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Determine-the-fourth-roots-of-16-giving-the-results-in-the-form-a-jb-




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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