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Question

If $${\left( {\frac{{1 + i}}{{1 - i}}} \right)^3} - {\left( {\frac{{1 - i}}{{1 + i}}} \right)^3} = x + iy$$, then find $$(x,\,y)$$,


Solution

$$ \left ( \dfrac{1+i}{1-i} \right )^{3} - \left ( \dfrac{1-i}{1+i} \right )^{3} = x+iy$$

Rationalizing, we get, 

$$ \left ( \dfrac{(1+i)^{2}}{2} \right )^{3} - \left ( \dfrac{(1-i)^{2}}{2} \right )^{3} = x + iy$$

$$ \left ( \dfrac{1+2i-1}{2} \right )^{3}-\left ( \dfrac{1-2i-1}{2} \right )^{3}=x+iy$$

$$ (i)^{3} - (-i)^{3} = x + iy $$

$$ -i - (+i) = x + iy$$

$$- 2i = x + iy$$

so, $$y = - 2$$
 
$$x = 0$$


Mathematics

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