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    Hi, I've got to evaluate (root3 - i)^4 divided by (-1 + i)^3, I've so far gotten to (-2 - 2root3) + (2 - 2root3)i, but I have to convert it to exponential form, and the question is non-calculator.
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    If your answer is supposed to be in the form r e^{i\theta}, then you need to convert \sqrt{3} - i and -1 + i into this form and then use DeMoivre. It's not difficult - they are all "standard" angles.
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    (Original post by jordanwu)
    Hi, I've got to evaluate (root3 - i)^4 divided by (-1 + i)^3, I've so far gotten to (-2 - 2root3) + (2 - 2root3)i, but I have to convert it to exponential form, and the question is non-calculator.
    Draw a diagram and convert each complex number to the form re^{i\theta}. That makes applying the powers, and dividing, a lot easier.
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    (Original post by DFranklin)
    If your answer is supposed to be in the form r e^{i\theta}, then you need to convert \sqrt{3} - i and -1 + i into this form and then use DeMoivre. It's not difficult - they are all "standard" angles.
    Ok, but I'm not sure if I've got the right exponential forms: 2e^i11/6pi and root2e^i3/4pi?
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    (Original post by jordanwu)
    Ok, but I'm not sure if I've got the right exponential forms: 2e^i11/6pi and root2e^i3/4pi?
    Yeah those are correct
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    (Original post by DFranklin)
    If your answer is supposed to be in the form r e^{i\theta}, then you need to convert \sqrt{3} - i and -1 + i into this form and then use DeMoivre. It's not difficult - they are all "standard" angles.
    Not too sure about how to use that on the exponential form
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    (Original post by jordanwu)
    Not too sure about how to use that on the exponential form
    Well, if \sqrt{3}-i=2e^{-\frac{1}{6}\pi i} as you found, then (\sqrt{3}-i)^4=....?
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    (Original post by RDKGames)
    Well, if \sqrt{3}-i=2e^{-\frac{1}{6}\pi i} as you found, then (\sqrt{3}-i)^4=....?
    Oh it's supposed to be -11/6pi? I've got 11/6pi
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    (Original post by jordanwu)
    Oh it's supposed to be -11/6pi? I've got 11/6pi
    if you have w = re

    then wk = rkeikΘ
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    (Original post by jordanwu)
    Oh it's supposed to be -11/6pi? I've got 11/6pi
    e^{\frac{11}{6}\pi i}=e^{-\frac{1}{6}\pi i}
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    (Original post by RDKGames)
    e^{\frac{11}{6}\pi i}=e^{-\frac{1}{6}\pi i}
    Anywhere I can look that rule up? I've definitely forgotten it. So I have to do 2^4 and (e^-1/6pi*i)^4? And same with the exponential on the denominator?
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    (Original post by jordanwu)
    Anywhere I can look that rule up? I've definitely forgotten it. So I have to do 2^4 and (e^-1/6pi*i)^4? And same with the exponential on the denominator?
    You can remove (or add) any number of integer multiples of 2\pi, as it's removing complete revolutions from the angle.
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    (Original post by jordanwu)
    Anywhere I can look that rule up? I've definitely forgotten it. So I have to do 2^4 and (e^-1/6pi*i)^4? And same with the exponential on the denominator?
    An exponential form complex number e^{i \theta} repeats itself every 2\pi so e^{i \theta}=e^{i (2\pi k + \theta)} for n \in \mathbb{Z}

    Pick \theta= \frac{11}{6} \pi and k=-1 and you arrive at the result I've shown. However, it is much more intuitive to mark down the complex number \sqrt{3}-i on the complex plane, draw a line to it (its modulus), then notice how your angle is simply measured anticlockwise from \mathbb{R}_{>0} while my angle is just measuring backwards (clockwise, hence the negative) to get my solution.


    Anywho, yes you simply exponentiate these numbers to 4, then repeat similarly for the other complex number. Then you're simply in a position where you have \displaystyle \frac{re^{i \theta}}{Re^{i \phi}} which should be straight-forward to evaluate.
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    (Original post by RDKGames)
    An exponential form complex number e^{i \theta} repeats itself every 2\pi so e^{i \theta}=e^{i (2\pi k + \theta)} for n \in \mathbb{Z}

    Pick \theta= \frac{11}{6} \pi and k=-1 and you arrive at the result I've shown. However, it is much more intuitive to mark down the complex number \sqrt{3}-i on the complex plane, draw a line to it (its modulus), then notice how your angle is simply measured anticlockwise from \mathbb{R}_{>0} while my angle is just measuring backwards (clockwise, hence the negative) to get my solution.


    Anywho, yes you simply exponentiate these numbers to 4, then repeat similarly for the other complex number. Then you're simply in a position where you have \displaystyle \frac{re^{i \theta}}{Re^{i \phi}} which should be straight-forward to evaluate.
    I'm not sure how to do (e^(11/6pi)i)^4
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    (Original post by jordanwu)
    I'm not sure how to do (e^(11/6pi)i)^4
    (e^a)^b = e^{ab}...
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    (Original post by DFranklin)
    (e^a)^b = e^{ab}...
    e^(22/3*pi^4*i^4)?
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    (Original post by jordanwu)
    e^(22/3*pi^4*i^4)?
    Ah dear, someone's forgot a GCSE rule.

    You have z=(\sqrt{3}-i)^4=(2e^{\frac{11}{6}\pi i})^4

    So... since (e^a)^b=e^{ab}, you simply have z=2^4e^{\frac{44}{6}\pi i}
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    (Original post by jordanwu)
    e^(22/3*pi^4*i^4)?
    No. Compare (e^{(11/6\pi)i})^4 with the form (e^a)^b.

    What is a here?
    What is b here?

    So, then what is ab?

    So what is e^{ab}?
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    (Original post by RDKGames)
    Ah dear, someone's forgot a GCSE rule.

    You have z=(\sqrt{3}-i)^4=(e^{\frac{11}{6}\pi i})^4

    So... since (e^a)^b=e^{ab}, you simply have z=e^{\frac{44}{6}\pi i}
    Wow, what's wrong with me... don't even know how I got that. Ok, 16e^44/6pi*i / 2root2e^9/4pi*i?
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    (Original post by jordanwu)
    Wow, what's wrong with me... don't even know how I got that. Ok, 16e^44/6pi*i / 2root2e^9/4pi*i?
    Yeah
 
 
 
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