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Thread: Complex numbers

  1. #1
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    Complex numbers

    Can someone help me with this problem? I already solved a =1/2 but I'm not sure how to find the roots.
    Complex numbers-photo-9-26-18-8.48-pm.jpg
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  2. #2
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    Re: Complex numbers

    Quote Originally Posted by anomaly3 View Post
    Can someone help me with this problem? I already solved a =1/2 but I'm not sure how to find the roots.
    Click image for larger version. 

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    Why do you think that $a=\dfrac{1}{2}~?$

    What is $\left[2\exp\left(\dfrac{\pi i}{12}\right)\right]^4=~?$
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  3. #3
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    Re: Complex numbers

    Sorry, not sure how to type symbols.Complex numbers-photo-9-26-18-9.55-pm.jpg
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  4. #4
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    Re: Complex numbers

    $$(re^{i\theta})^n = r^ne^{i n \theta} = r^n(\cos (n\theta) + i \sin (n\theta))$$
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  5. #5
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    Re: Complex numbers

    In this case, how will I calculate r?
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  6. #6
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    Re: Complex numbers

    Quote Originally Posted by anomaly3 View Post
    In this case, how will I calculate r?
    @anomaly, I don't think that you understand this question.
    In any case, you did not answer my question.
    $\left[2\exp\left(\dfrac{\pi i}{12}\right)\right]^4=16\exp\left(\dfrac{\pi i}{3}\right)=16\left(\dfrac{1}{2}+i~\dfrac{\sqrt3} {2}\right)$

    I think that you should post the original question in exact wording.
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  7. #7
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    Re: Complex numbers

    Sorry that I got confused by the question and to take your time.
    I'll keep that in mind.
    Can you please take a look at my work and see if it is correct?
    Thank you very much for your help!Complex numbers-photo-9-26-18-11.21-pm.jpg
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  8. #8
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    Re: Complex numbers

    You are taking the third root, but you should be taking the fourth root of $z^4$. The first root is $z$. All four roots can be found like this:

    $$\left(16e^{\tfrac{i\pi}{3}+2ni\pi}\right)^{1/4}, n=0,1,2,3$$

    For $n=0$, you have $z=2e^{\tfrac{i\pi}{12}}$.
    For $n=1$, you have $2e^\tfrac{7i\pi}{12}$.
    For $n=2$, you have ...
    For $n=3$, you have ...

    Complete the last two.
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  9. #9
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    Re: Complex numbers

    For n=2, it's 2e13iπ12
    for n=3, it's 2e19ni12

    Edit:Complex numbers-photo-9-26-18-11.46-pm.jpg
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  10. #10
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    Re: Complex numbers

    Quote Originally Posted by anomaly3 View Post
    For n=2, it's 2e13iπ12
    for n=3, it's 2e19ni12

    Edit:Click image for larger version. 

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    That is correct. Really, there are an infinite number of values for $n$, but because both sine and cosine are $2\pi$-periodic, I restricted answers to where $0 \le \dfrac{2n\pi}{4} < 2\pi$, which is when $n=0,1,2,3$.
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  11. #11
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    Re: Complex numbers

    Thank you to both of you for taking the time to teach me! I think my understanding towards complex numbers has improved a lot. Now I can move on to the next question with a much better understanding, lol.
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