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Math Help - Proving Equivalence of certain Complex numbers

  1. #1
    Junior Member EliteAndoy's Avatar
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    Proving Equivalence of certain Complex numbers

    Hello Everyone! Today we are asked to prove that both this expressions are true: \left | e^{\theta i } \right |=1 and \overline{e^{\theta i}}=e^{-\theta i}. To be quite honest, I am not very sure as to how I can prove these, but my attempt for the first one, I was able to prove that \sin^2(\theta)+cos^2(\theta)=1 but was unsure as to how I would be able to turn it into \sin(\theta)+cos(\theta)=1 which is what I need to prove the first equality.

    As for the second one I did something like this:

    e^{\theta i}=\cos(\theta)+i\sin(\theta)

    So that:

    \overline{e^{\theta i}}=\cos(\theta)-i\sin(\theta)

    and by properties of odd and even function:

    \overline{e^{\theta i}}=\cos(-\theta)+i\sin(-\theta)

    Thus, it should be true to say that :

    \overline{e^{\theta i}}=e^{-\theta i}

    The proof above, for me, is quite sloppy, but I think I could live with that. But for the first one, I really have no idea as to how to prove it. Anyone got some ideas as to how I should prove both equivalence? Thanks everyone in advance!
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    Re: Proving Equivalence of certain Complex numbers

    Surely you know that \displaystyle \begin{align*} \sqrt{ \sin^2{(\theta)} + \cos^2{(\theta)}} \end{align*} is NOT \displaystyle \begin{align*} \sin{(\theta)} + \cos{(\theta)} \end{align*}...
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    Re: Proving Equivalence of certain Complex numbers

    Quote Originally Posted by EliteAndoy View Post
    \left | e^{\theta i } \right |=1 and \overline{e^{\theta i}}=e^{-\theta i}. To be quite honest, I am not very sure as to how I can prove these, but my attempt for the first one, I was able to prove that \sin^2(\theta)+cos^2(\theta)=1 but was unsure as to how I would be able to turn it into \sin(\theta)+cos(\theta)=1 which is what I need to prove the first equality.
    You must know that \sqrt{a^2+b^2}\ne a+b~! and \sqrt{1}=1


    Quote Originally Posted by EliteAndoy View Post
    \overline{e^{\theta i}}=e^{-\theta i}.
    As for the second one I did something like this:
    e^{\theta i}=\cos(\theta)+i\sin(\theta)
    So that:
    \overline{e^{\theta i}}=\cos(\theta)-i\sin(\theta)
    and by properties of odd and even function:
    \overline{e^{\theta i}}=\cos(-\theta)+i\sin(-\theta)
    Thus, it should be true to say that :
    \overline{e^{\theta i}}=e^{-\theta i}
    Yours above is the standard proof for this.
    You might add that this works because cosine is an even function and sine is an odd function.
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    Junior Member EliteAndoy's Avatar
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    Re: Proving Equivalence of certain Complex numbers

    Yeah, that's absolutely true,that's why I was unable to prove the equality. Here's how my little failed proof goes:
    \left |e^{i \theta} \right |= \sin^2(\theta)+\cos^2(\theta)

    From the equation above, we can see that the modulus is 1 so that:

    \sqrt{\sin^2(\theta)+\cos^2(\theta)}=1

    For the equation above to be true, the radicand should equal to 1, so that

    \sin^2(\theta))+\cos^2(\theta)=1

    From here, I tried to play with a lot of trig identities which all I was able to do is prove that this equation is true. But yeah, like Plato and Prove it said, this is not equal to
    \sin(\theta)+\cos(\theta)=1. So, anyone here have any idea as to how I should proceed with the proof? Thanks again to both of you!
    Last edited by EliteAndoy; April 28th 2013 at 08:47 AM.
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    Re: Proving Equivalence of certain Complex numbers

    Quote Originally Posted by EliteAndoy View Post
    So, anyone here have any idea as to how I should proceed with the proof? Thanks again to both of you!

    What in the world are you on about?

    \left|e^{i\theta}\right|=\left|\cos(\theta)+i\sin(  \theta)\right|=\sqrt{\cos^2(\theta)+\sin^2(\theta)  }=\sqrt1=1

    ALL of those are equals. What don't you get?
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    Junior Member EliteAndoy's Avatar
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    Re: Proving Equivalence of certain Complex numbers

    Damn it, I forgot about that. Well thanks for verifying that.
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    Re: Proving Equivalence of certain Complex numbers

    Quote Originally Posted by EliteAndoy View Post
    it, I forgot about that. Well thanks for verifying that.
    Watch the language, please.

    -Dan
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    Junior Member EliteAndoy's Avatar
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    Re: Proving Equivalence of certain Complex numbers

    Oh sorry about that hehe.
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