1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Polynomials and complex numbers

  1. Apr 8, 2014 #1
    1. The problem statement, all variables and given/known data
    Suppose that u and v are real numbers for which u + iv has modulus 3. Express the imaginary part of (u + iv)^−3 in terms of a polynomial in v.


    2. Relevant equations



    3. The attempt at a solution
    |u+iv|=3 then sort(u^2+i^2) = 3 then
    u = 3 and v=0 or u=0 and v=3


    (0+3i)^-3

    i swear i am missing something with these equations we've only being doing a topic on roots but wow it's different what i've been seeing in class
     
    Last edited: Apr 8, 2014
  2. jcsd
  3. Apr 8, 2014 #2

    Mark44

    Staff: Mentor

    If |u + iv| = 3, then u + iv is an arbitrary point on a circle of radius 3, centered at the origin in the complex plane. What effect does subtracting 3 from u + iv have on this circle?
     
  4. Apr 8, 2014 #3

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Those are two solutions, but there are infinitely many more.

    By the way, is there a typo? The imaginary part of ##(u+iv) - 3## is simply ##v##, which is already a polynomial in ##v##.
     
  5. Apr 8, 2014 #4

    Mark44

    Staff: Mentor

    The way I'm interpreting the problem, u + iv - 3 doesn't represent just a single point.
     
  6. Apr 8, 2014 #5
    Yes definitely a type it was meant to be (u+iv)^3
     
  7. Apr 8, 2014 #6

    Mark44

    Staff: Mentor

    In the OP you changed it to (u + iv)-3 and above you have (u + iv)3. Which is it?

    Really, you need to be more careful. If we don't know what the problem is, we can't help you.
     
  8. Apr 8, 2014 #7
    Sorry it is (u+iv)^-3 my bad
     
  9. Apr 8, 2014 #8

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    OK, as hint I would recommend starting with
    $$(u+iv)(u-iv) = |u+iv|^2$$
    Rearrange this to get a formula for ##(u+iv)^{-1}## and use what you know about the modulus.
     
  10. Apr 8, 2014 #9
    Thank you so much that is a great hint haha
     
  11. Mar 27, 2015 #10
    @jbunniii can you please explain?
    So (u + iv)^-1 = (u-iv)/9 then how do i solve it?
     
  12. Mar 27, 2015 #11

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Can you really not see how to get ##(u + i v)^{-3}## if you know ##(u + iv)^{-1}##?
     
  13. Mar 27, 2015 #12
    (U + iv)^-3 = (u - iv)/9(u + iv)^2 ? I seriously have no clue how to express the imaginary part in terms of polynomial in v...
     
  14. Mar 27, 2015 #13

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    $$(u+iv)^{-3} = [(u+iv)^{-1}]^3 = \left(\frac{u-iv}{9}\right)^3 = ???$$
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted