Finding a complex number from an equation

Click For Summary

Homework Help Overview

The problem involves finding all complex numbers \( z \) such that \( z^6 = -64 \). Participants are exploring methods to solve this equation, particularly through polar form and geometric interpretations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using polar form and the sum of cubes to approach the problem. There is an exploration of the angles involved and the geometric representation of the roots on a circle.

Discussion Status

Some participants have identified potential errors in their calculations and are reflecting on their reasoning. There is an ongoing examination of the relationship between the roots and their geometric properties.

Contextual Notes

Participants are working under the constraints of homework guidelines, which may limit the methods they can use or the information they can assume. There is a noted discrepancy between the solutions found and those provided in a reference book.

Temp0
Messages
79
Reaction score
0

Homework Statement



Find all complex numbers z such that z^6 = -64

Homework Equations



I tried it in two ways, once with the sum of cubes (a + b)(a^2 - ab + b^2)
as well as turning it into polar form and attempting it that way, z = re^(iθ)

The Attempt at a Solution



I completely couldn't solve it at all with the sum of cubes, however, I managed to get six solutions by using the polar form and then saying that θ = ∏/3 (3 + k), where k = 0, 1, 2, 3, 4, 5

By subbing in k into the equation z = r(cosθ + isinθ), where r = 2, I get the answers: -2i, +-(-1 - sqrt(3)i), +- (1 + sqrt(3)i), and +2i. However, the book I'm using gives the answers +- 2i, and +- (sqrt(3) + i), and +- (-sqrt(3) - i). Thank you guys in advance.
 
Physics news on Phys.org
Temp0 said:

Homework Statement



Find all complex numbers z such that z^6 = -64

Homework Equations



I tried it in two ways, once with the sum of cubes (a + b)(a^2 - ab + b^2)
as well as turning it into polar form and attempting it that way, z = re^(iθ)

The Attempt at a Solution



I completely couldn't solve it at all with the sum of cubes, however, I managed to get six solutions by using the polar form and then saying that θ = ∏/3 (3 + k), where k = 0, 1, 2, 3, 4, 5

By subbing in k into the equation z = r(cosθ + isinθ), where r = 2, I get the answers: -2i, +-(-1 - sqrt(3)i), +- (1 + sqrt(3)i), and +2i. However, the book I'm using gives the answers +- 2i, and +- (sqrt(3) + i), and +- (-sqrt(3) - i). Thank you guys in advance.

Well, (-1 - sqrt(3)i)^6=64, not -64. Something clearly went wrong with your polar solution. Can you show us what you did?
 
We used to solve these in a geometrically way noting that the roots will be on a circle equally spaced at every 60 degrees about 0+0i and with a radius of 2. So given one root you can find the others.
 
Last edited:
Dick said:
Well, (-1 - sqrt(3)i)^6=64, not -64. Something clearly went wrong with your polar solution. Can you show us what you did?

Well, r^6 = 64 so r = 2, and then because a / r = - 32, and b / r = 0, I just assumed that the angle I start at is pi, so my angle equation becomes 6θ = ∏ + 2∏k, and oh wait I pretty much just found my error as soon as I typed it out LOL, thanks a lot.
 

Similar threads

  • · Replies 22 ·
Replies
22
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
12
Views
2K
Replies
6
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
4K
  • · Replies 9 ·
Replies
9
Views
6K
  • · Replies 27 ·
Replies
27
Views
2K