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Math Help - COmplex number uestion

  1. #1
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    COmplex number uestion

    Hi again,

    I need help with this question,

    Find the complex number z that satisfies the equation 50/z* - 10/z = 2 + 9i given | z | = 2sqrt(10).

    I had put 50 and 10 to the same denominator ie. zz* which then gives me |z|^2, and then I subbed in the value for |z| so my denominator was 40.
    I then brought 40 over to RHS and got 80+360i.
    The problem comes with the LHS, I am not sure how to deal with 50z-10z*..
    I thought since z+ is a conjugate, I can turn it into z by changing the sign thus 50z+10z,
    but I do not think I'm right, the answer key proved me wrong anyway.



    Thank you so much!
    J.

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  2. #2
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    Re: COmplex number uestion

    what we are given:

    |z| = 2√(10)

    50/z* - 10/z = 2 + 9i

    what i like to do is get everything over a common denominator:

    50z/(zz*) - 10z*/(zz*) = 2 + 9i

    50z - 10z* = (2 + 9i)(zz*)

    now zz* = |z|2 = (2√(10))2 = 40 so we have:

    50z - 10z* = (2 + 9i)(40)

    take out a common factor of 10:

    5z - z* = 8 + 36i

    let's write this in terms of a and b, where z = a + bi

    5(a + bi) - (a - bi) = 8 + 36i

    (5a - a) + (5b + b)i = 8 + 36i

    4a + 6bi = 8 + 36i

    compare real and imaginary parts:

    4a = 8 ---> a = 2
    6b = 36 ---> b = 6.

    so z = 2 + 6i. does this work?

    first, we check |z| = √(a2 + b2) = √(4 + 36) = √40 = √4√(10) = 2√(10). check.

    next, let's find 1/z = 1/(2 + 6i) = (2 - 6i)/(4 + 36) = (1/40)(2 - 6i)

    = (1/20)(1 - 3i).

    finally, let's find 1/z* = z/|z|2 = (1/40)(2 + 6i) = (1/20)(1 + 3i).

    so 50/z* = (50/20)(1 + 3i) = (5/2)(1 + 3i), and 10/z = (10/20)(1 - 3i) = (1/2)(1 - 3i).

    subtract the second from the first: (5/2)(1 + 3i) - (1/2)(1 - 3i) =

    (5/2 - 1/2) + (15/2 + 3/2)i = 4/2 + (18/2)i = 2 + 9i. check.
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  3. #3
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    Re: COmplex number uestion

    Thank you!

    How did you get from
    1/(2 + 6i) to (2 - 6i)/(4 + 36) ?

    Thanks loads!
    J.
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  4. #4
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    Re: COmplex number uestion

    Quote Originally Posted by Tutu View Post
    Thank you!

    How did you get from
    1/(2 + 6i) to (2 - 6i)/(4 + 36) ?

    Thanks loads!
    J.
    Multiply top and bottom by the bottom's conjugate...
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  5. #5
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    Re: COmplex number uestion

    Quote Originally Posted by Tutu View Post
    Thank you!

    How did you get from
    1/(2 + 6i) to (2 - 6i)/(4 + 36) ?

    Thanks loads!
    J.

    for any complex number a+bi:

    1/(a+bi) = (a-bi)/(a2+b2).

    how do we prove this?

    suppose xy = 1. then y = 1/x. so if we want to show y = 1/x, we can instead show xy = 1.

    so let's prove (a+bi)[(a-bi)/(a2+b2)] = 1

    the denominator in the second factor is just a real number, so we can "pull it out front":

    (a+bi)[(a-bi)/(a2+b2)] = (1/(a2+b2))[(a+bi)(a-bi)]

    = (1/(a2+b2))(a2-abi+abi-(bi)2)

    = (1/(a2+b2))(a2-b2i2)

    = ((1/(a2+b2))(a2-b2(-1))

    = (1/(a2+b2))(a2+b2) = 1.

    a shorter proof:

    zz* = |z|2 so

    zz*/|z|2 = 1, that is:

    z(z*/|z|2) = 1, so 1/z = z*/|z|2.
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