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!

Solving z^2+pz+qiz+a+bi=0

  1. Feb 17, 2017 #1
    • Thread moved from a technical forum, so homework template missing
    ok having major problems. i can easily solve z^2 + pz +a+bi=0 solutions but that extra qiz is really annoying me.

    z^2 + 3z+4iz-1+5i=0

    (z+2i)^2+3z-5+5i=0
    z+2i = w, z=w-2i

    w=-3(w-2i)+5-5i

    then im not getting anything sensible for solving x and yi. what am i doing wrong?
     
  2. jcsd
  3. Feb 17, 2017 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    The real and imaginary parts need to be zero at the same time ... simultaneous equations.
    If z is complex, then put z = a+ib.
     
  4. Feb 17, 2017 #3
    that doesnt give any answers.
     
  5. Feb 17, 2017 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Just use the quadratic formula on the equation ##z^2 + (3+4i) z +(5i-1) = 0.## This has the form ##z^2 + bz + c=0##, with ##b = 3+4i## and ##c = 5i-1##. The only real complication is that the discriminant ##b^2 - 4 a c## can be complex---it equals ##-3 + 4i## in your case---so you need to be able to take square roots of complex numbers.
     
  6. Feb 17, 2017 #5
    nevermind. got it. thanks a lot.
     
  7. Feb 17, 2017 #6

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Great! What did you end up doing?
     
  8. Feb 18, 2017 #7
    basically took the root out of ((p/2)^2-q)^(1/2)

    (p/2)^2-q = x+iy

    (a+bi)(a+bi)=(x+iy)

    a^2-b^2=x (if x is positive a^2>b^2 if not b^2>a^2)
    2abi=iy = > 2b=y (if y is positive then a and b both must either be negative or positive, if y is negative then either x or y has to be negative or vice versa and the other positive)

    this way you can figure out your roots which will always be 2 roots since its a z^2 equation.

    then just add. z=(p/2)+-(x+iy)
     
    Last edited: Feb 18, 2017
  9. Feb 18, 2017 #8

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Please write down the final solutions (to your numerical example) explicitly. We cannot tell if you have made an error if you will not show us what you got.
     
  10. Feb 20, 2017 #9
    no worries got the right answer. thanks.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted