Question and Answers Forum

All Questions      Topic List

Number Theory Questions

Previous in All Question      Next in All Question      

Previous in Number Theory      Next in Number Theory      

Question Number 94080 by Rio Michael last updated on 16/May/20

$$3.\left(\mathrm{a}\right)\mathrm{Find}\mathrm{the}\mathrm{complex}\mathrm{number}z\mathrm{which}\mathrm{satisfy}$$$$\mathrm{the}\mathrm{equation}{z}^3=8i,\mathrm{giving}\mathrm{your}\mathrm{answer}\mathrm{in}\mathrm{the}$$$$\mathrm{form}a+bi\mathrm{where}a\mathrm{and}b\mathrm{are}\mathrm{real}.$$

Answered by mr W last updated on 17/May/20

$$sayz={re}^{i\theta }=r(\cos \theta +i\sin \theta )$$$$r^3{e}^{3\theta i}=8i=2^3{e}^{\frac{\pi }{2}i}$$$$\Rightarrow r=2$$$$\Rightarrow 3\theta =\frac{\pi }{2}+2k\pi \Rightarrow \theta =\frac{\pi }{6}+\frac{2k\pi }{3}withk=0,1,2$$$$\Rightarrow z=2e^{(\frac{\pi }{6}+\frac{2k\pi }{3})i}=2[\cos (\frac{\pi }{6}+\frac{2k\pi }{3})+i\sin (\frac{\pi }{6}+\frac{2k\pi }{3})]$$$$=\sqrt{3}+i,-\sqrt{3}+i,-2i$$

Commented by Rio Michael last updated on 17/May/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com