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Math Help - [SOLVED] Complex numbers?

  1. #1
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    [SOLVED] Complex numbers?

    Determine the value of a if (√3 + ai)/(1 - √3i) is a real number, and find this number. The answer they gave is a = 3 and the real number is √3.

    If (√3 + ai)/(1 - √3i) is a real number, then doesn't it mean that (√3 + ai)/(1 - √3i) is x in x + yi? So, is it correct to rearrange it in this way,

    (√3 + ai)/(1 - √3i) + yi = z

    But then, if I arranged it in that way, I cannot see any possible way to find "a". Can you help me? Thanks.
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    (delete)
    Last edited by bobak; June 3rd 2009 at 10:38 AM.
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  3. #3
    MHF Contributor Amer's Avatar
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    \frac{\sqrt{3}+ai}{1-\sqrt{3}i}

    \frac{\sqrt{3}+ai}{1-\sqrt{3}i}\times\frac{1+\sqrt{3}i}{1+\sqrt{3}i}

    now after you modify it you should but the value of a which make the number real (make the i(0))
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  4. #4
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by mark1950 View Post
    Determine the value of a if (√3 + ai)/(1 - √3i) is a real number, and find this number. The answer they gave is a = 3 and the real number is √3.

    If (√3 + ai)/(1 - √3i) is a real number, then doesn't it mean that (√3 + ai)/(1 - √3i) is x in x + yi? So, is it correct to rearrange it in this way,

    (√3 + ai)/(1 - √3i) + yi = z

    But then, if I arranged it in that way, I cannot see any possible way to find "a". Can you help me? Thanks.
    Hi mark1950.

    The answer they give should be a=-3.

    What you want to do is to rationalize the denominator.

    \frac{\sqrt3+ai}{1-\sqrt3i}\ =\ \frac{(\sqrt3+ai)(1+\sqrt3i)}{(1-\sqrt3i)(1+\sqrt3i)}\ =\ \frac{\sqrt3-\sqrt3a+(a+3)i}{1+3}\ =\ \frac{\sqrt3-\sqrt3a}4+\frac{a+3}4i

    Now you set the imaginary part to be 0 and find a, then substitute this into the real part to find the real number.
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  5. #5
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    \frac{{\left( {\sqrt 3  + ai} \right)}}<br />
{{\left( {1 - \sqrt 3 i} \right)}} = \frac{{\left( {\sqrt 3  + ai} \right)\left( {1 + \sqrt 3 i} \right)}}<br />
{4} = \frac{{\left( {\sqrt 3  - a\sqrt 3 } \right) + i\left( {3 + a} \right)}}<br />
{4}

    From the above it is clear that a=-3 gives a real number.
    So, the given answer in not correct.
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  6. #6
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    Oh, thanks people! Many mighty thanks. Huh...should have asked you guys before I wasted half of my pen's ink on this question. Thanks again, dudes! By the way, what does it mean by (√3 + ai)/(1 - √3i) is a real number? If what you guys did above is true ( which I believe it really is) then, wouldn't it make (√3 + ai)/(1 - √3i) a complex number but not a real number? Because if it was a real number, then the equation (√3 + ai)/(1 - √3i) + yi = z would have been correct, right? Sorry, as you can see my tag name, I'm a newbie in polynomials.

    And, the Abstractionist,

    (a + 3)/4 = 0
    a = -3

    Why do I have to set the imaginary part to 0? Is there any condition that I missed out?
    Last edited by mark1950; June 3rd 2009 at 10:37 AM.
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  7. #7
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    Quote Originally Posted by mark1950 View Post
    Why do I have to set the imaginary part to 0? Is there any condition that I missed out?
    Any complex number is made like this: z = \Re (z) + i\Im (z).
    It consists of real and imaginary parts.
    Here are some examples: \begin{gathered}<br />
  \Re (3 - 2i) = 3\;\& \,\Im (3 - 2i) =  - 2 \hfill \\<br />
  \Re ( - 1 + 2i) =  - 1\;\& \,\Im ( - 1 + 2i) = 2 \hfill \\ <br />
\end{gathered} .

    Thus \Im \left( {\frac{{\left( {\sqrt 3  - a\sqrt 3 } \right) + i\left( {3 + a} \right)}}<br />
{4}} \right) = \frac{{3 + a}}{4}.

    A number is real if the imaginary part is zero.
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  8. #8
    MHF Contributor Amer's Avatar
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    in general we write the complex number like

     z=x+iy

    when we say \frac{\sqrt3+ai}{1-\sqrt3i}

    real number that's mean the factor of i should be zero because if we have a factor instead of zero that mean it is not a real number it is imaginary I think you know that

    i=(-1)^{\frac{1}{2}}

    for example if they ask you is \frac{6+\sqrt{2}}{i-\sqrt{2}}
    real you should simplify make no i in the denominator bu multiply it with i+\sqrt{2}
    like that we remove i from the denominator you will see that it is not a real because we find i in the number try to evaluate it

    I think it is clear now
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  9. #9
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    Thanks! Your description have cleared me doubts!
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