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Question Number 122903 by Khalmohmmad last updated on 20/Nov/20

$${\left(\frac{1+i}{1-i}\right)}^{50}=?$$

Answered by Dwaipayan Shikari last updated on 20/Nov/20

$${\left(\frac{1+i}{1-i}\right)}^{50}={\left(\frac{{(1+i)}^2}{{(1-i)}^2}\right)}^{25}={\left(\frac{2i}{-2i}\right)}^{25}=-1$$

Answered by Bird last updated on 20/Nov/20

$$1+i=\sqrt{2}{e}^{\frac{i\pi }{4}}and1-i=\sqrt{2}{e}^{-\frac{i\pi }{4}}\Rightarrow$$$$\frac{1+i}{1-i}=e^{\frac{i\pi }{2}}=i\Rightarrow {\left(\frac{1+i}{1-i}\right)}^{50}=i^{50}$$$$={\left(i^2\right)}^{25}={(-1)}^{25}=-1$$

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