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  1. #1
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    complex numbers question

    hi guys, this is my first post.

    wondered is you could help me with these two questions, its the methods i need more than the answers

    1) the complex numbers z and w satisfy the simultaneous equations

    2z + iw = -1
    z - w = 3 + 3i

    a) use algebra to find z, giving your answer in the form a+ib, where a and b are real

    b) carculate arg z, giving your answer in radians to 2dp.

    2) the complex numbers z and w are given by

    z = A / (1-i)
    w = B / (1-3i)

    where A and B are real numbers. given that z + w = i

    a) find the value of A and the value of B

    b) for these values of A and B, find tan[arg(w-z)]
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  2. #2
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    Quote Originally Posted by djr8793 View Post
    hi guys, this is my first post.

    wondered is you could help me with these two questions, its the methods i need more than the answers

    1) the complex numbers z and w satisfy the simultaneous equations

    2z + iw = -1
    z - w = 3 + 3i

    a) use algebra to find z, giving your answer in the form a+ib, where a and b are real
    1. 2z + iw = -1
     z - w = 3 + 3i.

    From equation 2 we can see that w = z - 3 - 3i.

    Substituting into equation 1 gives

    2z + i(z - 3 - 3i) = -1

    2z + iz - 3i + 3 = -1

    (2 + i)z = -4 + 3i

    z = \frac{-4 + 3i}{2 + i}

    z = \frac{(-4 + 3i)(2 - i)}{(2 + i)(2 - i)}

    z = \frac{-8 + 4i + 6i + 3}{4 + 1}

    z = \frac{-5 + 10i}{5}

    z = -1 + 2i.


    To find \arg{z}, notice that \arg{z} is in the second quadrant.

    So work out \theta for the same complex number in the first quadrant and then use symmetry:

    \theta = \arctan{\frac{2}{1}}

     = \arctan{2}.


    So this means that \arg{z} = \pi - \arctan{2}.
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    Quote Originally Posted by djr8793 View Post
    2) the complex numbers z and w are given by

    z = A / (1-i)
    w = B / (1-3i)

    where A and B are real numbers. given that z + w = i

    a) find the value of A and the value of B

    b) for these values of A and B, find tan[arg(w-z)]
    z + w = i, so

    \frac{A}{1 - i} + \frac{B}{1 - 3i} = i

    \frac{A(1 + i)}{(1 - i)(1 + i)} + \frac{B(1 + 3i)}{(1 - 3i)(1 + 3i)} = 0 + i

    \frac{A + Ai}{2} + \frac{B + 3Bi}{10} = 0 + i

    \frac{5A + 5Ai + B + 3Bi}{10} = 0 + i

    5A + B + 5Ai + 3Bi = 10(0 + i)

    5A + B + (5A + 3B)i = 0 + 10i.


    Now equate real and complex coefficients:

    5A + B = 0 and 5A + 3B = 10.


    From equation 1 we can see B = -5A.

    Substituting into equation 2 gives:

    5A + 3(-5A) = 10

    5A - 15A = 10

    -10A = 10

    A = -1.


    And since B = -5A

    B = -5(-1)

    B = 5.



    b) z - w = -\frac{1}{1 - i} - \frac{5}{1 - 3i}

     = -\frac{1(1 + i)}{(1 - i)(1 + i)} - \frac{5(1 + 3i)}{(1 - 3i)(1 + 3i)}

     = \frac{-1 - i}{2} + \frac{-5 - 15i}{10}

     = \frac{-5 - 5i - 5 - 15i}{10}

     = \frac{-10 - 20i}{10}

     = -1 - 2i.


    Working out angle \theta in the first quadrant gives:

    \theta = \arctan{\frac{2}{1}}

    \theta = \arctan{2}.


    And since \arg{(z - w)} will be in the third quadrant, and tangent is positive in the third quadrant, this means

    \tan{\arg{(z - w)}} = \tan{\theta}

     = \tan{\arctan{2}}

     = 2.
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    thanks prove it, i understannd it all now
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