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Math Help - anybody with any idea please help!!!

  1. #1
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    Post anybody with any idea please help!!!

    i have 3 questions if anyone can help it would be greatly appreciated.

    1) Have to use de Moivre's theorem to calculate
    (1+√3i)3 / (1-i) 4
    Expressing the answer in z=a+bi form.

    I have found both the polar forms for them which could possibly be wrong, its just that get confused when i convert it to a cartesian. I need to calrify my first answer and see how to do it correctly!

    2) If z= cosŲ + i sinŲ, show that
    cos nŲ = ½ (zn + 1/n)
    sin nŲ = ½i (zn + 1/zn)

    3) Prove that sin5Ų + sinŲ = 2 sin 3Ų cosŲ


    with the other two question, i pretty much don't have a clue on what to do!!!
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  2. #2
    MHF Contributor kalagota's Avatar
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    Taguig City, Philippines
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    Quote Originally Posted by softdrink_jr View Post
    i have 3 questions if anyone can help it would be greatly appreciated.

    1) Have to use de Moivre's theorem to calculate
    (1+√3i)3 / (1-i) 4
    Expressing the answer in z=a+bi form.

    I have found both the polar forms for them which could possibly be wrong, its just that get confused when i convert it to a cartesian. I need to calrify my first answer and see how to do it correctly!
    so you have

    \frac{(1 + \sqrt3 i)^3}{(1 - i)^4}

    can you convert it to the form z=r \, cis\theta

    i mean the numerator without the power and the denominator also without the power, then apply de moivre's theorem..
    for the numerator part:
    \theta = \frac{\pi}{3} and r=2
    for the denominator:
    \theta = \frac{7\pi}{4} \, \, and \, \, r=\sqrt 2

    then apply the theorem on each:
    (r cis\theta)^n = r^n cis n\theta
    also, apply the laws of addition and multiplication/division of complex numbers..


    Quote Originally Posted by softdrink_jr View Post
    2) If z= cosŲ + i sinŲ, show that
    cos nŲ = ½ (zn + 1/n)
    sin nŲ = ½i (zn + 1/zn)
    hmm, is that z^n

    Quote Originally Posted by softdrink_jr View Post
    3) Prove that sin5Ų + sinŲ = 2 sin 3Ų cosŲ


    with the other two question, i pretty much don't have a clue on what to do!!!
    \sin 5\theta + \sin \theta = \sin (4\theta + \theta) + \sin \theta
     = \sin 4\theta \cos \theta + \sin \theta \cos 4\theta + \sin \theta
     = 2\sin 2\theta \cos 2\theta \cos \theta + \sin \theta (\cos^2 2\theta - \sin^2 2\theta) + \sin \theta
     = 2\sin 2\theta \cos 2\theta \cos \theta + \sin \theta (\cos^2 2\theta - \sin^2 2\theta + 1)
     = 2\sin 2\theta \cos 2\theta \cos \theta + \sin \theta (2\cos^2 2\theta)
     = 2\sin 2\theta \cos 2\theta \cos \theta + 2\sin \theta \cos^2 2\theta
     = 2\cos 2\theta (\sin 2\theta \cos \theta + \sin \theta \cos 2\theta)
     = 2\cos 2\theta \sin 3\theta

    aww, i got i wrong conclusion..
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  3. #3
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    correction

    hmm, is that z^n

    yeh thats what its meant to be z^n!!!
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  4. #4
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    Can u please show how to work it out proply using the laws???
    [quote=kalagota;81295]so you have

    \frac{(1 + \sqrt3 i)^3}{(1 - i)^4}

    can you convert it to the form z=r \, cis\theta

    i mean the numerator without the power and the denominator also without the power, then apply de moivre's theorem..
    for the numerator part:
    \theta = \frac{\pi}{3} and r=2
    for the denominator:
    \theta = \frac{7\pi}{4} \, \, and \, \, r=\sqrt 2

    then apply the theorem on each:
    (r cis\theta)^n = r^n cis n\theta
    also, apply the laws of addition and multiplication/division of complex numbers..
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  5. #5
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    Quote Originally Posted by softdrink_jr View Post
    i have 3 questions if anyone can help it would be greatly appreciated.

    1) Have to use de Moivre's theorem to calculate
    (1+√3i)3 / (1-i) 4
    Expressing the answer in z=a+bi form.

    I have found both the polar forms for them which could possibly be wrong, its just that get confused when i convert it to a cartesian. I need to calrify my first answer and see how to do it correctly!
    okay, so
    numerator:
    z_1 = x + yi = 1 + \sqrt3 i
     r = \sqrt{x^2 + y^2} = 2
     \theta = \tan^{-1} \frac{y}{x} = \tan^{-1} \sqrt3
     \theta = \frac{\pi}{3} \, (why?)
    therefore
    z_1 = 2 \, cis \frac{\pi}{3}
    so that (by De Moivre's)
    (z_1)^3 = 8 \, cis \pi

    denominator:
    z_2 = x + yi = 1 - i
     r = \sqrt{x^2 + y^2} = \sqrt2
     \theta = \tan^{-1} \frac{y}{x} = \tan^{-1} (-1)
     \theta = \frac{-\pi}{4} \, (why?)
    therefore
    z_2 = \sqrt2 \, cis {(-\pi)}
    so that (by De Moivre's)
    (z_2)^4 = 4 \, cis (-4\pi)

    now

    \frac{(z_1)^3}{(z_2)^4} = \frac{8 \, cis \pi}{4 \, cis -4\pi} = \frac{2(\cos \pi + i \sin \pi)}{\cos (-4\pi) + i\sin (-4\pi)} = \frac{2(-1 + i (0))}{1 + 0} = -2


    have you checked number 3? i had a different conclusion..
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