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Question Number 14588 by tawa tawa last updated on 02/Jun/17

Determine the fourth roots of − 16, giving the results in the form a + jb.

$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{fourth}\:\mathrm{roots}\:\mathrm{of}\:\:\:−\:\mathrm{16},\:\:\:\mathrm{giving}\:\mathrm{the}\:\mathrm{results}\:\mathrm{in}\:\mathrm{the}\:\mathrm{form}\:\:\mathrm{a}\:+\:\mathrm{jb}. \\ $$

Answered by mrW1 last updated on 03/Jun/17

z^4 =−16 z=x+iy=r(cos θ+i sin θ) z^4 =r^4 (cos 4θ+isin 4θ)=−16=16(−1+i0) ⇒r^4 =16⇒r=2 ⇒cos 4θ=−1 ⇒sin 4θ=0 ⇒4θ=(2n−1)π, n=1,2,3,4 θ_1 =(π/4) θ_2 =((3π)/4) θ_3 =((5π)/4) θ_4 =((7π)/4) z_1 =2(cos (π/4)+isin (π/4))=(√2)+i(√2) z_2 =2(cos ((3π)/4)+isin ((3π)/4))=−(√2)+i(√2) z_3 =2(cos ((5π)/4)+isin ((5π)/4))=−(√2)−i(√2) z_4 =2(cos ((7π)/4)+isin ((7π)/4))=(√2)−i(√2)

$${z}^{\mathrm{4}} =−\mathrm{16} \\ $$$${z}={x}+{iy}={r}\left(\mathrm{cos}\:\theta+{i}\:\mathrm{sin}\:\theta\right) \\ $$$${z}^{\mathrm{4}} ={r}^{\mathrm{4}} \left(\mathrm{cos}\:\mathrm{4}\theta+{i}\mathrm{sin}\:\mathrm{4}\theta\right)=−\mathrm{16}=\mathrm{16}\left(−\mathrm{1}+{i}\mathrm{0}\right) \\ $$$$\Rightarrow{r}^{\mathrm{4}} =\mathrm{16}\Rightarrow{r}=\mathrm{2} \\ $$$$\Rightarrow\mathrm{cos}\:\mathrm{4}\theta=−\mathrm{1} \\ $$$$\Rightarrow\mathrm{sin}\:\mathrm{4}\theta=\mathrm{0} \\ $$$$\Rightarrow\mathrm{4}\theta=\left(\mathrm{2}{n}−\mathrm{1}\right)\pi,\:{n}=\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4} \\ $$$$\theta_{\mathrm{1}} =\frac{\pi}{\mathrm{4}} \\ $$$$\theta_{\mathrm{2}} =\frac{\mathrm{3}\pi}{\mathrm{4}} \\ $$$$\theta_{\mathrm{3}} =\frac{\mathrm{5}\pi}{\mathrm{4}} \\ $$$$\theta_{\mathrm{4}} =\frac{\mathrm{7}\pi}{\mathrm{4}} \\ $$$${z}_{\mathrm{1}} =\mathrm{2}\left(\mathrm{cos}\:\frac{\pi}{\mathrm{4}}+{i}\mathrm{sin}\:\frac{\pi}{\mathrm{4}}\right)=\sqrt{\mathrm{2}}+{i}\sqrt{\mathrm{2}} \\ $$$${z}_{\mathrm{2}} =\mathrm{2}\left(\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{4}}+{i}\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{4}}\right)=−\sqrt{\mathrm{2}}+{i}\sqrt{\mathrm{2}} \\ $$$${z}_{\mathrm{3}} =\mathrm{2}\left(\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{4}}+{i}\mathrm{sin}\:\frac{\mathrm{5}\pi}{\mathrm{4}}\right)=−\sqrt{\mathrm{2}}−{i}\sqrt{\mathrm{2}} \\ $$$${z}_{\mathrm{4}} =\mathrm{2}\left(\mathrm{cos}\:\frac{\mathrm{7}\pi}{\mathrm{4}}+{i}\mathrm{sin}\:\frac{\mathrm{7}\pi}{\mathrm{4}}\right)=\sqrt{\mathrm{2}}−{i}\sqrt{\mathrm{2}} \\ $$

Commented by tawa tawa last updated on 04/Jun/17

Wow, i really appreciate sir. God bless you.

$$\mathrm{Wow},\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

Commented by RasheedSoomro last updated on 04/Jun/17

∈×cellent!

$$\in×\mathrm{cellent}! \\ $$

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