Question Number 187857 by thean last updated on 23/Feb/23
Answered by a.lgnaoui last updated on 23/Feb/23
![a)z^2 =−i=cos (−(π/2))+isin (−(π/2)) z=[cos (−(π/2))+isin (−(π/2))]^(1/2) = cos (−(π/4))+isin (−(π/4))= arg(z)=(−(π/4);((3π)/4)) z={+((√2)/2)−i((√2)/2);−((√2)/2)+i((√2)/2)} b)z^5 =1−i(√3) =2((1/2)−i((√3)/2))= z^5 =[2,−(π/3)mod(2π)]⇒ z=[2^(1/5) ,−(π/(15))+((k2π)/(15))(mod2π)] z_1 =^5 (√2) [cos (−(π/(15)))+isin (−(π/(15)))] z_2 =^5 (√2) [cos ((π/(15)))+isin ((π/(15)))] z3=^5 (√2)[cos ((π/5))isin( (π/5))] z_4 =^5 (√2) [cos (((4π)/(15)))+isin (((4π)/(15)))] z_5 =^5 (√2) [cos (((7π)/(15)))+isin (((7π)/(15)))] c)z^3 =4(√2) (1+i) z^3 =8(((√2)/2)+i((√2)/2)) z=2(cos (π/(12))+isin (π/(12)))mod((k2π)/(12)) z_1 =2(cos (π/(12))+isin (π/(12))) z_2 =2(((√2)/2)+i((√2)/2)) z_3 =2(cos ((5π)/(12))+isin ((5π)/(12))) d)z^2 =8−6i =10((4/5)+i(3/5)) z=a+ib a^2 −b^2 +i(2ab)=8−6i Methode[ algebrique: { ((a^2 −b^2 =8)),((2ab =−6)) :} ab=−3⇒a^2 b^2 =9 X^2 −8X−9=0 64+36=10^2 X=((8±10)/2) ⇒(a,b)=(3,i) z_1 =−3+i z_2 =3−i^](https://www.tinkutara.com/question/Q187884.png)
$$\left.{a}\right){z}^{\mathrm{2}} =−{i}=\mathrm{cos}\:\left(−\frac{\pi}{\mathrm{2}}\right)+{i}\mathrm{sin}\:\left(−\frac{\pi}{\mathrm{2}}\right) \\ $$$${z}=\left[\mathrm{cos}\:\left(−\frac{\pi}{\mathrm{2}}\right)+{i}\mathrm{sin}\:\left(−\frac{\pi}{\mathrm{2}}\right)\right]^{\frac{\mathrm{1}}{\mathrm{2}}} = \\ $$$$\mathrm{cos}\:\left(−\frac{\pi}{\mathrm{4}}\right)+{i}\mathrm{sin}\:\left(−\frac{\pi}{\mathrm{4}}\right)= \\ $$$$\:\:\:\:{arg}\left({z}\right)=\left(−\frac{\pi}{\mathrm{4}};\frac{\mathrm{3}\pi}{\mathrm{4}}\right) \\ $$$$\:\:\:\:{z}=\left\{+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−{i}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}};−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right\} \\ $$$$\left.{b}\right){z}^{\mathrm{5}} =\mathrm{1}−{i}\sqrt{\mathrm{3}}\:=\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}−{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)= \\ $$$$\:\:{z}^{\mathrm{5}} =\left[\mathrm{2},−\frac{\pi}{\mathrm{3}}{mod}\left(\mathrm{2}\pi\right)\right]\Rightarrow \\ $$$$\:\:\:{z}=\left[\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{5}}} ,−\frac{\pi}{\mathrm{15}}+\frac{{k}\mathrm{2}\pi}{\mathrm{15}}\left({mod}\mathrm{2}\pi\right)\right] \\ $$$${z}_{\mathrm{1}} =^{\mathrm{5}} \sqrt{\mathrm{2}}\:\left[\mathrm{cos}\:\left(−\frac{\pi}{\mathrm{15}}\right)+{i}\mathrm{sin}\:\left(−\frac{\pi}{\mathrm{15}}\right)\right] \\ $$$${z}_{\mathrm{2}} =^{\mathrm{5}} \sqrt{\mathrm{2}}\:\left[\mathrm{cos}\:\left(\frac{\pi}{\mathrm{15}}\right)+{i}\mathrm{sin}\:\left(\frac{\pi}{\mathrm{15}}\right)\right] \\ $$$${z}\mathrm{3}=^{\mathrm{5}} \sqrt{\mathrm{2}}\left[\mathrm{cos}\:\left(\frac{\pi}{\mathrm{5}}\right){i}\mathrm{sin}\left(\:\frac{\pi}{\mathrm{5}}\right)\right] \\ $$$${z}_{\mathrm{4}} =^{\mathrm{5}} \sqrt{\mathrm{2}}\:\left[\mathrm{cos}\:\left(\frac{\mathrm{4}\pi}{\mathrm{15}}\right)+{i}\mathrm{sin}\:\left(\frac{\mathrm{4}\pi}{\mathrm{15}}\right)\right] \\ $$$${z}_{\mathrm{5}} =^{\mathrm{5}} \sqrt{\mathrm{2}}\:\left[\mathrm{cos}\:\left(\frac{\mathrm{7}\pi}{\mathrm{15}}\right)+{i}\mathrm{sin}\:\left(\frac{\mathrm{7}\pi}{\mathrm{15}}\right)\right] \\ $$$$\left.{c}\right){z}^{\mathrm{3}} =\mathrm{4}\sqrt{\mathrm{2}}\:\left(\mathrm{1}+{i}\right) \\ $$$$\:\:\:{z}^{\mathrm{3}} =\mathrm{8}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right) \\ $$$${z}=\mathrm{2}\left(\mathrm{cos}\:\frac{\pi}{\mathrm{12}}+{i}\mathrm{sin}\:\frac{\pi}{\mathrm{12}}\right){mod}\frac{{k}\mathrm{2}\pi}{\mathrm{12}} \\ $$$${z}_{\mathrm{1}} =\mathrm{2}\left(\mathrm{cos}\:\frac{\pi}{\mathrm{12}}+{i}\mathrm{sin}\:\frac{\pi}{\mathrm{12}}\right) \\ $$$$\:{z}_{\mathrm{2}} =\mathrm{2}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right) \\ $$$$\mathrm{z}_{\mathrm{3}} =\mathrm{2}\left(\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{12}}+{i}\mathrm{sin}\:\frac{\mathrm{5}\pi}{\mathrm{12}}\right) \\ $$$$\left.{d}\right){z}^{\mathrm{2}} =\mathrm{8}−\mathrm{6}{i}\:=\mathrm{10}\left(\frac{\mathrm{4}}{\mathrm{5}}+{i}\frac{\mathrm{3}}{\mathrm{5}}\right) \\ $$$${z}={a}+{ib} \\ $$$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} +{i}\left(\mathrm{2}{ab}\right)=\mathrm{8}−\mathrm{6}{i} \\ $$$${Methode}\left[\:{algebrique}:\right. \\ $$$$\begin{cases}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{8}}\\{\mathrm{2}{ab}\:\:\:\:\:\:\:=−\mathrm{6}}\end{cases} \\ $$$$\:\:{ab}=−\mathrm{3}\Rightarrow{a}^{\mathrm{2}} {b}^{\mathrm{2}} =\mathrm{9} \\ $$$${X}^{\mathrm{2}} −\mathrm{8}{X}−\mathrm{9}=\mathrm{0} \\ $$$$\mathrm{64}+\mathrm{36}=\mathrm{10}^{\mathrm{2}} \\ $$$${X}=\frac{\mathrm{8}\pm\mathrm{10}}{\mathrm{2}}\:\:\:\Rightarrow\left({a},{b}\right)=\left(\mathrm{3},{i}\right) \\ $$$${z}_{\mathrm{1}} =−\mathrm{3}+{i}\:\:\:\:{z}_{\mathrm{2}} =\mathrm{3}−\overset{} {{i}} \\ $$
Commented by a.lgnaoui last updated on 23/Feb/23
$${we}\:{can}\:{solve}\:{auther}\:{questions} \\ $$$${with}\:{methode}\:{algebrique} \\ $$$${we}\:{pout}\:{z}={x}+{iy} \\ $$$${and}\:{develope}\:{z}\:^{\mathrm{2}} {or}\:{z}^{\mathrm{3}^{} } \:{or}\:{z}^{\mathrm{5}} \\ $$$${parrie}\:{reelle}\:{et}\:{partie}\left[{umag}\right. \\ $$$${z}^{\mathrm{3}} =\left({x}^{\mathrm{3}} −\mathrm{3}{xy}^{\mathrm{3}} \right)+{i}\left({y}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} {y}\right)={a}+{ib} \\ $$$$\begin{cases}{{x}^{\mathrm{3}} −\mathrm{3}{xy}^{\mathrm{2}} ={a}}\\{{y}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} {y}=\mathrm{2}{b}}\end{cases} \\ $$$$.\Rightarrow\left({x},{y}\right)={f}\left({a},{b}\right) \\ $$$$ \\ $$