If the roots of the equation z^2-i=0 are alph | vedclass

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Question

(C) Given the equation $z^2-i=0$,we have $z^2=i$. Expressing $i$ in polar form,$i = \cos(\frac{\pi}{2}) + i \sin(\frac{\pi}{2}) = e^{i\frac{\pi}{2}}$.

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$\pi$

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(C) Given the equation $z^2-i=0$,we have $z^2=i$. Expressing $i$ in polar form,$i = \cos(\frac{\pi}{2}) + i \sin(\frac{\pi}{2}) = e^{i\frac{\pi}{2}}$. The roots are $z = \pm e^{i\frac{\pi}{4}}$. Thus,the roots are $z_1 = e^{i\frac{\pi}{4}}$ and $z_2 = e^{i(\frac{\pi}{4} + \pi)} = e^{i\frac{5\pi}{4}}$. Let $\alpha = e^{i\frac{\pi}{4}}$ and $\beta = e^{i\frac{5\pi}{4}}$. Then $\operatorname{Arg} \alpha = \frac{\pi}{4}$ and $\operatorname{Arg} \beta = \frac{5\pi}{4}$. Therefore,$|\operatorname{Arg} \beta - \operatorname{Arg} \alpha| = |\frac{5\pi}{4} - \frac{\pi}{4}| = |\pi| = \pi$.

If the roots of the equation $z^2-i=0$ are $\alpha$ and $\beta$,then $|\operatorname{Arg} \beta-\operatorname{Arg} \alpha|=$

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