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Complex Numbers

If i23+1/i292...

Question

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

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Solution

We have,
$i^{23} = \frac{(i^{4})^{6}}{i} = \frac{1}{i} = - i$

and
${(\frac{1}{i})}^{29} = \frac{1}{i^{28} \cdot i} = \frac{1}{(i^{4})^{7} \times i} = \frac{1}{i} = - i$

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

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