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Dividing complex numbersss

  1. Nov 27, 2008 #1
    1. The problem statement, all variables and given/known data
    The complex number z is defined by

    [tex]z = \frac{a+2i}{a-i}[/tex]

    Given that the real part of z is 1/2, find the value of a


    2. Relevant equations
    3. The attempt at a solution
    Well first of all i multiplied the numerator and denominator of z by (a+i)
    which gave me

    [tex]\frac{a^{2} + 3ai + 1}{a^{2}-1}[/tex]

    Now the real part is going to be a² + 1

    so i set

    [tex]\frac{a^{2} + 1}{a^{2} - 1} = \frac{1}{2}[/tex]

    however I get a = [tex]\sqrt{-3}[/tex]

    Have I gone the right way about solving this question?

    Thanks :)
     
  2. jcsd
  3. Nov 27, 2008 #2

    nicksauce

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    The method looks good, however I believe the numerator should be a^2 + 3ai -2, not a^2 + 3ai + 1.
     
  4. Nov 27, 2008 #3

    gabbagabbahey

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    and the denominator should be a^2+1 not a^2-1
     
  5. Nov 28, 2008 #4
    Ahh yes i can see i made a istake in the numerator however I dont see how the denominator can be a² - 1?:
    (a-i)(a+i)
    = a² + ai - ai - i²
    = a² -1

    Does i x i always = 1 regardless of what the sign before it is?

    Thanks :)
     
  6. Nov 28, 2008 #5

    tiny-tim

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    No … (a-i)(a+i)
    = a² + ai - ai - i²
    = a² + 1.

    i x i always = -1 regardless of what the sign before it is! :smile:
     
  7. Nov 29, 2008 #6
    ahh i remeber asking one of my maths teachers this question

    i x i is always -1

    but, as surds:

    [tex] \sqrt{-1} . \sqrt{-1} = \sqrt{-1 \times -1} = \sqrt{1}[/tex]

    does it not?

    Thanks
     
  8. Nov 29, 2008 #7

    tiny-tim

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    hee hee! :biggrin:

    nooo … it doesn't work that way …

    √ (or 1/2) is ambiguous, like arcsin, and it's not a good idea to use anything ambiguous in a general formula! :smile:
     
  9. Nov 29, 2008 #8

    Mark44

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    The rule for square roots is that [tex]\sqrt{ab} = \sqrt{a}\sqrt{b}[/tex] provided that a and be are nonnegative.
     
  10. Nov 29, 2008 #9
    ah cool :)
     
  11. Nov 29, 2008 #10

    Mentallic

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    hehe thomas49th I like the way you think.

    While I always looked at [tex]\sqrt{-1}^2=-1[/tex] since I just cancel the square with the root, I never thought about expressing it your way. I too would've been a little confused if I ever did end up doing it that way. :smile:
     
  12. Nov 30, 2008 #11

    Mark44

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    I hope this is the way you used to do things, but don't do them that way now! Your equation [tex]\sqrt{-1}^2=-1[/tex] is incorrect on at least two counts:
    1. [tex]-1^2=-1[/tex], so you're attempting to take the square root of -1. If you meant the square of -1, rather than the negative of 1 squared, put parentheses around -1.
    2. [tex]\sqrt{(-1)^2}=+1, not -1[/tex]
     
  13. Nov 30, 2008 #12

    Mentallic

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    well I meant [tex](\sqrt{-1})^2[/tex]. Of course when I see i2 I instantly think -1, and don't get into the nitty gritty of it :smile:
     
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