Find two solutions to equation: w2 = ( -1 -i )

let , w2=-1-i⇒w=-1-i12 .......in general w=x+iy12 so x=-1 and y=-1 Now polar form of w is…

Answered: Find two solutions to equation: w2 = (…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

See similar textbooks

Related questions

Question

100%

Find two solutions to equation: w² = ( -1 -i )

Expert Solution

Step 1

let ,

$w^{2} = (- 1 - i) \Rightarrow w = {(- 1 - i)}^{\frac{1}{2}} . . . . . . . (i n g e n e r a l w = {(x + i y)}^{\frac{1}{2}} s o x = - 1 a n d y = - 1)$

Now polar form of w is ,

$l e t u s t a k e, x = r \cos (θ) a n d y = r \sin (θ) a s i n o u r c a s e x = - 1, y = - 1 . s o \frac{y}{x} = \frac{r \sin (θ)}{r \cos (θ)} = \tan θ \Rightarrow \tan θ = 1 b u t x a n d y a r e b o t h n e g a t i v e s o θ = π + \frac{π}{4} = \frac{5 π}{4} and r = \sqrt{x^{2} + y^{2}} = \sqrt{1 + 1} = \sqrt{2} h e n c e w b e c o m e s, w = {(\sqrt{2} \cos (\frac{5 π}{4} + 2 n π) + i \sqrt{2} \sin (\frac{5 π}{4} + 2 n π))}^{\frac{1}{2}} w h e r e n = 0, 1 .$