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Question Number 8336 by Rasheed Soomro last updated on 08/Oct/16

$$\mathrm{Determine}\mathrm{smallest}\mathrm{n}(\ne 0),\mathrm{for}\mathrm{which}$$$${(\omega +\mathrm{i})}^{\mathrm{n}}=1.$$

Commented by prakash jain last updated on 08/Oct/16

$$w=\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$$$$w+i=-\frac{1}{2}+i(\frac{\sqrt{3}}{2}+1)$$$$\mid w+i\mid =\sqrt{{\left(\frac{1}{2}\right)}^2+{(\frac{\sqrt{3}}{2}+1)}^2}$$$$=\sqrt{\frac{1}{4}+1+\frac{3}{4+2\sqrt{3}}}=\sqrt{2+\sqrt{3}}$$$$\mathrm{Since}\mid w+i\mid =\sqrt{2+\sqrt{3}}$$$$\mathrm{of}{(w+i)}^n=1\Rightarrow n=0$$$$\mathrm{There}\mathrm{is}\mathrm{no}\mathrm{value}\mathrm{of}n\ne 0\mathrm{for}\mathrm{which}$$$${(w+i)}^n=1$$

Commented by Rasheed Soomro last updated on 09/Oct/16

$$\mathrm{Thanks}\mathrm{a}\mathrm{lot}!$$

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