If ( 1+i 1-i )^3- ( 1-i 1+i )^3=x+iy, find (x, y) - Vidyadip

Question

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

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Answer

$\Big(\frac{1+\text{i}}{1-\text{i}}\Big)^3-\Big(\frac{1-\text{i}}{1+\text{i}}\Big)^3=\text{x}+\text{iy,}$
$\Rightarrow\Big(\frac{(1+\text{i})(1+\text{i})}{(1-\text{i})(1+\text{i})}\Big)^3-\Big(\frac{(1-\text{i})(1-\text{i})}{(1+\text{i})(1-\text{i})}\Big)^3=\text{x}+\text{iy}$ [Rationalizing the denominator]
$\Rightarrow\Big(\frac{1+2\text{i}-1}{1+1}\Big)^3-\Big(\frac{1-2\text{i}-1}{1+1}\Big)^3=\text{x}+\text{iy}$
$\Rightarrow\Big(\frac{2\text{i}}{2}\Big)^3-\Big(\frac{-2\text{i}}{2}\Big)^3=\text{x}+\text{iy}$
$\Rightarrow\text{i}^3-(-\text{i})^3=\text{x}+\text{iy}$
$\Rightarrow-\text{i}-\text{i}=\text{x}+\text{iy}$
$\Rightarrow-2\text{i}=\text{x}+\text{iy}$
Comparing the real and imaginary parts,
$(\text{x},\text{y})=(0,2)$

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