Find all the complex number in the rectangular form such that (z−1)^4 =−1


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 38284 by NECx last updated on 23/Jun/18

$F i n d a l l t h e c o m p l e x n u m b e r i n t h e$ $r e c t a n g u l a r f o r m s u c h t h a t$ ${(z - 1)}^{4} = - 1$
Commented by MrW3 last updated on 23/Jun/18

$l e t z - 1 = r (\cos θ + i \sin θ)$ $\Rightarrow r (\cos 4 θ + i \sin 4 θ) = - 1$ $\Rightarrow r = 1$ $\Rightarrow \cos 4 θ = - 1, \sin 4 θ = 0$ $\Rightarrow 4 θ = (2 n + 1) π, n = 0, 1, 2, 3$ $\Rightarrow θ = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$ $(\cos θ, \sin θ) = (\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}), (- \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}), (- \frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2}), (\frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2})$ $z = (1 + \cos θ) + i \sin θ$ $a l l v a l u e s o f z a r e :$ $z = (1 + \frac{\sqrt{2}}{2}) + i \frac{\sqrt{2}}{2}$ $z = (1 - \frac{\sqrt{2}}{2}) + i \frac{\sqrt{2}}{2}$ $z = (1 - \frac{\sqrt{2}}{2}) - i \frac{\sqrt{2}}{2}$ $z = (1 + \frac{\sqrt{2}}{2}) - i \frac{\sqrt{2}}{2}$
Commented by abdo.msup.com last updated on 23/Jun/18

$l e t p u t z - 1 = x (e) \Leftrightarrow x^{4} = e^{i π} i f x = r e^{i θ}$ $w e g e t r^{4} e^{i 4 θ} = e^{i (2 k + 1) π} \Rightarrow r = 1 a n d$ $4 θ = (2 k + 1) π \Rightarrow θ_{k} = \frac{(2 k + 1) π}{4}$ $w i t h k \in [[0, 3]] a n d x_{k} = e^{i \frac{(2 k + 1) π}{4}}$ $\Rightarrow z_{k} = x_{k} + 1 = 1 + e^{i \frac{(2 k + 1) π}{4}} \Rightarrow$ $z_{0} = 1 + e^{i \frac{π}{4}}$ $z_{1} = 1 + e^{i \frac{3 π}{4}}$ $z_{2} = 1 + e^{i \frac{5 π}{4}}$ $z_{3} = 1 + e^{i \frac{7 π}{4}}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 23/Jun/18
	$z−1=(−1)^(1/4) cosΠ=−1 z−1={cos(2nΠ+Π)+isin(2nΠ+Π)}^(1/4) x+iy−1=cos(((2n+1)/4))+isin(((2n+1)/4)) (x−1)^2 +y^2 =cos^2 (((2n+1)/4))+sin^2 (((2n+1)/4)) (x−1)^2 +y^2 =1$
	$z - 1 = {(- 1)}^{\frac{1}{4}}$ $c o s Π = - 1$ $z - 1 = {c o s (2 n Π + Π) + i s i n (2 n Π + Π)}^{\frac{1}{4}}$ $x + i y - 1 = c o s (\frac{2 n + 1}{4}) + i s i n (\frac{2 n + 1}{4})$ ${(x - 1)}^{2} + y^{2} = {c o s}^{2} (\frac{2 n + 1}{4}) + {s i n}^{2} (\frac{2 n + 1}{4})$ ${(x - 1)}^{2} + y^{2} = 1$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum