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Question Number 139608 by mohammad17 last updated on 29/Apr/21

$$findthefirstroot{(-8i)}^{\frac{1}{2}}$$

Answered by floor(10²Eta[1]) last updated on 29/Apr/21

$$\sqrt{-8\mathrm{i}}=\mathrm{a}+\mathrm{bi}$$$$-8\mathrm{i}={\mathrm{a}}^2-{\mathrm{b}}^2+2\mathrm{abi}$$$$\Rightarrow {\mathrm{a}}^2-{\mathrm{b}}^2=0\wedge -8=2\mathrm{ab}$$$$\mid \mathrm{a}\mid =\mid \mathrm{b}\mid \wedge \mathrm{ab}=-4$$$${\mathrm{b}}^2=4\Rightarrow \mathrm{b}=\pm 2,\mathrm{a}=\pm 2$$$$\sqrt{-8\mathrm{i}}=2-2\mathrm{i}\mathrm{or}-2+2\mathrm{i}$$

Answered by mr W last updated on 29/Apr/21

$$-8i=8e^{(2k\pi -\frac{\pi }{2})i}$$$${(-8i)}^{\frac{1}{2}}=2\sqrt{2}{e}^{(k-\frac{1}{4})\pi i},k=0,1$$$$=2\sqrt{2}(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i)=2-2i$$$$or=2\sqrt{2}(-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)=-2+2i$$

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