Find the least positive integral value of n for which ( 1+i 1-i )… - Vidyadip

Question

Find the least positive integral value of n for which $\Big(\frac{1+\text{i}}{1-\text{i}}\Big)^{\text{n}}$ is real.

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Answer

For n = 1, we have, $\Big(\frac{1+\text{i}}{1-\text{i}}\Big)^1=\frac{1+\text{i}}{1-\text{i}}$ $=\frac{1+\text{i}}{1-\text{i}}\times\frac{1+\text{i}}{1+\text{i}}$ $=\frac{(1+\text{i})^2}{1^2+1^2}$ $=\frac{1^2+\text{i}^2+2\times1\times\text{i}}{2}$ $=\frac{2\text{i}}{2} \ \big(\therefore \ \text{i}^2=-1\big)$ $=\text{i},$ which is not real For n = 2, we have $\Big(\frac{1+\text{i}}{1-\text{i}}\Big)^2=\text{i}^2 \ \big(\because \ \frac{1+\text{i}}{1-\text{i}}=1 \ \text{from above}\big)$ $=-1,$ which is real Hence the least positive integral value of n is 2.

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