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Please help with this complex number qn

  1. Apr 22, 2012 #1
    find all the solutions to z^2+4z ̅+4=0 where z is a complex number.
     
  2. jcsd
  3. Apr 22, 2012 #2

    Mentallic

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    Is the quadratic [itex]z^2+4z+4[/itex]? Simply use the quadratic formula.
     
  4. Apr 22, 2012 #3
    i think i'm getting confused so is the answer for z=2 or do i have to represent it as complex 2isquared? because the question asks for all solution so i don't think 2 is right... and what does the bar on top of the z mean?
     
  5. Apr 22, 2012 #4

    Mentallic

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    Oh I see, so you were trying to solve [itex]z^2+4\overline{z}+4=0[/itex] then? Because in your OP I'm just seeing a box that is being replaced by whatever you wanted to write down.

    If it's just [itex]z^2+4z+4[/itex] then you can simply factorize it into [itex](z+2)^2=0[/itex] which would give you the answer of z=-2, but this can't be right because the question already stated that z is a complex number (imaginary part is not 0 by assumption).

    If it's [itex]z^2+4\overline{z}+4=0[/itex] then you'll need to use the fact that [itex]z=a+ib[/itex] and [itex]\overline{z}=a-ib[/itex] and plug these values into the expression and then simplify and equate the real and imaginary parts.
    What I mean by equating them is if you end up with something like [itex]ab+2+(a-1)i=0[/itex] then what you can take away from this is that [itex]ab+2=0[/itex] and [itex]a-1=0[/itex] because the LHS is equal to the RHS thus the real part on the LHS is equal to 0 on the RHS, and the imaginary part on the LHS is equal to the imaginary part on the RHS (the RHS is equal to 0+0i).
     
  6. Apr 22, 2012 #5
    so i partially understand what you have
    so i did something like this:
    z = a+ib and zbar=a-ib
    because i split them like you said (z+2i)(z-2i)=0
    so i made a+ib=z+2 and a-ib=zbar+2
    so z=a+ib-2 and zbar=a-ib-2
    i put these back into th eqn (a+ib-2)squared +4(a-ib-2)+4=0
    then i expand and simplify i get a squared+2aib-8ib-b-4=0
    and so (a squared -b-4)+2i(ab-4b)=0?
    then a squared -b=4 and ab=4b
    so a=4 and b=12?
    then so i get z=4+12i and zbar=4-12i ???
    i don't get why we have to split it as z and zbar...
     
  7. Apr 22, 2012 #6

    Mentallic

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    Whew that was a tough one to follow.

    Ok stop right there. Where did I say to "split them" as such? If [itex]z=a+ib[/itex] and [itex]\overline{z}=a-ib[/itex] then substituting these values into

    [tex]z^2+4\overline{z}+4=0[/tex]

    gives us

    [tex](a+ib)^2+4(a-ib)+4=0[/tex]

    Now expand, simplify, and collect the real and imaginary parts and equate them.
     
  8. Apr 22, 2012 #7
    i reread your explanation and redid it
    so what i did was sub a+ib and a-ib straight into z^2+4zbar+4=0
    and i got a squared -b squared +4 + 4a +2bi(a-2) = o
    and then a-2=0 a=2 and a squared -b squared +4+4a=0
    and i get a=2 and b=+/-4
    do i sub these back into z and z bar or something????
     
  9. Apr 22, 2012 #8
    so my solutions are: z=2+4i and z=2-4i??
     
  10. Apr 22, 2012 #9

    Curious3141

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    And z=-2. Real numbers are a subset of the complex plane.
     
  11. Apr 22, 2012 #10

    Curious3141

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    So far so good.

    OR b=0.

    That's also a valid solution.

    When you sub b=0 back into this, you get a2 + 4a + 4=0, which will give you a=-2, b=0; the only real solution z=-2 which I mentioned in my last post.

    So there are three solutions in total.
     
  12. Apr 22, 2012 #11
    yep... so my solutions are z=2+4i, z=2-4i and z=-2
    thankyou!!!
     
  13. Apr 22, 2012 #12
    In the future use punctuation and be neater so people can help you easily!
     
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