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Complex Numbers

If i23+ 1i292...

Question

If ${(i^{23} + {(\frac{1}{i})}^{29})}^{2} = k,$ then the value of $k^{2}$ is

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Solution

We have,
$i^{23} = \frac{(i^{4})^{6}}{i} = \frac{1}{i} = - i {(\frac{1}{i})}^{29} = \frac{1}{i^{28} \cdot i} = \frac{1}{(i^{4})^{7} \times i} = \frac{1}{i} = - i$

$∴ {(i^{23} + {(\frac{1}{i})}^{29})}^{2} = (- i - i)^{2} = - 4 \Rightarrow k = - 4 ∴ k^{2} = 16$

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