Finding the solutions of a complex number

In summary, the homework statement is trying to find three solutions to the equation ei\pi/3z3=1/(1+i). Using polar form, the student gets z3=\stackrel{1}{\sqrt{2}}ei-\pi/12. However, for only one solution, they need to add n\cdot 2\pi i to the exponent of the complex number describing z^3.
  • #1
EmmaK
26
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?
 
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  • #2
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]
 
  • #3
but that is only one solution and it asks for 3
 
  • #4
You can add any multiple of 2 pi i to theta.
 
  • #5
ahh, of course. thank you!
 
  • #6
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:
  • #7
haha, oh yea.
where do you get n2/pi i from?
 
  • #8
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.
 
  • #9
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.
 

1. What is a complex number?

A complex number is a number that contains both a real part and an imaginary part. It is written in the form a + bi, where a and b are real numbers and i is the imaginary unit equal to the square root of -1.

2. How do you find the solutions of a complex number?

The solutions of a complex number can be found by solving the equation it is a part of. This could include using algebraic methods or graphing the equation on a complex plane.

3. What is a conjugate complex number?

A conjugate complex number is a complex number that has the same real part but the imaginary part has the opposite sign. For example, the conjugate of 3 + 4i is 3 - 4i.

4. Can a complex number have more than one solution?

Yes, a complex number can have multiple solutions. This is because the solutions of a complex number are found by solving an equation, and equations can have multiple solutions.

5. How are complex numbers used in real life?

Complex numbers are used in a variety of fields, including engineering, physics, and mathematics. They are often used to represent quantities with both a magnitude and direction, such as in electrical circuits and vector calculations.

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