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Question
If \[\left( \frac{1 + i}{1 - i} \right)^3 - \left( \frac{1 - i}{1 + i} \right)^3 = x + iy\] find (x, y).
Solution
\[\left( \frac{1 + i}{1 - i} \right) = \frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i}\]
\[ = \frac{\left( 1 + i \right)^2}{1^2 - i^2}\]
\[ = \frac{1^2 + i^2 + 2i}{1 + 1} [ \because i^2 = - 1]\]
\[ = \frac{1 - 1 + 2i}{2}\]
\[ = \frac{2i}{2}\]
\[ = i . . . . (1)\]
Also,
\[\left( \frac{1 - i}{1 + i} \right) = \frac{1 - i}{1 + i} \times \frac{1 - i}{1 - i}\]
\[ = \frac{\left( 1 - i \right)^2}{1^2 - i^2}\]
\[ = \frac{1^2 + i^2 - 2i}{1 + 1} [ \because i^2 = - 1]\]
\[ = \frac{1 - 1 - 2i}{2}\]
\[ = \frac{- 2i}{2}\]
\[ = - i . . . . (2)\]
It is given that,
\[\left( \frac{1 + i}{1 - i} \right)^3 - \left( \frac{1 - i}{1 + i} \right)^3 = x + iy\]
\[ \Rightarrow (i )^3 - ( - i )^3 = x + iy [\text {From (1) and (2)}]\]
\[ \Rightarrow i^3 + i^3 = x + iy\]
\[ \Rightarrow 2 i^3 = x + iy\]
\[ \Rightarrow 0 - 2i = x + iy [ \because i^3 = - i]\]
\[ \Rightarrow x = 0 \text { and } y = - 2\]
Thus, (x, y) = (0, −2).
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