Polar form

  • Thread starter Mathysics
  • Start date
  • #1
37
0

Homework Statement


solve below using de Moivre's theorem, in polar form
z^3 = i


Homework Equations


r CIS theta

Answer:CIS pi/6, CIS 5pi/6, CIS (-pi/6)

The Attempt at a Solution


r^3 = sqrt(0^2+1^2)
r^3 = sqrt (1)
r = sqrt (1)

no idea how to get the angle
 

Answers and Replies

  • #2
danago
Gold Member
1,122
4

Homework Statement


solve below using de Moivre's theorem, in polar form
z^3 = i


Homework Equations


r CIS theta

Answer:CIS pi/6, CIS 5pi/6, CIS (-pi/6)

The Attempt at a Solution


r^3 = sqrt(0^2+1^2)
r^3 = sqrt (1)
r = sqrt (1)

no idea how to get the angle
It is not true that i = 1, which is essentially what you have done in solving the equation. 1 is simply the modulus of the complex number i.

Try writing the complex number i in polar form i.e. in the form [tex]R cis \theta[/tex], and then taking the cube root of both sides of the equation.
 

Related Threads on Polar form

  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
6
Views
20K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
6
Views
937
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
14K
Replies
4
Views
1K
Replies
9
Views
335
Top