If-z-1-z-2-and-z-3-are-vertices-of-an-equilateral-traingle-inscribed-in-the-circle-z-2-and-if-z-1-1-i-3-then-find-the-value-of-z-2-and-z-3-

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Question Number 229 by ssahoo last updated on 25/Jan/15

$$\mathrm{If}{z}_1,z_2\mathrm{and}{z}_3\mathrm{are}\mathrm{vertices}\mathrm{of}\mathrm{an}\mathrm{equilateral}$$$$\mathrm{traingle}\mathrm{inscribed}\mathrm{in}\mathrm{the}\mathrm{circle}\mid z\mid =2\mathrm{and}\mathrm{if}$$$$z_1=1+i\sqrt{3},\mathrm{then}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}{z}_2\mathrm{and}{z}_3.$$

Answered by 123456 last updated on 16/Dec/14

$$\mid z_1\mid =2$$$$z_1=\frac{2}{2}(1+i\sqrt{3})$$$$=2(\frac{1}{2}+i\frac{\sqrt{3}}{2})$$$$=2(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3})$$$$=2e^{\frac{\pi }{3}i}$$$$\mathrm{since}{z}_1,z_2,z_3\mathrm{are}\mathrm{vertice}\mathrm{of}\mathrm{a}\mathrm{equilatery}\mathrm{triangle}\mathrm{escribed}\mathrm{into}\mathrm{a}\mathrm{circle}\mathrm{at}\mathrm{origin},\mathrm{then}\mathrm{the}\mathrm{other}\mathrm{points}\mathrm{are}\mathrm{defased}\mathrm{by}\frac{2\pi }{3}$$$$\mathrm{then}$$$$z_2=z_1{e}^{\frac{2\pi }{3}i}$$$$=2e^{(1+2)\frac{\pi }{3}i}$$$$=2e^{\pi i}$$$$=2(\cos \pi +i\sin \pi )$$$$=-2$$$$z_3=z_2{e}^{\frac{2\pi }{3}i}=z_1{e}^{-\frac{2\pi }{3}i}$$$$=2e^{\pi i}{e}^{\frac{2\pi }{3}i}=2e^{\frac{\pi }{3}i}{e}^{-\frac{2\pi }{3}i}$$$$=2e^{(1+\frac{2}{3})\pi i}=2e^{(1-2)\frac{\pi }{3}i}$$$$=2e^{\frac{5\pi }{3}i}=2e^{-\frac{\pi }{3}i}$$$$=2(\cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3})=2(\cos -\frac{\pi }{3}+i\sin -\frac{\pi }{3})$$$$=2(\frac{1}{2}-i\frac{\sqrt{3}}{2})=2(\cos \frac{\pi }{3}-i\sin \frac{\pi }{3})$$$$=1-i\sqrt{3}$$$$\mathrm{so}$$$$\{z_1,z_2,z_3\}=\{1+i\sqrt{3},-2,1-i\sqrt{3}\}$$$$\mathrm{solution}\mathrm{of}(a\ne 0)$$$$a[z-(1+i\sqrt{3})][z-(-2)][z-(1-i\sqrt{3})]=0$$$$a[(z-1)-i\sqrt{3}][(z-1)+i\sqrt{3}](z+2)=0$$$$a[{(z-1)}^2+3](z+2)=0$$$$a(z^2-2z+1+3)(z+2)=0$$$$a(z^2-2z+4)(z+2)=0$$$$a(z^3-2z^2+4z+2z^2-4z+8)=0$$$$a(z^3+8)=0$$

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