Finding the Smallest n for Real z^n in Complex Numbers

Click For Summary

Homework Help Overview

The discussion revolves around expressing a complex number in polar form and determining the smallest positive integer n such that raising the complex number to the power of n results in a real number. The subject area includes complex numbers and their properties, particularly in relation to their polar representation.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the expression of the complex number in polar form and the conditions under which its powers yield real numbers. There is a focus on the argument of the complex number and its relationship to real values.

Discussion Status

Some participants have expressed uncertainty about how to approach part b of the problem, while others have suggested considering the properties of real numbers in relation to the argument of the complex number. A potential direction has been identified regarding the conditions for the imaginary part to equal zero.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the information they can provide or the methods they can explore. There is an emphasis on understanding the relationship between the argument of the complex number and its powers.

nayfie
Messages
50
Reaction score
0
I have solved part a, I just have no idea how to go about doing part b. If anybody could point me in the right direction, that would be greatly appreciated!

Homework Statement



a. Express z = \frac{1 + \sqrt{3}i}{-2 -2i} in the form rcis\theta

b. What is the smallest positive integer n such that z^{n} is a real number? Find z^{n} for this particular n.


Homework Equations



z^{n} = r^{n}cis(n\theta)

\frac{z_{1}}{z_{2}} = \frac{r_{1}}{r_{2}}cis(\theta_{1} - \theta_{2})

The Attempt at a Solution



Part A

z_{1} = 1 + \sqrt{3}i

r = |z_{1}| = \sqrt{4} = 2

\theta = arccos(\frac{1}{2}) = \frac{\pi}{3}

z_{1} = 2cis(\frac{\pi}{3})

z_{2} = -2 -2i

r = |z_{2}| = \sqrt{8} = 2\sqrt{2}

\theta = arccos(\frac{-2}{2\sqrt{2}}) = \frac{3\pi}{4}

z_{2} = 2\sqrt{2}cis(\frac{3\pi}{4})

\frac{z_{1}}{z_{2}} = \frac{2}{2\sqrt{2}}cis(\frac{\pi}{3} - \frac{3\pi}{4}) = \frac{1}{\sqrt{2}}cis(\frac{-5\pi}{12})

Part B

No idea :(
 
Physics news on Phys.org
nayfie said:
I have solved part a, I just have no idea how to go about doing part b. If anybody could point me in the right direction, that would be greatly appreciated!

Homework Statement



a. Express z = \frac{1 + \sqrt{3}i}{-2 -2i} in the form rcis\theta

b. What is the smallest positive integer n such that z^{n} is a real number? Find z^{n} for this particular n.


Homework Equations



z^{n} = r^{n}cis(n\theta)

\frac{z_{1}}{z_{2}} = \frac{r_{1}}{r_{2}}cis(\theta_{1} - \theta_{2})

The Attempt at a Solution



Part A

z_{1} = 1 + \sqrt{3}i

r = |z_{1}| = \sqrt{4} = 2

\theta = arccos(\frac{1}{2}) = \frac{\pi}{3}

z_{1} = 2cis(\frac{\pi}{3})

z_{2} = -2 -2i

r = |z_{2}| = \sqrt{8} = 2\sqrt{2}

\theta = arccos(\frac{-2}{2\sqrt{2}}) = \frac{3\pi}{4}

z_{2} = 2\sqrt{2}cis(\frac{3\pi}{4})

\frac{z_{1}}{z_{2}} = \frac{2}{2\sqrt{2}}cis(\frac{\pi}{3} - \frac{3\pi}{4}) = \frac{1}{\sqrt{2}}cis(\frac{-5\pi}{12})

Part B

No idea :(

Hint: What property does a real number have (what is its imaginary component equal to)?

Based on an argand diagram (a two dimensional graph with real and imaginary part), what is the argument (angle) equal to?
 
Oh right. So they want me to find the smallest value of n such that arg(z) = k\pi (so that the imaginary part = 0)?
 
Okay it turns out that's the way to solve the question.

Thank you for the help!
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
27
Views
5K
Replies
4
Views
3K
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K