$= r^{2} [e^{i θ_{1}} + e^{(\frac{2 k π}{2024} + θ_{2}) i}] [e^{- i θ_{1}} + e^{- i (\frac{2 k π}{2024}) + θ_{2})}]$
$= r^{2} [2 + 2 \cos (θ_{1} - θ_{2} - \frac{K π}{1012})] = 2 r^{2} [1 + \cos (θ_{1} - θ_{2} - \frac{K π}{1012})]$
$∴ \sum_{K = 0}^{2023} {|Z_{1} + α^{K} Z_{2}|}^{2} = 2 r^{2} [2024 + \underset{0}{\underset{⏟}{\cos θ + \cos (θ - α) + ... + \cos (θ - 2023)]}}$

$(∵ θ_{1} - θ_{2} = θ)$ $= 2 (\frac{1}{1012}) \times 2024$ = 4