pgt math past question paper solution

Question : how to solve following problem of pgt math past question paper solutionSolution :\[\left(-i\right)^\frac{1}{2}=\left(\cos{\left(\frac{\pi}{2}\right)}-isin\left(\frac{\pi}{2}\right)\right)^\frac{1}{2}\] By Demoiver’s theorem \[\left(\cos{\left(\frac{\pi}{2}\right)}-isin\left(\frac{\pi}{2}\right)\right)^\frac{1}{2}\] \[=\cos{\left(\frac{\left(\frac{\pi}{2}+2\pi k\right)}{2}\right)}-isin\left(\frac{\left(\frac{\pi}{2}+2\pi k\right)}{2}\right)\ \ ,\ \ \ k=0,1\] When k=0 \[\cos{\left(\frac{\pi}{4}\right)}-isin\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt2}-i\frac{1}{\sqrt2}=\frac{1}{\sqrt2}\left(1-i\right)\] When k=1 \[\cos{\left(\frac{5\pi}{4}\right)}-isin\left(\frac{5\pi}{4}\right) -\frac{1}{\sqrt2}+\frac{1}{\sqrt2}i=-\frac{1}{\sqrt2}\left(1-i\right)\] Thus \[\left(-i\right)^\frac{1}{2}=\pm\frac{1}{\sqrt2}\left(1-i\right)\] So the correct option…

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