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Complex numbers : quadratic equation

  1. May 26, 2016 #1
    1. The problem statement, all variables and given/known data
    Showing all necessary working solve the equation ##iz^2+2z-3i=0## giving your answer in the form ##x+iy## where x and y are real and exact,


    2. Relevant equations


    3. The attempt at a solution
    ##iz^2+2z-3i=0, z^2+(2/i)z-3=0##,using quadratic formula →##(-2/i± √8)/2 , z= √2+1/i## and ##√2-1/i## is this correct?
     
    Last edited by a moderator: May 26, 2016
  2. jcsd
  3. May 26, 2016 #2

    SteamKing

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    I think in order to answer in the form x + i y, you need to rationalize your answers above.

    https://mathway.com/examples/Linear.../Rationalizing-with-Complex-Conjugates?id=272
     
    Last edited by a moderator: May 26, 2016
  4. May 26, 2016 #3
    ok we have ## √2+1/i= (√2i+1)/i ## multiplying numerator and denominator by i we get ## (-√2+1i)/-1 = √2-i ##and → ##√2+i## respectively
     
  5. May 26, 2016 #4

    SteamKing

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    Double check your arithmetic for the z-values which you calculated.
     
    Last edited by a moderator: May 26, 2016
  6. May 26, 2016 #5
    still getting ## (-√2+i)/-1= √2-i##
     
  7. May 26, 2016 #6

    ehild

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    Up to z=(-2/i± √8)/2 it is correct. Dividing by 2, you get -1/i ±√2. What is -1/i? Remember i was defined so i2=-1.
     
  8. May 26, 2016 #7

    Ray Vickson

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    If I were doing it I would first multiply the original equation by ##-i## to get ##z^2 -2 i z -3=0##, which is a tinier bit simpler to work with.
     
  9. May 26, 2016 #8

    Math_QED

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    In fact you can check your answers by plugging in the solutions you found for z. Another way is to let wolframalpha solve it for you.
     
  10. May 27, 2016 #9
    Ray your approach to me looks good.
    ##z^2-2iz-3=0, ⇒((-2i±√(-4+12))/2⇒(-2i±2√2)/2⇒√2±i##
    ##z=√2+i, √2-i##
    which is still the same as the solution i got earlier.
     
  11. May 27, 2016 #10

    SteamKing

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    Ooooh! It's so subtle, you're still missing it!

    You got ##z = \frac{(-2i\, ±\, 2\sqrt{2})}{2}## which you say is the same as ##z = \sqrt{2}\, ±\, i##
     
  12. May 27, 2016 #11
    lol##z=-i+√2, z=-i-√2## ⇒##z=√2-i, z=-√2-i##
    Mathway gives solution as ##z=√2-i##
     
  13. May 27, 2016 #12
    on another point, if i hadn't multiplied the original equation by ##-i##, are we saying that we can't get alternative solution(s)? just a thought, like the way i had done it, or are we going to get the same solution as found in post11?
     
  14. May 27, 2016 #13
    post 9 looks wrong to me ......using the quadratic formula:, ##z=(2i±2√2)/2##
    and subsequently everything following from it.
     
  15. May 27, 2016 #14
    from post 13:, ##z=i±√2## which agrees with my earlier post number 3. Advice please?
     
  16. May 27, 2016 #15

    SteamKing

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    ##z = i\, ±\, \sqrt{2} ## (Post 14) is not the same as ##z=\sqrt{2}\,±\,i## (Post 3).

    You can't flip where the ± goes in the expression.
     
  17. May 27, 2016 #16
    ok we have:, either ##z=√2+i, z=-√2+i##
     
  18. May 27, 2016 #17
    look at my post 1;, it should be correct, was my rationalization wrong? this is how i did it,
    the original equation, ##iz^2+2z-3i=0,## i divided each term by ##i##, i got ,## z^2+(2/i)z-3=0##, on using the quadratic formula i got ## z=((-2/i)±√(-4-4(1)(-3))/2, →((-2/i)±2√2)/2→(-1/i)±√2##⇒
    ##z=(-1/i)+√2, z=(-1/i)-√2##,
    ⇒##z=(-1+√2 i)/i, z=(-1-√2 i)/i##,
    on multiplying the numerator and denominator by ##i##, we have
    ##z=-i-√2, z=-i+√2##⇒ ##z=-√2-i, z=√2-i##
    mathway gives solution as ##z=√2-i##, that is why i had mentioned earlier in post (12), if there is a possibility of many solutions to this problem dependant on the approach used.
     
    Last edited: May 27, 2016
  19. May 27, 2016 #18

    SteamKing

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    There's nothing wrong with rationalizing ##\frac{1}{i}## by multiplying that term by ##\frac{i}{i}##.

    That shouldn't change the fact that the quadratic formula gave ##z = (\frac{\frac{-2}{i}±\sqrt{8}}{2})##.
    You can rationalize the imaginary term without affecting the sign of the real term.
     
  20. May 27, 2016 #19
    steam thanks a lot, kindly look at my post number 1 and post number 3. Which means i was correct from that point.
     
  21. May 27, 2016 #20

    SteamKing

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    That's just it. The problem starts in Post #1 and carries over into Post #3.

    You had:
    which is correct. Then you wrote

    which is not correct, and not because of how you rationalized the imaginary term in z above.

    It's not correct because you moved the ± from in front of the square root and put it in front of the imaginary term.

    It's like saying 4 ± 6 = 6 ± 4. You can check this example and see that while 4 + 6 = 6 + 4, the converse 4 - 6 ≠ 6 - 4. It's a very subtle distinction, but one which nevertheless must be made.
     
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