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Question Number 187857 by thean last updated on 23/Feb/23

	Answered by a.lgnaoui last updated on 23/Feb/23
	$a)z^2 =−i=cos (−(π/2))+isin (−(π/2)) z=[cos (−(π/2))+isin (−(π/2))]^(1/2) = cos (−(π/4))+isin (−(π/4))= arg(z)=(−(π/4);((3π)/4)) z={+((√2)/2)−i((√2)/2);−((√2)/2)+i((√2)/2)} b)z^5 =1−i(√3) =2((1/2)−i((√3)/2))= z^5 =[2,−(π/3)mod(2π)]⇒ z=[2^(1/5) ,−(π/(15))+((k2π)/(15))(mod2π)] z_1 =^5 (√2) [cos (−(π/(15)))+isin (−(π/(15)))] z_2 =^5 (√2) [cos ((π/(15)))+isin ((π/(15)))] z3=^5 (√2)[cos ((π/5))isin( (π/5))] z_4 =^5 (√2) [cos (((4π)/(15)))+isin (((4π)/(15)))] z_5 =^5 (√2) [cos (((7π)/(15)))+isin (((7π)/(15)))] c)z^3 =4(√2) (1+i) z^3 =8(((√2)/2)+i((√2)/2)) z=2(cos (π/(12))+isin (π/(12)))mod((k2π)/(12)) z_1 =2(cos (π/(12))+isin (π/(12))) z_2 =2(((√2)/2)+i((√2)/2)) z_3 =2(cos ((5π)/(12))+isin ((5π)/(12))) d)z^2 =8−6i =10((4/5)+i(3/5)) z=a+ib a^2 −b^2 +i(2ab)=8−6i Methode[ algebrique: { ((a^2 −b^2 =8)),((2ab =−6)) :} ab=−3⇒a^2 b^2 =9 X^2 −8X−9=0 64+36=10^2 X=((8±10)/2) ⇒(a,b)=(3,i) z_1 =−3+i z_2 =3−i^$
	$a) z^{2} = - i = \cos (- \frac{π}{2}) + i \sin (- \frac{π}{2})$ $z = {[\cos (- \frac{π}{2}) + i \sin (- \frac{π}{2})]}^{\frac{1}{2}} =$ $\cos (- \frac{π}{4}) + i \sin (- \frac{π}{4}) =$ $a r g (z) = (- \frac{π}{4}; \frac{3 π}{4})$ $z = {+ \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}; - \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}}$ $b) z^{5} = 1 - i \sqrt{3} = 2 (\frac{1}{2} - i \frac{\sqrt{3}}{2}) =$ $z^{5} = [2, - \frac{π}{3} m o d (2 π)] \Rightarrow$ $z = [2^{\frac{1}{5}}, - \frac{π}{15} + \frac{k 2 π}{15} (m o d 2 π)]$ $z_{1} =^{5} \sqrt{2} [\cos (- \frac{π}{15}) + i \sin (- \frac{π}{15})]$ $z_{2} =^{5} \sqrt{2} [\cos (\frac{π}{15}) + i \sin (\frac{π}{15})]$ $z 3 =^{5} \sqrt{2} [\cos (\frac{π}{5}) i \sin (\frac{π}{5})]$ $z_{4} =^{5} \sqrt{2} [\cos (\frac{4 π}{15}) + i \sin (\frac{4 π}{15})]$ $z_{5} =^{5} \sqrt{2} [\cos (\frac{7 π}{15}) + i \sin (\frac{7 π}{15})]$ $c) z^{3} = 4 \sqrt{2} (1 + i)$ $z^{3} = 8 (\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2})$ $z = 2 (\cos \frac{π}{12} + i \sin \frac{π}{12}) m o d \frac{k 2 π}{12}$ $z_{1} = 2 (\cos \frac{π}{12} + i \sin \frac{π}{12})$ $z_{2} = 2 (\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2})$ $z_{3} = 2 (\cos \frac{5 π}{12} + i \sin \frac{5 π}{12})$ $d) z^{2} = 8 - 6 i = 10 (\frac{4}{5} + i \frac{3}{5})$ $z = a + i b$ $a^{2} - b^{2} + i (2 a b) = 8 - 6 i$ $M e t h o d e [a l g e b r i q u e :$ ${\begin{cases} a^{2} - b^{2} = 8 \\ 2 a b = - 6 \end{cases}$ $a b = - 3 \Rightarrow a^{2} b^{2} = 9$ $X^{2} - 8 X - 9 = 0$ $64 + 36 = 10^{2}$ $X = \frac{8 \pm 10}{2} \Rightarrow (a, b) = (3, i)$ $z_{1} = - 3 + i z_{2} = 3 - \overset{}{i}$
	Commented by a.lgnaoui last updated on 23/Feb/23
	$we can solve auther questions with methode algebrique we pout z=x+iy and develope z^2 or z^3^ or z^5 parrie reelle et partie[umag z^3 =(x^3 −3xy^3 )+i(y^3 +3x^2 y)=a+ib { ((x^3 −3xy^2 =a)),((y^3 +3x^2 y=2b)) :} .⇒(x,y)=f(a,b)$
	$w e c a n s o l v e a u t h e r q u e s t i o n s$ $w i t h m e t h o d e a l g e b r i q u e$ $w e p o u t z = x + i y$ $a n d d e v e l o p e z^{2} o r z^{3^{}} o r z^{5}$ $p a r r i e r e e l l e e t p a r t i e [u m a g$ $z^{3} = (x^{3} - 3 {x y}^{3}) + i (y^{3} + 3 x^{2} y) = a + i b$ ${\begin{cases} x^{3} - 3 {x y}^{2} = a \\ y^{3} + 3 x^{2} y = 2 b \end{cases}$ $. \Rightarrow (x, y) = f (a, b)$
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