Finding the solutions of a complex number

Click For Summary

Homework Help Overview

The problem involves finding the three solutions of the equation \( e^{i\pi/3} z^3 = \frac{1}{1+i} \). The context is within complex numbers and polar forms.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss converting the expression into polar form and identifying the magnitude and angle for \( z^3 \). There is a focus on deriving multiple solutions from the complex exponential form.

Discussion Status

Participants are actively exploring the implications of adding multiples of \( 2\pi i \) to the angle in order to find all unique solutions. Some guidance has been provided regarding the need to account for these multiples to derive the additional solutions.

Contextual Notes

There is a mention of confusion regarding the notation used for the multiples of \( 2\pi i \) and how they relate to the solutions of the equation. The discussion reflects a need to clarify the approach to finding all three solutions without resolving the problem itself.

EmmaK
Messages
24
Reaction score
0

Homework Statement



Find the 3 solutions of ei[tex]\pi[/tex]/3z3=1/(1+i)

Homework Equations


ei[tex]\theta[/tex]=cos([tex]\theta[/tex])+isin([tex]\theta[/tex])


The Attempt at a Solution



i have put i/(1+i) into polar form,1/[tex]\sqrt{2}[/tex] ei[tex]\stackrel{\pi}{4}[/tex]

So i get z3 = [tex]\stackrel{1}{\sqrt{2}}[/tex]ei-[tex]\pi/12[/tex]

Then i got stuck... z3=r3ei[tex]\theta[/tex]

So shouldn't r3=1/[tex]\sqrt{2}[/tex] and [tex]\theta[/tex]=-[tex]\pi[/tex]/36 ..but that's only 1 solution?
 
Physics news on Phys.org
well.
if [itex]z^3=\frac{1}{\sqrt{2}} e^{-\frac{i \pi}{12}}[/itex]
then [itex]z=(\frac{1}{\sqrt{2}})^{\frac{1}{3}} e^{-\frac{i \pi}{36}}[/itex]
 
but that is only one solution and it asks for 3
 
You can add any multiple of 2 pi i to theta.
 
ahh, of course. thank you!
 
EmmaK said:
ahh, of course. thank you!

Note that for unique solutions, you need to add [tex]n\cdot 2\pi i[/tex] to the exponent of the complex number describing [tex]z^3[/tex]

Otherwise you're just describing the same number over and over!
 
Last edited:
haha, oh yea.
where do you get n2/pi i from?
 
EmmaK said:
haha, oh yea.
where do you get n2/pi i from?

That's how much you need to add to the angle of the exponent, [tex]r e^{i\theta}[/tex] so that you get the same value.

[tex]re^{i\theta}=re^{i(\theta+2\pi)}[/tex]

You can easily see this using Euler's identity since the sine and cosine both have a period of [tex]2\pi[/tex] radians.

When you take the cube root, you use De-Moivre and divide the angle by 3. Note that you get different angles depending on whether you add [tex]2\pi[/tex] once, twice, or three times to the original exponent's angle.
 
yes i should have mentioned that. sorry.
 
  • #10
so the final answer is [tex]\stackrel{n2\pi}{3}[/tex]? i think i just misread your post as [tex]\stackrel{2n}{\pi}[/tex] or something :)
 
  • #11
well, is [itex](\frac{2n \pi}{3})^3=\frac{1}{\sqrt{2}}e^{-\frac{1 \pi}{12}}[/itex]?

go to my first line of working in post 2, the other two solutions will correspond to [itex]e^{-\frac{25 i \pi}{12}}[/itex] and [itex]e^{-\frac{49 i \pi}{12}}[/itex].
then of course, you have to do the division by 3 etc as before to get to the final answer.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 22 ·
Replies
22
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
3
Views
2K
Replies
32
Views
4K