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Math Help - solving complex numbers

  1. #1
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    solving complex numbers

    This may sound like silly question, but I've made multiple attempts and can't figure out what do do with this question:

    Find all solutions to z2 + 4z' + 4 = 0 where z is a complex number.
    *z' being the conjugate
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  2. #2
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    Re: solving complex numbers

    Quote Originally Posted by cellae View Post
    This may sound like silly question, but I've made multiple attempts and can't figure out what do do with this question:

    Find all solutions to z2 + 4z' + 4 = 0 where z is a complex number.
    *z' being the conjugate
    Let z=a+bi , then :

    (a^2-b^2)+2abi+4(a-bi)+4=0

    Hence :

    \begin{cases}a^2-b^2+4a+4=0 \\2ab-4b=0 \end{cases}
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  3. #3
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    Re: solving complex numbers

    Why, thank you !!
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  4. #4
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    Re: solving complex numbers

    In this case z=z*.
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  5. #5
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    Re: solving complex numbers

    .. Could you extrapolate?
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  6. #6
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    Re: solving complex numbers

    Extrapolate?

    (z+2)^2=z^2+4z+4

    So with Im(z)=0, z=z' and you have the solution z=2.
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  7. #7
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    Re: solving complex numbers

    Quote Originally Posted by cellae View Post
    .. Could you extrapolate?
    If a polynomial has real coefficients then if z_0 is a root then so is \overline{z_0}.
    Thus all that needs to be done is to solve z^2+4z+4=0~.
    Thanks from a tutor
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  8. #8
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    Re: solving complex numbers

    Ok I get that.
    I was confused initially because I wasn't really sure how to go about the conjagate.
    That being because it's the imaginary number that get's its sign swapped. So Wasn't sure if i could change the sign of a whole complex number.

    Brilliant
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