1-i-1-i-50-

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

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

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

(((1+i)/(1−i)))^(50) =((((1+i)^2 )/((1−i)^2 )))^(25) =(((2i)/(−2i)))^(25) =−1

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

Answered by Bird last updated on 20/Nov/20

1+i=(√2)e^((iπ)/4) and 1−i =(√2)e^(−((iπ)/4)) ⇒ ((1+i)/(1−i))=e^((iπ)/2) =i ⇒(((1+i)/(1−i)))^(50) =i^(50) =(i^2 )^(25) =(−1)^(25) =−1

$$\mathrm{1}+{i}=\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \:{and}\:\mathrm{1}−{i}\:=\sqrt{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\Rightarrow \\ $$$$\frac{\mathrm{1}+{i}}{\mathrm{1}−{i}}={e}^{\frac{{i}\pi}{\mathrm{2}}} \:={i}\:\Rightarrow\left(\frac{\mathrm{1}+{i}}{\mathrm{1}−{i}}\right)^{\mathrm{50}} \:={i}^{\mathrm{50}} \\ $$$$=\left({i}^{\mathrm{2}} \right)^{\mathrm{25}} \:=\left(−\mathrm{1}\right)^{\mathrm{25}} \:=−\mathrm{1} \\ $$

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1-i-1-i-50-

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