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Math Help - Complex Number

  1. #1
    Senior Member slevvio's Avatar
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    Complex Number

    In order to start complex differentiation we need to understand some properties of complex numbers. However I am stuck here and would appreciate some help if possible. thanks very much

    If  0 < \theta < \frac{\pi}{4}, find the modulus and principal value of the argument of  e^{i\theta} + e^{5i \theta}.

     |e^{i\theta} + e^{5i \theta}|^2 = (\cos \theta + \cos 5 \theta)^2 + (\sin \theta + \sin 5 \theta)^2 = 2 + 2 \cos 4 \theta so the modulus is  \sqrt{ 2 + 2\cos 4 \theta} but I don't really see how to calculate the argument and how to involve the restrictions on theta. Thanks for any help
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  2. #2
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    This may not be very pleasant (simplified) but it is nonetheless fact.
    \text{Arg}(e^{i\theta }  + e^{5i\theta } ) = \arctan \left( {\frac{{\sin (\theta ) + \sin (5\theta )}}{{\cos (\theta ) + \cos (5\theta )}}} \right)
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  3. #3
    Senior Member slevvio's Avatar
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    Ahh i thought about writing that but it looked horrific ! does it matter that theta is between zero and pi/4?
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    Quote Originally Posted by slevvio View Post
    does it matter that theta is between zero and pi/4?
    We do need this to be true \cos(\theta)+\cos(5\theta)>0.
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    Quote Originally Posted by slevvio View Post
    In order to start complex differentiation we need to understand some properties of complex numbers. However I am stuck here and would appreciate some help if possible. thanks very much

    If  0 < \theta < \frac{\pi}{4}, find the modulus and principal value of the argument of  e^{i\theta} + e^{5i \theta}.

     |e^{i\theta} + e^{5i \theta}|^2 = (\cos \theta + \cos 5 \theta)^2 + (\sin \theta + \sin 5 \theta)^2 = 2 + 2 \cos 4 \theta so the modulus is  \sqrt{ 2 + 2\cos 4 \theta} but I don't really see how to calculate the argument and how to involve the restrictions on theta. Thanks for any help
    A little trigonometry comes in handy here:

    1+\cos 2\theta=2\cos^2\theta\,,\,\,so\,\,\,\sqrt{2+2\cos 4\theta}=2\cos 2\theta ...and this looks nicer, doesn't it? (**)

    As for the argument: using \sin\alpha+\sin\beta=2\sin\frac{\alpha+\beta}{2}\c  os\frac{\alpha-\beta}{2}\,,\,\,\cos\alpha+\cos\beta=2\cos\frac{\a  lpha+\beta}{2}\cos\frac{\alpha-\beta}{2} , we get:

    \sin\theta+\sin 5\theta=2\sin 3\theta\cos 2\theta

    \cos\theta+\cos 5\theta=2\cos 3\theta\cos 2\theta

    So for the argument we have \arctan\left(\frac{2\sin 3\theta\cos 2\theta}{2\cos 3\theta\cos 2\theta}\right)=\arctan\left(\tan 3\theta\tan 2\theta\right)

    Tonio
    Last edited by tonio; January 26th 2010 at 02:32 AM. Reason: Correcting a typo: the cosine isn't squared in (**)
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    Quote Originally Posted by tonio View Post
    So for the argument we have \arctan\left(\frac{2\sin 3\theta\cos 2\theta}{2\cos 3\theta\cos 2\theta}\right)=\arctan\left(\tan 3\theta\tan 2\theta\right)
    But be careful. For a principal argument using arctangent alone we need the real part to be positive.
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  7. #7
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     e^{i\theta} + e^{5i\theta}

     = e^{3i\theta}( e^{-2i\theta} + e^{2i\theta})

     = e^{3i\theta} 2 \cos(2\theta)

    the argument is  3\theta

    the modulus is  | e^{3i\theta} 2 \cos(2\theta) | = 2 |\cos(2\theta)|
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