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Math Help - Complex numbers-finding real number pairs

  1. #1
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    Complex numbers-finding real number pairs

    Hello,
    I am having trouble with this question:

    "Find all possible real number pairs p, q such that 3+5i/1+pi =q+4i"

    Im sure it's easy but I think I am overlooking something. I multiplied both sides by the conjugate of 1+pi....ie.(1-pi) but I think it's wrong.

    Any help would be appreciated.

    Regards,
    Neverquit
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  2. #2
    Super Member Deadstar's Avatar
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    Multiply both sides by 1+pi

    Then you get...

    3 + 5i = (1 + pi)(q + 4i) = q + pqi + 4i - 4p

    => 3 + 5i = (q - 4p) + 4i + pqi

    => 3 + i = (q - 4p) + pqi

    So you want to solve...

    q - 4p = 3

    pq = 1
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  3. #3
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    found 1 solution but not the other

    I found the solutions 0.25, -1 after re-arranging pq = 1 to q = 1/p to get 1/p - 4p = 3 to get quadratic 4p^2+3p-1=0.


    There is also apparently another solution of 4, -1 which I canít find.

    How do I find it?
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  4. #4
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    This is a tricky question.
    \frac{{3 + 5i}}{{1 + pi}} = \frac{{\left( {3 + 5i} \right)\left( {1 - pi} \right)}}{{1 + p^2 }} = \frac{{3 + 5p}}{{1 + p^2 }} + \frac{{\left( {5 - p} \right)i}}{{1 + p^2 }}
    Now set real part equal to real part and imaginary equal to imaginary.
    \frac{3+5p}{1+p^2}=q~\&~\frac{5-3p}{1+p^2}=4
    From that we get p=\frac{1}{4}~\&~p=-1.
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  5. #5
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    Quote Originally Posted by Neverquit View Post
    I found the solutions 0.25, -1 after re-arranging pq = 1 to q = 1/p to get 1/p - 4p = 3 to get quadratic 4p^2+3p-1=0.

    There is also apparently another solution of 4, -1 which I canít find.

    How do I find it?
    I think you are misinterpreting your own answer. When you solve the system of equations, you will find that

    p=1/4 ~~\mbox{  or  }~~ p=-1

    These are two separate solutions, not a single solution. You need to find the value of q that pairs with each of these solutions for p. So, you need to plug each value of p back into the system of equations and find the corresponding values of q. Since pq=1, it's a pretty straightforward calculation:

    p=1/4 \implies q=4

    p=-1 \implies q=-1


    Therefore, the solutions are:

    First solution: p=1/4 \mbox{ and } q=4

    Second solution: p=-1 \mbox{ and } q=-1


    You might also write this as:

    (p,q) = (1/4,4) \mbox{ or } (p,q) = (-1,-1)


    But you definitely would not say that the solutions are (1/4,-1) and (4,-1).
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by Neverquit View Post
    Hello,
    I am having trouble with this question:

    "Find all possible real number pairs p, q such that 3+5i/1+pi =q+4i"

    Im sure it's easy but I think I am overlooking something. I multiplied both sides by the conjugate of 1+pi....ie.(1-pi) but I think it's wrong.

    Any help would be appreciated.

    Regards,
    Neverquit
    \frac{3+5i}{1+pi} =q+4i

    Multiply top and bottom of LHS by 1-pi to give:

    \frac{(3+5p)+(5-3p)i}{1+p^2}=q+4i

    Equate real and imaginary parts to get:

    3+5p=(1+p^2)q

    and:

    5-3p=4(1+p^2)

    Now the problem is to find all solutions to this pair of equations.

    CB
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  7. #7
    Super Member Deadstar's Avatar
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    Having Plato and CP post in this thread has made me question whether my answer is wrong. Is it?

    Solving the equations I arrived at...

    q - 4p = 3

    pq = 1

    gives

    q=1/p,

    So subbing that into the first equation will give you a polynomial (if you multiply both sides by p) which gives you p = 1/4 and -1.

    Hence q = 4 and -1.

    Solutions are (1/4,4) and (-1, -1)

    Are these the only solutions? Why must you multiply the top and bottom by the conjugate?
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  8. #8
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    Quote Originally Posted by Deadstar View Post
    Having Plato and CP post in this thread has made me question whether my answer is wrong. Is it?
    Your solution was correct. Multiplying a complex fraction by the conjugate of the denominator is probably just a habit for them. But it's not necessary at all in this particular problem.
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  9. #9
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    Using the conjugate

    Quote Originally Posted by Plato View Post
    This is a tricky question.
    \frac{{3 + 5i}}{{1 + pi}} = \frac{{\left( {3 + 5i} \right)\left( {1 - pi} \right)}}{{1 + p^2 }} = \frac{{3 + 5p}}{{1 + p^2 }} + \frac{{\left( {5 - p} \right)i}}{{1 + p^2 }}
    Now set real part equal to real part and imaginary equal to imaginary.
    \frac{3+5p}{1+p^2}=q~\&~\frac{5-3p}{1+p^2}=4
    From that we get p=\frac{1}{4}~\&~p=-1.
    I think the solution that Plato gives using the congugate is what the text books author had in mind as the question is shortly after conjugates of complex numbers is explained.

    Deastar, your solution still gives the same answer in the text book so it must be correct.

    ...........
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  10. #10
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    generalisation

    Quote Originally Posted by Neverquit View Post
    Hello,
    I am having trouble with this question:

    "Find all possible real number pairs p, q such that 3+5i/1+pi =q+4i"

    Im sure it's easy but I think I am overlooking something. I multiplied both sides by the conjugate of 1+pi....ie.(1-pi) but I think it's wrong.

    Any help would be appreciated.

    Regards,
    Neverquit
    the basic idea behind questions of your type is equaslising real and imaginary parts.
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