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Thread: Imaginary, quadratic

  1. #1
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    Imaginary, quadratic

    Trying to solve this, not allowed to use polar form or Eulers.
    $\displaystyle z^2-4z+4+2i=0$

    Here is my work: $\displaystyle (z-2)^2 = -2i $
    then I have to take the root of $\displaystyle -2i$ and that didnīt work very well.

    I tried this too:
    rewritting $\displaystyle z^2 = x^2-y^2+2xyi$
    and $\displaystyle -4z = -4x-4yi$
    but then it is a $\displaystyle x^4$ equationsystem with two unknows and one x^3 term etc so that didnīt work either.
    and here I am, any ideas?
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  2. #2
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    Quote Originally Posted by Henryt999 View Post
    Trying to solve this, not allowed to use polar form or Eulers.
    $\displaystyle z^2-4z+4+2i=0$

    Here is my work: $\displaystyle (z-2)^2 = -2i $
    then I have to take the root of $\displaystyle -2i$ and that didnīt work very well.

    Why? What's the problem? You'll have to do the same if you use the well-known formula for the roots of a quadratic equation: you need the square root of $\displaystyle -2i$ , so you can put:

    $\displaystyle (x+yi)^2=-2i\Longleftrightarrow x^2-y^2+2xyi=-2i\Longrightarrow x^2-y^2=0$ , $\displaystyle 2xy=-2\Longrightarrow xy=-1\Longrightarrow y=-\frac{1}{x}$ , so $\displaystyle x^2=y2\Longleftrightarrow x^2=\frac{1}{x^2}\Longrightarrow x^4=1\Longrightarrow x=1\,,\,y=-1\,,\,\,x=-1\,,\,y=1$ are the solutions.

    Tonio


    I tried this too:
    rewritting $\displaystyle z^2 = x^2-y^2+2xyi$
    and $\displaystyle -4z = -4x-4yi$
    but then it is a $\displaystyle x^4$ equationsystem with two unknows and one x^3 term etc so that didnīt work either.
    and here I am, any ideas?
    .
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  3. #3
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    First of all thank you for your help.
    But are you sure this is correct!
    Entering $\displaystyle x = 1; y=(-1)$ into the $\displaystyle z^2-4z+4+2i = 0$ doesnt = 0
    or maybe I made some mistake?:
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  4. #4
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    Quote Originally Posted by Henryt999 View Post
    First of all thank you for your help.
    But are you sure this is correct!
    Entering $\displaystyle x = 1; y=(-1)$ into the $\displaystyle z^2-4z+4+2i = 0$ doesnt = 0
    or maybe I made some mistake?:

    Who said $\displaystyle x=1\,,\,y=-1\Longleftrightarrow z=1-i$ is a solution to your question?!? . It is a solution to $\displaystyle z^2=-2i$ !! Now take this, input it in the right place and THEN find the correcto solutions! Just re-read your own intent of solution ...

    Tonio
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