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Help me factorize a quadratic equation in a complex variable

  1. Sep 18, 2014 #1
    The equation is: (appears while solving a trigonometric integral using residue theorem)

    2Z2+iZ2-6Z+2-i
    =(2+i)Z2-6Z+(2-i)


    The roots are:
    Z1=(2-i) and Z2=(2-i)/5

    I can't write the equation in factored form.
    If I simply write it like this:
    {z-(2-i)}{5z-(2-i)}
    It doesn't give the same equation back i.e.
    {z-(2-i)}{5z-(2-i)}=5Z2-12Z+6iZ+3-4i
    which is not the equation that I started with.

    Can anyone please write the equation in factored form...
     
  2. jcsd
  3. Sep 18, 2014 #2

    BvU

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    Did you notice PF has a template ?

    In the first place, I don't see an equation. But I suppose you want to solve 2Z2+iZ2-6Z+2-i = 0 ?

    You claim a root is z = 2-i. How do you know ?

    Let me give you a hint, anyway: what is 5/(2+i) ?
     
  4. Sep 18, 2014 #3

    ehild

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    The roots 2-i and (2-i)/5 are correct. They are the same as 5/(2+i) and 1/(2+i).

    ehild
     
  5. Sep 18, 2014 #4

    ehild

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    az2+bz+c=a(z-z1)(z-z2)

    You forgot the coefficient of z2.

    ehild
     
  6. Sep 18, 2014 #5

    BvU

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    Dear ehild, what a splurge of posts. Why don't we let poor old Ari contemplate a bit ?
     
  7. Sep 18, 2014 #6

    ehild

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    The OP solved the quadratic equation corresponding to the original formula correctly. Why did you try to confuse him?

    ehild
     
  8. Sep 18, 2014 #7

    BvU

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    Oh boy, that wasn't the intention. The intention was to let Ari see that a simple multiplication of the coefficients with (2+i)/5 would let the one eq go over into the other.

    Maybe it would even have worked! Or does Ari have the feeling I tried to confuse him/her ?
     
  9. Sep 18, 2014 #8

    ehild

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    You asked the OP how did he know that 2-i was a solution. Was not it confusing?

    ehild
     
  10. Sep 18, 2014 #9
    Thank you for the replies...

    Well .. my bad .. that certainly wasn't a good way to ask for help and my equation looked weird. I've bookmarked the guidelines thread .. will refer to it before making a new thread in future. Thank you for that...

    And I did get a root that was 5/(2+i) but just wanted to get rid of i in the denominator.

    Thank you .. Didn't even realize I was making this mistake. Had completely forgotten that you add the coefficient of Z2 while factorizing a quadratic.
     
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