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!

Complex limit help

  1. May 14, 2015 #1

    jjr

    User Avatar

    1. The problem statement, all variables and given/known data
    Calculate the following limit if it exists:

    ##\lim_{z\to i} = \frac{z^3+i}{z-i}##

    2. Relevant equations
    Possibly relevant:
    ## \lim_{z\to\infty} f(z) = \omega_0 \hspace{5mm} \text{if} \hspace{5mm} \lim_{z\to 0} f\left(\frac{1}{z}\right) = \omega_0##

    3. The attempt at a solution
    The problem is obviously that the denominator goes to zero, so the solution likely has something to do with rewriting the limit so that this does not happen.

    I tried rewriting the first equation to fit the form of the RHS of the possibly relevant equation written above.
    ##\lim_{z\to i} = \frac{z^3+i}{z-i} = \lim_{z\to 0} \frac{(z^3+i^3)+i}{(z+i)-i} = \omega_0##
    so that
    ## \omega_0 = \lim_{z\to\infty} \frac{(1/z^3 + i^3)+i}{(1/z + i) - i} ##
    I still wind up with a zero in the denominator.

    I also tried multiplying by the complex conjugate of the expression in the denominator, making the denominator real, but ## \lim_{z\to i} ## implies that the real part goes to zero, so the denominator again goes to zero.

    I think I need to factor the expression ## (z^3+i) ## so that I can cancel a term in both the numerator and denominator, but I am having some trouble. The only root I can find is ## z = i ## and I'm not sure how to find the other terms using this information.

    Any hints?

    Thanks,
    J
     
  2. jcsd
  3. May 14, 2015 #2

    jjr

    User Avatar

    Never mind! Made en error in the polynomial division. Managed to solve it. Disregard thread.
     
  4. May 14, 2015 #3

    Mark44

    Staff: Mentor

    @jjr, what did you get? When you use polynomial division you get ##z^2 + z + 1 + \frac{2i}{z - i}##, and there is still a problem with the denominator in the remainder fraction.

    Edit: My division is incorrect, as pointed out by Dick in a later post...

    L'Hopital's Rule might be useful here.
     
    Last edited: May 14, 2015
  5. May 14, 2015 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    ##z^3+i=(z-i)(z^2+iz-1)##. Another error in polynomial division??
     
  6. May 14, 2015 #5

    Mark44

    Staff: Mentor

    Oops! You caught me!
     
  7. May 14, 2015 #6

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Expand out and simplify ##(z+i)^3 + i## before taking ##z \to 0##.
     
  8. May 14, 2015 #7

    Svein

    User Avatar
    Science Advisor

    Factoring z3+i: Start by solving [itex] z^{3}=-i=e^{\frac{3\pi}{2}}[/itex], which gives you [itex] z=e^{\frac{i\pi}{2}}=i[/itex], [itex] z=e^{\frac{7\cdot i\pi}{6}}[/itex] and [itex]z=e^{\frac{11\cdot i\pi}{6}} [/itex]. Now dividing by [itex] z-i[/itex] is easy.
     
  9. May 14, 2015 #8
    I would like to solve this question.
    the numerator tends to 1+i while the denominator tends to 0. so shouldn't the answer be infinite?
     
  10. May 14, 2015 #9
    Hey, check carefully
    Numerator is tending to zero not 1+i
     
  11. May 14, 2015 #10
    sorry. i got confused with ##\omega^3## and ##i^3##. Can i directly go for LH rule?
     
  12. May 14, 2015 #11
    Yeah, sure why not. It's the most easy way. A one step answer.
     
  13. May 14, 2015 #12

    Svein

    User Avatar
    Science Advisor

    Sorry. As z→i, z3→-i. As I said above, the quotient is [itex](z-e^{\frac{7\pi i}{6}})\cdot (z-e^{\frac{11\pi i}{6}}) [/itex]. Insert z = i and you are done.
     
  14. May 14, 2015 #13
    i wanted to solve without using it.
    You can write it as ##\frac{(z+i)(z^2-iz-1)}{z-i}##. I dont know how to proceed.
     
  15. May 14, 2015 #14
    Can you explain why you took ##e^{\frac{7\pi}{6}}##?
     
  16. May 14, 2015 #15

    jjr

    User Avatar

    @AdityaDev
    You get ## \lim_{z\to i} \frac{(z-i)(z^2+iz-1)}{(z-i)} ##, simplify, insert, solved

    Edit: Typo
     
  17. May 14, 2015 #16

    jjr

    User Avatar

    Edit: Double post
     
  18. May 14, 2015 #17
    ##a^3+b^3=(a+b)(a^2-ab+b^2)##. so isnt it -iz?
     
  19. May 14, 2015 #18

    jjr

    User Avatar

    ## (z-i)(z^2+iz-1) = z^3 + iz^2 - z -iz^2 -i^2z+i = z^3+i ##
     
  20. May 14, 2015 #19
    But in question the numerator is z3 + i
     
  21. May 14, 2015 #20

    jjr

    User Avatar

    Sorry, made a mistake. Edited now.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Complex limit help
  1. Complex limit (Replies: 1)

  2. Complex Limits (Replies: 1)

  3. Complex limit (Replies: 3)

  4. Complex Limit (Replies: 3)

  5. Complex Limits (Replies: 3)

Loading...