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Finding the solutions to a quadratic equation with a complex conjugate

  1. Apr 24, 2012 #1
    1. The problem statement, all variables and given/known data
    Find all solutions to z^2 + 4conjugate[z] + 4 = 0 where z is a complex number.

    2. Relevant equations
    Alternate form: 4conjugate[z] + z^2 = -4

    3. The attempt at a solution
    I have tried solving this solution using the quadratic formula.
    However, √b^2 - 4ac = √16 - 4x1x4 = 0. Therefore, as the square root is not negative, there are no imaginary numbers and the solution cannot be complex, right? Although, I am also confused with solving this, given that there is a conjugate in the equation. So, would I have to solve the equation twice, one as z^2+4z+4=0 and the other as z^2-4z+4=0?
    I also put the equation into wolfram alpha and got the real solution as z=-2 and the complex solutions as z=2-4i, z=2+4i. Is that the right answer? How would you get the real solution and the complex solution from the equation then?
     
    Last edited: Apr 24, 2012
  2. jcsd
  3. Apr 24, 2012 #2

    tiny-tim

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    Hi Cottontails! :smile:

    (try using the X2 button just above the Reply box :wink:)
    hint: the complex conjugate of that equation (indeed, any true equation) is also true :wink:
     
  4. Apr 24, 2012 #3
    Sorry, I don't understand. What do you mean by saying that the complex conjugate is true?
     
  5. Apr 24, 2012 #4

    Mark44

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    Strictly speaking, your equation is not quadratic in z (because of the presence of [itex]\overline{z}[/itex]. It would be a quadratic if it were z2 + 4z + 4 = 0.

    The way to go here is to write z = a + bi so [itex]\overline{z}[/itex] = a - bi. Substitute in the original equation and separate into real and imaginary parts, and then solve for a and b. This works to produce the same answers that wolframalpha provided.
     
  6. Apr 24, 2012 #5

    tiny-tim

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    [itex]\bar{z}^2 + 4z + 4 = 0[/itex] :wink:

    (for the same z)
     
  7. Apr 24, 2012 #6
    Yes, I have tried that method as well however, was unable to finish solving it so I assumed I was solving it incorrectly.
    z = a + bi conjugate[z]= a - bi
    (a+bi)^2 + 4(a-bi) + 4 = 0
    -> a^2 - bi^2 - 2abi + 4 =0
    That is all I got up to. I do not know how to solve the equation from there.
     
  8. Apr 24, 2012 #7

    Mark44

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    You have a couple of mistakes above, possibly due to skipping steps.
    For one mistake, (bi)2 = -b2, not -bi2.

    Start by carefully expanding this equation: (a+bi)2 + 4(a-bi) + 4 = 0
     
  9. Apr 24, 2012 #8
    Yes, sorry. I missed adding in the expansion of 4(a-bi). I also thought about bi^2 = -b^2 (as i^2=-1) yet, got confused as to whether it should be left or not. Thanks for the correction there.
    Okay, so now I have:
    a^2 + 2abi - b^2 + 4a - 4bi + 4 = 0
    So, from here, would I have to factorise to reduce the equation further? As, I can't find any common factors.
     
  10. Apr 24, 2012 #9

    Mark44

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    No, split this equation into its real part and its imaginary part, which will result in two equations. At that point you will need to do some easy factoring. Then solve for a and b.
     
  11. Apr 24, 2012 #10
    Okay, thanks. I understand that you solve for a and b and those values will the be substituted into z = a + bi and z = a - bi to get the two complex solutions.
    However, how you find the real solution (in finding z = -2)? I obtained a = -2 from solving the imaginary parts. So, would you then use b = 0 and then substitute that into z = (-2) + (0)i to then give z = -2?
    As for solving the real parts, I have gotten stuck with it.
    So, the equation: a^2 - b^2 + 4a + 4 = 0
    a^2 = b^2 - 4a - 4
    a = √b^2 - 4a - 4
    a = b - 2√a - 2
    b^2 = a^2 + 4a + 4
    b = √a^2 + 4a + 4
    b = a + 2√a + 2
    Then substituting b into a:
    a = (a + 2√a + 2) - 2√a - 2
    Yet, that then equals a = a
    With doing the same for substituting a into b, I also get b = b.
    I know I'm doing something wrong here so, what is my mistake?
    (I previously solved the equation, mistakingly using the equation a^2 - b^2 + 4 =0 and I was able to understand how it was able to give the values of a and b (yet, I know that's not right as the equation is wrong). So, I do understand the method required but just can't recognise where I went wrong with my working.
     
  12. Apr 24, 2012 #11

    Mark44

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    Write this as a2 + 4a + 4 - b2 = 0.
    or (a + 2)2 - b2 = 0.
    Can you continue from there?

     
  13. Apr 25, 2012 #12
    (a + 2)^2 - b^2 = 0
    b^2 = (a + 2)^2
    b = a + 2
    However, how would I solve it to get the value of a? I have tried substituting b = a + 2 back into the equation but then I get 0, which I know isn't right...
     
  14. Apr 25, 2012 #13

    tiny-tim

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    Hi Cottontails! :smile:

    Slow dowwwwn, and go back to …
    (try using the X2 button just above the Reply box :wink:)

    a2 + 2abi - b2 + 4a - 4bi + 4 = 0

    is two simultaneous equations

    so (general strategy :wink:) solve the easy one :tongue2: first, then substitute that into the other one

    in this case, the easy one is 2abi - 4bi = 0 :smile:
     
  15. Apr 25, 2012 #14
    Okay, thanks for pointing that out. I solved 2abi - 4bi = 0
    2abi = 4bi
    -> a = 2
    b = a + 2
    b = 2 + 2
    -> b = 4
    z = a + bi, z = a - bi
    z = 2 + 4i, z = 2 - 4i are the imaginary/complex solutions.
    However, how would you find the real solution (z = -2)? Would you just use the original equation and solve it using the quadratic formula?
     
  16. Apr 25, 2012 #15

    tiny-tim

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    aha! wrooong!! :biggrin:

    the correct statement is …

    2abi = 4bi

    -> a = 2 or b = 0​

    … innit? o:)

    (the moral: never divide by 0 and expect the universe not to notice :rolleyes:)
     
  17. Apr 25, 2012 #16
    Yes, that makes sense. However, if a = 2 and b = 0, then if you substitute it back into z = a + bi, then you get z = 2, which doesn't satisfy the imaginary or real solutions. Yet, if you substitute b = 0 into b = a + 2, you get a = -2. Substituting that into z = a + bi, it would then equal z = -2 (real solution).
    Sorry, I'm really confused now...
     
  18. Apr 25, 2012 #17

    tiny-tim

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    "or" … "and" :rolleyes:

    get some sleep! :zzz:​
     
  19. Apr 25, 2012 #18
    Oh, right. So, from solving the imaginary parts, you get b = 0. You substitute that into b = a + 2, to get a = -2. Therefore, substituting that into z = a + bi, z = -2 (real solution).
    For the imaginary solution, it is how I had solved it previously.
    Is that now all correct?
     
  20. Apr 25, 2012 #19

    tiny-tim

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  21. Apr 25, 2012 #20
    Okay, great. Thank you for all your help!
     
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