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Math Help - Complex Numbers Help - Again (Sorry!)

  1. #1
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    Complex Numbers Help again - Just 1 more question (Sorry!)

    Sorry to bother you again so soon but this is another question I became stuck on.

    ===========================================

    Consider the complex number:

    z = \frac {\left(\cos \frac {\pi}{4} - isin \frac {\pi}{4}\right)^2 \left(\cos \frac {\pi}{3} + isin \frac {\pi}{3}\right)^3}{\left(\cos \frac {\pi}{24} - isin \frac {\pi}{24}\right)^4}


    Find the |z| and arg z

    ===========================================

    Man, that LaTeX thing is hard to use!

    Anyway this is what I have done so far:

    z = \frac {\left(\cos \frac {\pi}{4} - isin \frac {\pi}{4}\right)^2 \left(\cos \frac {\pi}{3} + isin \frac {\pi}{3}\right)^3}{\left(\cos \frac {\pi}{24} - isin \frac {\pi}{24}\right)^4}


    z = \frac {\left(\frac {\sqrt 2}{2} - \frac {\sqrt 2}{2}i\right)^2 \left(\frac {1}{2} + \frac {\sqrt 3}{2}i\right)^3}{\left(\cos \frac {\pi}{24} - isin \frac {\pi}{24}\right)^4}


    No idea what \frac {\pi}{24} is so I can't continue...


    Maybe this is the wrong way - perhaps instead of tranforming them into numbers, I use cis instead:


    \left(\cos \frac {\pi}{3} + isin \frac {\pi}{3}\right)^3 becomes cis \frac {\pi}{3} and from there, use cis properties...

    Anyway, this question has got me completely confused and so I'm finding it rather difficult to continue.

    ----------------------------------------------------

    Could somebody please help me please? Thanks - all help is, of course, appreciated.

    PS: How do you guys manage to type up that LaTeX thing smoothly? It takes me ages to get it right! (my first time!)
    Last edited by sqleung; July 3rd 2008 at 02:48 AM.
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  2. #2
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    Hi again

    Quote Originally Posted by sqleung View Post
    Sorry to bother you again so soon but this is another question I became stuck on.
    No problem for that, it's enjoyable to read what you have done

    Consider the complex number:

    z = \frac {\left(\cos \frac {\pi}{4} + isin \frac {\pi}{4}\right)^2 \left(\cos \frac {\pi}{3} + isin \frac {\pi}{3}\right)^3}{\left(\cos \frac {\pi}{24} + isin \frac {\pi}{24}\right)^4}



    z = \frac {\left(\cos \frac {\pi}{4} - isin \frac {\pi}{4}\right)^2 \left(\cos \frac {\pi}{3} + isin \frac {\pi}{3}\right)^3}{\left(\cos \frac {\pi}{24} - isin \frac {\pi}{24}\right)^4}
    Which one is it ?
    There are - and + sign that differ...



    For the resolution, remember : \boxed{\cos x+i \sin x=e^{ix}}

    Similarly, you can show that \boxed{\cos x{\color{red}-}i \sin x=e^{{\color{red}-}ix}}
    This is why I ask you to clarify whether it is - or +

    PS: How do you guys manage to type up that LaTeX thing smoothly? It takes me ages to get it right! (my first time!)
    well, I would say that this one would have been a pain for me to write... ^^
    It's pretty well written for a first time !
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    Oh, sorry! I mistyped the question! It should be:

    z = \frac {\left(\cos \frac {\pi}{4} - isin \frac {\pi}{4}\right)^2 \left(\cos \frac {\pi}{3} + isin \frac {\pi}{3}\right)^3}{\left(\cos \frac {\pi}{24} - isin \frac {\pi}{24}\right)^4}

    I edited my original post - thanks for correcting me.
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    Quote Originally Posted by sqleung View Post
    Oh, sorry! I mistyped the question! It should be:

    z = \frac {\left(\cos \frac {\pi}{4} - isin \frac {\pi}{4}\right)^2 \left(\cos \frac {\pi}{3} + isin \frac {\pi}{3}\right)^3}{\left(\cos \frac {\pi}{24} - isin \frac {\pi}{24}\right)^4}

    I edited my original post - thanks for correcting me.
    Ok, so like I said, replace with the exponential form :

    \cos \frac {\pi}{4} - isin \frac {\pi}{4}=e^{-i\frac{\pi}4}

    \cos \frac {\pi}{24} - isin \frac {\pi}{24}=e^{-i\frac{\pi}{24}}

    \cos \frac {\pi}{3} + isin \frac {\pi}{3}=e^{i\frac{\pi}{3}}


    Then, the rules of exponents... :

    (e^a)^b=e^{ab}

    e^a e^b=e^{a+b}

    \tfrac{e^a}{e^b}=e^{a-b}


    \boxed{z=|z|e^{i arg(z)}}
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  5. #5
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    Quote Originally Posted by sqleung View Post
    Anyway this is what I have done so far:

    z = \frac {\left(\cos \frac {\pi}{4} - isin \frac {\pi}{4}\right)^2 \left(\cos \frac {\pi}{3} + isin \frac {\pi}{3}\right)^3}{\left(\cos \frac {\pi}{24} - isin \frac {\pi}{24}\right)^4}

    z = \frac {\left(\frac {\sqrt 2}{2} - \frac {\sqrt 2}{2}i\right)^2 \left(\frac {1}{2} + \frac {\sqrt 3}{2}i\right)^3}{\left(\cos \frac {\pi}{24} - isin \frac {\pi}{24}\right)^4}


    No idea what \frac {\pi}{24} is so I can't continue...
    (\cos \theta + i \sin \theta)^n \equiv \cos n \theta + i \sin n\theta
    Last edited by Simplicity; July 3rd 2008 at 01:40 AM. Reason: Fixing Typo
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    Quote Originally Posted by Air View Post
    (\cos \theta + i \sin \theta)^n \equiv \cos n \theta + i \sin {\color{red}n} \theta
    Typo
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    Quote Originally Posted by sqleung View Post
    Oh, sorry! I mistyped the question! It should be:

    z = \frac {\left(\cos \frac {\pi}{4} - isin \frac {\pi}{4}\right)^2 \left(\cos \frac {\pi}{3} + isin \frac {\pi}{3}\right)^3}{\left(\cos \frac {\pi}{24} - isin \frac {\pi}{24}\right)^4}

    I edited my original post - thanks for correcting me.
    z = \frac{ \left[ \text{cis} \left( -\frac{\pi}{4}\right) \right]^2 \left[ \text{cis} \left(\frac{\pi}{3}\right) \right]^3}{\left[ \text{cis} \left(-\frac{\pi}{24}\right) \right]^4}


    Apply de Mooivre's Theorem:


    = \frac{\text{cis} \left[ 2\left( -\frac{\pi}{4}\right) \right]\, \text{cis} \left[3 \left(\frac{\pi}{3}\right)\right]}{\text{cis} \left[ 4 \left(- \frac{\pi}{24}\right)\right]} = \frac{\text{cis} \left( -\frac{\pi}{2}\right) \, \text{cis} (\pi)}{\text{cis} \left(- \frac{\pi}{6}\right)}


    Apply the usual theorems for multiplying and dividing complex numbers expressed in polar form:


    = \text{cis} \left( -\frac{\pi}{2} + \pi - \left[-\frac{\pi}{6} \right]\right) = \text{cis} \left( \frac{2\pi}{3} \right).


    And of course you can easily express this in Cartesian form if desired.
    Last edited by mr fantastic; July 3rd 2008 at 03:00 AM. Reason: Had a 6 instead of a 4 as the power
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    Thank you all for your help - you guys are truly amazing!
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    There's another part to this question:

    Using De Moivre's theorem, show that z = \sqrt [3]{1}

    ----------------------

    Also, Mr Fantastic, thank you for your answer but there's still something I don't quite understand - how did you get:

    \left(\cos \frac {\pi}{24} - isin \frac {\pi}{24}\right)^4 to become \left[cis \frac {\pi}{4} \right]^6

    Your final answer is cis \frac {\pi}{4} but I seem to have gotten cis \frac {5\pi}{6}

    Is there something I'm not understanding?

    ----------------------

    Also, how do I show z = \sqrt [3]{1}?

    How can I apply De Moivre's theorem into this?

    Thank you - just this last one!
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  10. #10
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    Quote Originally Posted by sqleung View Post
    [snip]
    Also, Mr Fantastic, thank you for your answer but there's still something I don't quite understand - how did you get:

    \left(\cos \frac {\pi}{24} - isin \frac {\pi}{24}\right)^4 to become \left[cis \frac {\pi}{4} \right]^6

    Your final answer is cis \frac {\pi}{4} but I seem to have gotten cis \frac {5\pi}{6}

    Is there something I'm not understanding?

    [snip]
    *Sigh* I got that because I'm a blockhead. I've edited my reply.

    The final answer should be z = \text{cis} \left( \frac{2 \pi}{3} \right).
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  11. #11
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    Quote Originally Posted by sqleung View Post
    Also, Mr Fantastic, thank you for your answer but there's still something I don't quite understand - how did you get:

    \left(\cos \frac {\pi}{24} - isin \frac {\pi}{24}\right)^4 to become \left[cis \frac {\pi}{4} \right]^6

    Your final answer is cis \frac {\pi}{4} but I seem to have gotten cis \frac {5\pi}{6}

    Is there something I'm not understanding?
    What can be not understood is why he made this mistake

    z=\frac{[cis(-\frac{\pi}{4})]^2 [cis(\frac{\pi}{3})]^3}{[cis(-\frac{\pi}{24})]^4}

    z=\frac{cis(-\frac{\pi}{2}) \cdot cis(\pi)}{cis(-\frac{\pi}{6})}

    z=cis\left(-\frac{\pi}{2}+\pi+\frac{\pi}6\right)=cis \left(\frac{2\pi}{3}\right)


    @Mr F : still not, because it's -\frac{\pi}{24} in the denominator
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    Quote Originally Posted by sqleung View Post
    There's another part to this question:

    Using De Moivre's theorem, show that z = \sqrt [3]{1}

    [snip] how do I show z = \sqrt [3]{1}?

    How can I apply De Moivre's theorem into this?

    Thank you - just this last one!
    z = \text{cis} \left( \frac{2 \pi}{3} \right) \Rightarrow z^3 = \left[ \text{cis} \left( \frac{2 \pi}{3} \right) \right]^3 = \text{cis} \left( 3 \left( \frac{2 \pi}{3} \right) \right) = \text{cis} (2 \pi) = 1.
    Last edited by mr fantastic; July 3rd 2008 at 03:13 AM. Reason: The blockhead strikes again.
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    Quote Originally Posted by Moo View Post
    What can be not understood is why he made this mistake

    z=\frac{[cis(-\frac{\pi}{4})]^2 [cis(\frac{\pi}{3})]^3}{[cis(-\frac{\pi}{24})]^4}

    z=\frac{cis(-\frac{\pi}{2}) \cdot cis(\pi)}{cis(-\frac{\pi}{6})}

    z=cis\left(-\frac{\pi}{2}+\pi+\frac{\pi}6\right)=cis \left(\frac{2\pi}{3}\right)


    @Mr F : still not, because it's -\frac{\pi}{24} in the denominator
    I fixed it while you were busy laughing at me


    (Actually I know how much you love finding my mistakes so I left it as a gift for you (not quite a gold necklace).

    )
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    Hehe, thank you very much for your help! I know I can always count on you guys
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