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Math Help - How do I write arccos 4 as a + ib (complex number)?

  1. #1
    Member ssadi's Avatar
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    How do I write arccos 4 as a + ib (complex number)?

    This a question from complex number chapter of fp3
    Find the following in the form a+ib, a,b member of real number set:
    (a) arccos 4 (b) arcsin 2 (c) arcsin i
    I tried to these sums by: x=arccos4, so cosx=4, so e^(ix)-e^-(ix)=8
    The answers:
    (a) My answer:+/- i arcosh 4
    Book's answer: 2m*pi +/- i arcosh 4
    (b) My answer: +/- arcosh 2(i am not sure how i get that)
    Book's answer: (4m+1)*pi/2 +/- i arcosh 2
    (c) My answer: i ln ((-2+/- sqrt 5)/2))
    Book's answer: n*pi + i arsinh [(-1)^n]

    So I gathered that I am missing something major here. Please anyone tell me the correct method to do these sums, thus enabling me to try it out myself. Just tell me the method, please This is my first sums of this kind:woo:
    P.s. i=imaginary.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by ssadi View Post
    This a question from complex number chapter of fp3
    Find the following in the form a+ib, a,b member of real number set:
    (a) arccos 4 (b) arcsin 2 (c) arcsin i
    I tried to these sums by: x=arccos4, so cosx=4, so e^(ix)-e^-(ix)=8
    The answers:
    (a) My answer:+/- i arcosh 4
    Book's answer: 2m*pi +/- i arcosh 4
    (b) My answer: +/- arcosh 2(i am not sure how i get that)
    Book's answer: (4m+1)*pi/2 +/- i arcosh 2
    (c) My answer: i ln ((-2+/- sqrt 5)/2))
    Book's answer: n*pi + i arsinh [(-1)^n]

    So I gathered that I am missing something major here. Please anyone tell me the correct method to do these sums, thus enabling me to try it out myself. Just tell me the method, please This is my first sums of this kind:woo:
    P.s. i=imaginary.
    The first thing you might want to check is in your first step it should be e^{ix}{\color{red}+}e^{-ix}=8

    Or what you might want to try is solving y=\frac{e^{ix}+e^{-ix}}{2} for y by letting e^{ix}=\varphi. But be sure to remember that e^{ix} is a many valued (many-to-one) function so there isn't going to be just one number as the answer.
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  3. #3
    Member ssadi's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    The first thing you might want to check is in your first step it should be e^{ix}{\color{red}+}e^{-ix}=8

    Or what you might want to try is solving y=\frac{e^{ix}+e^{-ix}}{2} for y by letting e^{ix}=\varphi. But be sure to remember that e^{ix} is a many valued (many-to-one) function so there isn't going to be just one number as the answer.
    oh i typed wrong here, in the sum i used e^ix + e^-ix :embarassed:
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  4. #4
    Member ssadi's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    The first thing you might want to check is in your first step it should be e^{ix}{\color{red}+}e^{-ix}=8

    Or what you might want to try is solving y=\frac{e^{ix}+e^{-ix}}{2} for y by letting e^{ix}=\varphi. But be sure to remember that e^{ix} is a many valued (many-to-one) function so there isn't going to be just one number as the answer.
    and from where the m and n comes into book's answer? I am really nonplussed.
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  5. #5
    Member ssadi's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    The first thing you might want to check is in your first step it should be e^{ix}{\color{red}+}e^{-ix}=8

    Or what you might want to try is solving y=\frac{e^{ix}+e^{-ix}}{2} for y by letting e^{ix}=\varphi. But be sure to remember that e^{ix} is a many valued (many-to-one) function so there isn't going to be just one number as the answer.
    Isn't there a general formula or something i can use to do this type of sum?
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by ssadi View Post
    and from where the m and n comes into book's answer? I am really nonplussed.
    That is because as I said, the imaginary exponential function is multi-valued, e^{ix}=\cos(x)+i\sin(x)\implies{e^{i(2\pi{n}+x)}=\  cos(x)+i\sin(x)}=e^{ix}
    Quote Originally Posted by ssadi View Post
    oh i typed wrong here, in the sum i used e^ix + e^-ix :embarassed:
    Dont worry! We all make typos!
    Quote Originally Posted by ssadi View Post
    Isn't there a general formula or something i can use to do this type of sum?
    Try my suggestion and see if it helps.

    There is no need to have three seperate posts
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  7. #7
    Member ssadi's Avatar
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    Smile

    Quote Originally Posted by Mathstud28 View Post
    That is because as I said, the imaginary exponential function is multi-valued, e^{ix}=\cos(x)+i\sin(x)\implies{e^{i(2\pi{n}+x)}=\  cos(x)+i\sin(x)}=e^{ix}

    Dont worry! We all make typos!

    Try my suggestion and see if it helps.

    There is no need to have three seperate posts
    My ideas were coming randomly, hence the array of posts
    I am not actually familiar with multi-valuation of imaginary exponential funtion, but I think i can solve the equation in the trigonometric form. But I am not sure how sin x=0 can lead to (4m+1)*pi/2 as the value of x, especially when 3/2 pi is a solution of x. Can you give me a site which addresses how to get the general formula for cos x=0, sinx =0 , sin x=c etc. I will be perfectly happy then
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  8. #8
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    The key to understanding these is understanding the multi-valued logarithm function. Until you understand that, you're going to have problems throughout Complex Analysis. Now \sin^{-1} z=-i\log\bigg[iz+\sqrt{1-z^2}\bigg]

    That formula is a big deal because both logarithm and the root object are multivalued:

    \sqrt{z}=r^{1/2}e^{i/2(\Theta_z+2k\pi)};\quad k=0,1

    \log(z)=\ln(r)+i(\Theta_z+2n\pi); n\in\mathbb{Z}

    Then:

    \begin{aligned}\sin^{-1} 2&=-i\log\bigg[2i+\sqrt{3}e^{i/2(\pi+2k\pi)}\bigg]\\&=-i\log\bigg[2i\pm i\sqrt{3}\bigg]<br />
\end{aligned}

    This then bifurcates into two infinite sets of values:

    \sin^{-1} 2=-i\bigg[\ln|2+\sqrt{3}|+i(\Theta_a+2n\pi)\bigg]

    \sin^{-1} 2=-i\bigg[\ln|2-\sqrt{3}|+i(\Theta_b+2n\pi)\bigg]

    Where in this case \Theta_a=\Theta_b=Arg(2\pm\sqrt{3}) and n\in\mathbb{Z}
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