Q59 / 160 · TS EAMCET 2013 · Mathematics

$\left(\frac{1+i}{1-i}\right)^4+\left(\frac{1-i}{1+i}\right)^4$ is equal to

  • A
    $0$
  • B
    $1$
  • Correct answer
    C
    $2$
  • D
    $4$

Solution

$\begin{aligned}\left(\frac{1+i}{1-i}\right)^4 & +\left(\frac{1-i}{1+i}\right)^4 \\ & =\left\{\frac{(1+i)(1+i)}{(1-i)(1+i)}\right\}^4+\left\{\frac{(1-i)(1-i)}{(1+i)(1-i)}\right\}^4 \\ & =\left\{\frac{(1+i)^2}{1-i^2}\right\}^4+\left\{\frac{(1-i)^2}{1-i^2}\right\}^4 \\ & =\left\{\frac{1+i^2+2 i}{1+1}\right\}^4+\left\{\frac{1+i^2-2 i}{1+1}\right\}^4 \\ & =\left\{\frac{1-1+2 i}{2}\right\}^4+\left\{\frac{1-1-2 i}{2}\right\}^4 \\ & =(i)^4+(-i)^4 \\ & =i^4+i^4=2 i^4=2(1)=2\end{aligned}$

Hence, the correct answer is option: 'C'.

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