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Thread: Complex Numbers in Polar Form

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    Complex Numbers in Polar Form

    I'm not sure where to start with this question at all, some help would be appreciated

    A point Z in an Argand diagram represents z = e^i \theta

    By writing z + 1 as e^(1/2i\theta) (e^(1/2i\theta) + e^(-1/2i\theta), find the modulus and argument of z+1.

    Hmm, the editing has gone badly, it should be:

    e^(itheta) in the first and
    e^(0.5itheta)(e^(0.5itheta) + e^(-0.5itheta)) in the second.
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  2. #2
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    Quote Originally Posted by alrightgeez View Post
    I'm not sure where to start with this question at all, some help would be appreciated

    A point Z in an Argand diagram represents z = e^i \theta

    By writing z + 1 as e^{(1/2)i\theta} (e^{(1/2)i\theta} + e^{-(1/2)i\theta}), find the modulus and argument of z+1.
    LaTeX tip: use braces { } rather than parentheses ( ) to enclose the superscripts.

    Write that expression for z + 1 with the factors in the opposite order: z+1 = (e^{(1/2)i\theta} + e^{-(1/2)i\theta})e^{(1/2)i\theta}. Then remember that a complex number has a unique polar form re^{i\theta} with r a positive real number. Notice also that e^{(1/2)i\theta} + e^{-(1/2)i\theta} = 2\cos\theta, which is real (though it is not always positive).
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    Quote Originally Posted by Opalg View Post
    LaTeX tip: use braces { } rather than parentheses ( ) to enclose the superscripts.

    Write that expression for z + 1 with the factors in the opposite order: z+1 = (e^{(1/2)i\theta} + e^{-(1/2)i\theta})e^{(1/2)i\theta}. Then remember that a complex number has a unique polar form re^{i\theta} with r a positive real number. Notice also that e^{(1/2)i\theta} + e^{-(1/2)i\theta} = 2\cos\theta, which is real (though it is not always positive).
    Thanks so much for that, I managed to work it out by myself in my head as I was out for a jog but thanks anyway.

    The next bit has me stuck though as it doesn't seem to cancel down correctly to the given answer.

    I'm told to use a similar method for 'z-1' (meaning e^{i\theta} - 1)

    So I write it out in the same way as: e^{1/2\theta i}(e^{1/2\theta i} - e^{-1/2\theta i})

    This is my working:

    e^{1/2\theta i}(e^{1/2\theta i} - e^{-1/2\theta i})
    = e^{1/2\theta i}(cos(1/2\theta) + isin(1/2\theta) - cos(-1/2\theta) - isin(-1/2\theta)))
    = e^{1/2\theta i}(cos(1/2\theta) + isin(1/2\theta) - cos(1/2\theta) + isin(1/2\theta)))
    = e^{1/2\theta i}(2isin(1/2\theta))
    = 2isin(1/2\theta)cos(1/2\theta) - 2sin^2(1/2\theta)

    Then try to find the modulus by squaring both terms:

    4sin^4(1/2\theta) - 4sin^2(1/2\theta)cos^2(1/2\theta)
    4sin^4(1/2\theta) - 4sin^2(1/2\theta)(1 - sin^2(1/2\theta))
    8sin^4(1/2\theta) - 4sin^2(1/2\theta)

    However the answer given in the book is 2sin(1/2\theta), which the terms above doesn't cancel to. Can anyone see where I've gone wrong?
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    Quote Originally Posted by alrightgeez View Post
    Thanks so much for that, I managed to work it out by myself in my head as I was out for a jog but thanks anyway.

    The next bit has me stuck though as it doesn't seem to cancel down correctly to the given answer.

    I'm told to use a similar method for 'z-1' (meaning e^{i\theta} - 1)

    So I write it out in the same way as: e^{1/2\theta i}(e^{1/2\theta i} - e^{-1/2\theta i})

    This is my working:

    e^{1/2\theta i}(e^{1/2\theta i} - e^{-1/2\theta i})
    = e^{1/2\theta i}(cos(1/2\theta) + isin(1/2\theta) - cos(-1/2\theta) - isin(-1/2\theta)))
    = e^{1/2\theta i}(cos(1/2\theta) + isin(1/2\theta) - cos(1/2\theta) + isin(1/2\theta)))
    = e^{1/2\theta i}(2isin(1/2\theta)) Stop right there! You're almost finished at this point.
    = 2isin(1/2\theta)cos(1/2\theta) - 2sin^2(1/2\theta)

    Then try to find the modulus by squaring both terms:

    4sin^4(1/2\theta) - 4sin^2(1/2\theta)cos^2(1/2\theta)
    4sin^4(1/2\theta) - 4sin^2(1/2\theta)(1 - sin^2(1/2\theta))
    8sin^4(1/2\theta) - 4sin^2(1/2\theta)

    However the answer given in the book is 2sin(1/2\theta), which the terms above doesn't cancel to. Can anyone see where I've gone wrong?
    The next line after "Stop right there!" should be
    = (2\sin((1/2)\theta))ie^{(1/2)\theta i}. Notice that 2\sin((1/2)\theta) is real, and ie^{(1/2)\theta i} = e^{(1/2)\pi i}e^{(1/2)\theta i} = e^{(1/2)(\pi+\theta)i}. So the modulus-argument form of z1 is 2\sin((1/2)\theta)e^{(1/2)(\pi+\theta)i}, from which you can read off the modulus as 2\sin((1/2)\theta) (provided that \theta is chosen so that 2\sin((1/2)\theta) is positive).
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