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Question Number 50806 by Tawa1 last updated on 20/Dec/18

$$\mathrm{Determine}\mathrm{the}\mathrm{fourth}\mathrm{roots}\mathrm{of}-16,\mathrm{giving}\mathrm{the}\mathrm{results}\mathrm{in}\mathrm{polar}$$$$\mathrm{form}\mathrm{and}\mathrm{in}\mathrm{exponential}\mathrm{form}$$$$\mathbf{Answers}:\sqrt{2}(1+\mathbf{j}),\sqrt{2}(-1+\mathbf{j}),\sqrt{2}(-1-\mathbf{j}),\sqrt{2}(1-\mathbf{j})$$

Answered by mr W last updated on 20/Dec/18

$$x=r(\cos \theta +j\sin \theta )$$$$x^4=-16$$$$r^4(\cos 4\theta +j\sin 4\theta )=2^4(-1+0j)$$$$\Rightarrow r=2$$$$\cos 4\theta =-1$$$$\sin 4\theta =0$$$$\Rightarrow 4\theta =\pi ,3\pi ,5\pi ,7\pi$$$$\Rightarrow \theta =\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$$$$x=2(\cos \frac{\pi }{4}+j\sin \frac{\pi }{4})=\sqrt{2}(1+j)$$$$x=2(\cos \frac{3\pi }{4}+j\sin \frac{3\pi }{4})=\sqrt{2}(-1+j)$$$$x=2(\cos \frac{5\pi }{4}+j\sin \frac{5\pi }{4})=\sqrt{2}(1-j)$$$$x=2(\cos \frac{7\pi }{4}+j\sin \frac{7\pi }{4})=\sqrt{2}(-1-j)$$

Commented by Tawa1 last updated on 20/Dec/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by Tawa1 last updated on 21/Dec/18

$$\mathrm{Sir},\mathrm{am}\mathrm{trying}\mathrm{to}\mathrm{understand}\mathrm{how}\mathrm{you}\mathrm{got}3\pi ,5\pi ,7\pi ...$$$$\mathrm{I}\mathrm{thought}\mathrm{the}\mathrm{angles}\mathrm{should}\mathrm{be}\mathrm{in}\mathrm{the}\mathrm{same}\mathrm{interval}\mathrm{sir}.\mathrm{like}\mathrm{in}$$$$\mathrm{interval}\mathrm{of}\frac{360}{4}=90\mathrm{since}\mathrm{it}\mathrm{is}4\mathrm{th}\mathrm{roots}.$$$$\mathrm{so},4\theta =180,270,360,450,\mathrm{etc}...$$$$$$$$\mathrm{Please}\mathrm{sir},\mathrm{i}\mathrm{want}\mathrm{to}\mathrm{know}\mathrm{why}\mathrm{we}\mathrm{use}\pi ,3\pi ,5\pi ,7\pi ,\mathrm{and}\mathrm{not}\mathrm{the}$$$$\mathrm{interval}\mathrm{of}4\mathrm{th}\mathrm{roots}\frac{360}{4}=90.$$$$$$$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by mr W last updated on 21/Dec/18

$$0⩽\theta ⩽2\pi$$$$\Rightarrow 0⩽4\theta ⩽8\pi$$$$inthisrangewehavefollowingvalues$$$$for4\theta :\pi ,3\pi ,5\pi ,7\pi$$$$whichfulfill$$$$\cos 4\theta =-1$$$$\sin 4\theta =0$$

Answered by peter frank last updated on 21/Dec/18

$$z^4=-16(\cos \pi +j\sin \pi )$$$$r=\sqrt{{16}^2}=16$$$$from$$$$z_k=r^{\frac{1}{n}}[\cos \frac{\theta +2\pi k}{n}+j\sin \frac{\theta +2\pi k}{n}]$$$$n=4\theta =\pi$$$$z_k=2[\cos \frac{\pi +2\pi k}{4}+j\sin \frac{\pi +2\pi k}{4}]$$$$k=0$$$$z_0=2(\cos \frac{\pi }{4}+j\sin \frac{\pi }{4})=\sqrt{2}(1+j)$$$$\mathrm{k}=1$$$$z_1=2(\cos \frac{3\pi }{4}+j\sin \frac{3\pi }{4})=-\sqrt{2}(1-j)$$$$k=2$$$$z_2=2(\cos \frac{5\pi }{4}+j\sin \frac{5\pi }{4})=\sqrt{2}(-1-j)$$$$k=3$$$$z_3=2(\cos \frac{7\pi }{4}+j\sin \frac{7\pi }{4})=\sqrt{2}(1-j)$$$$$$$$$$

Commented by Tawa1 last updated on 21/Dec/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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