Question
What is the smallest positive integer n for which $(1+\text{i})^{2\text{n}}=(1-\text{i})^{2\text{n}} \ ?$
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Answer
$(1+\text{i})^{2\text{n}}=(1-\text{i})^{2\text{n}}$
$\Rightarrow\Big(\frac{1+\text{i}}{1-\text{i}}\Big)^{2\text{n}}=1$
$\Rightarrow\Big(\frac{(1+\text{i})(1+\text{i})}{(1-\text{i})(1+\text{i})}\Big)^{2\text{n}}=1$ [Rationalizing the denominator]
$\Rightarrow\Big(\frac{1+2\text{i}-1}{1+1}\Big)^{2\text{n}}=1$
$\Rightarrow\Big(\frac{2\text{i}}{2}\Big)^{2\text{n}}=1$
$\Rightarrow\text{i}^{2\text{n}}=1$
$\therefore\text{n}=2$
$\Rightarrow\Big(\frac{1+\text{i}}{1-\text{i}}\Big)^{2\text{n}}=1$
$\Rightarrow\Big(\frac{(1+\text{i})(1+\text{i})}{(1-\text{i})(1+\text{i})}\Big)^{2\text{n}}=1$ [Rationalizing the denominator]
$\Rightarrow\Big(\frac{1+2\text{i}-1}{1+1}\Big)^{2\text{n}}=1$
$\Rightarrow\Big(\frac{2\text{i}}{2}\Big)^{2\text{n}}=1$
$\Rightarrow\text{i}^{2\text{n}}=1$
$\therefore\text{n}=2$
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