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(b) (8211 (se ) (2ein/2) 4e5in /2

Hi. I come across this in a textbook. I don't understand why the answer is not 16e^(5i*pi/2). What am I missing here?

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Answer #1

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\small 8e^{2i\pi}=8(cos(2\pi)+isin(2\pi))=8(1+0i)=8\\ 2e^{\frac{i\pi}{2}}=2(cos(\frac{\pi}{2})+isin(\frac{\pi}{2}))=2(0+1i)=2i\\ (8e^{2i\pi})(2e^{\frac{i\pi}{2}})=8(2i)=16i\\ $Now we are going to express in polar form of 16i $\\ Let, rcos(\theta)+irsin(\theta)=16i\\ $Equating both side we have$\\ rcos(\theta)=0,rsin(\theta)=16\Rightarrow r=\sqrt{0^2+16^2}=16\\ So, cos(\theta)=0,sin(\theta)=1\Rightarrow \theta=\frac{\pi}{2}\\ 16i=16(cos(\frac{\pi}{2})+isin(\frac{\pi}{2}))=16e^{\frac{i\pi}{2}}\\ RHS\\ 4e^{\frac{5i\pi}{2}}\\ =4(cos(\frac{5\pi}{2})+isin(\frac{5\pi}{2}))\\ =4(0+1i)=4i\ne 16i\\ $Clearly the given expression is wrong$\\

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