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Solve z^5 = 1 + j. Can someone help walk me through this? watch

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    z^5 = 1 + j. Just a guide on how to approach and solve this. Thanks. Please walkthrough it in baby steps i am completely new to this.
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    (Original post by MatthewTG)
    z^5 = 1 + j. Just a guide on how to approach and solve this. Thanks.
    What have you tried?
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    (Original post by MatthewTG)
    z^5 = 1 + j. Just a guide on how to approach and solve this. Thanks.
    Express both sides as complex-exponentials
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    (Original post by Mr M)
    What have you tried?
    I think the de moivre theorum is used but im not sure
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    if z = re

    then z5 = r5ei5Θ

    and 1 + i = √2e iπ/4

    so r5ei5Θ = √2e iπ/4

    now you should be able to find values for r and
    Θ
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    (Original post by MatthewTG)
    I think the de moivre theorum is used but im not sure
    What's the modulus and argument of 1 + j ?
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    (Original post by MatthewTG)
    I think the de moivre theorum is used but im not sure
    De Moivre's theorem states that (\cos\theta + i\sin \theta)^n = \cos(n\theta) + i \sin(n\theta) which doesn't really help with it. It might, though it seems a bit long winded.

    Anyhow, the best way to approach this is to express your complex number in the form re^{i\theta} and then take the 5th root of both sides - keep in mind that you should have 5 solutions as you consider the period of the solutions.
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    (Original post by Mr M)
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    Good evening, Mr M. Do you know if this is a question I could get in FP1 (OCR MEI)?
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    (Original post by karmacrunch)
    Good evening, Mr M. Do you know if this is a question I could get in FP1 (OCR MEI)?
    No you couldn't. It's FP2.
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    (Original post by Mr M)
    No you couldn't. It's FP2.
    Thank you!
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    (Original post by the bear)
    if z = re

    then z5 = r5ei5Θ

    and 1 + i = √2e iπ/4

    so r5ei5Θ = √2e iπ/4

    now you should be able to find values for r and
    Θ
    You also need to take general form of the argument into account - the argument can be altered by any multiple of 2\pi without changing the complex number. So we want:

    z^5 = \sqrt{2}e^{i(\pi/4+2n\pi)} for n \in \mathbb{Z}

    else we won't get all of the solutions - there are 5.
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    (Original post by atsruser)
    You also need to take general form of the argument into account - the argument can be altered by any multiple of 2\pi without changing the complex number. So we want:

    z^5 = \sqrt{2}e^{i(\pi/4+2n\pi)} for n \in \mathbb{Z}

    else we won't get all of the solutions - there are 5.
    i did not want to make it too easy.
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    (Original post by the bear)
    i did not want to make it too easy.
    I see that the bear is a harsh and demanding taskmaster.
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    (Original post by atsruser)
    I see that the bear is a harsh and demanding taskmaster.

    :whip2:

    also bear™ is not a huge fan of j

    :mad:
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    (Original post by the bear)
    :whip2:

    also bear™ is not a huge fan of j

    :mad:
    The standard desensitisation therapy for that ailment is to force yourself to read a large number of electronic and electrical engineering texts - in severe cases, the treatment also involves a straitjacket, eyelid separators, and a CD of Beethoven's Ninth symphony.
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    (Original post by atsruser)
    The standard desensitisation therapy for that ailment is to force yourself to read a large number of electronic and electrical engineering texts - in severe cases, the treatment also involves a straitjacket, eyelid separators, and a CD of Beethoven's Ninth symphony.
    that is just weird... like a mechanical mandarin or something ?

    :afraid:

    j = :poo:
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    (Original post by the bear)
    that is just weird... like a mechanical mandarin or something ?
    Or maybe like an automated tangerine?

    Anyway, it's late - time for a little spatchka.
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    (Original post by atsruser)
    Or maybe like an automated tangerine?

    Anyway, it's late - time for a little spatchka.
    Horrorshow

    :borat:
 
 
 
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