Find-all-the-complex-number-in-the-rectangular-form-such-that-z-1-4-1-

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)}^{\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