A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

HOW TO find the imaginary solution for x^4+1=0?? pLEASE HELP....

  • This Question is Closed
  1. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    To get it I would factor it into (x^2+1)(x^2-1) [sum of perfect squares] and then factor these to (x+1)(x-1)(x+i)(x-i) so your imaginary solutions are i and -i

  2. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it is not factorable , I thought

  3. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oops I just realized my first factors are in fact incorrect. I misread the original as x^4-1

  4. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    x^4=-1 x=fourth root of -1

  5. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the answer is 1+i/square root of 2,, but I dont know why

  6. sgadi
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Since equation is of order 4, we get four different solutions in complex domain. 1+0i -1+0i 0+1i 0-1i or simply \[\pm i\mbox{ , }\pm 1\]

  7. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    square root of -1 is i square root of i is -i and i

  8. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Instead factor it to (x^2 + i)(x^2 - i)

  9. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes.. I did, but ten what's square root of I ??

  10. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So from the quadratic formula we get (1+√i)/2, (1-√i)/2, (-1+√i)/2, and (-1-√i)/2 but as for √i..... I suppose you could rewrite it as (-1)^{1/4}

  11. sgadi
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    please see my answer above ... those are roots of 1.

  12. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I agree that \[\pm{i}\] are solutions but 1 and -1 are not. If you go back to the original problem and substitute them in you get an incorrect statement. 1^4 + 1≠0

  13. sgadi
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I am sorry ... I have done a terrible mistake in the previous calculation. given equation \[x^4=-1\] since \[e^{i\theta}=\cos{\theta}+i\sin{\theta}\] So, we can write \[-1=e^{i\pi}\] So, we can say \[x=(-1)^{1/4}=e^{\frac{i\pi}{4}}=\cos{\frac{\pi}{4}}+i\sin{\frac{\pi}{4}}\] \[\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\] Similarly taking diffrent values of \[\theta\] we can get \[\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}\] \[-\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\] \[-\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}\] are also roots for the given equation.

  14. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.