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Math Help - imaginary part

  1. #1
    Super Member Random Variable's Avatar
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    imaginary part

    I'm having a hard time seeing why the following is true:

     \text{Im} \  i (1+e^{i\theta})^{\alpha} e^{i(\beta+1)\theta} = \Big(2 \cos \frac{\theta}{2} \Big)^{\alpha} \cos \Big( \big(\beta + \frac{a}{2} +1 \big) \theta \Big) \ \ \alpha, \beta \ge 0



     i (1+e^{i\theta})^{\alpha} e^{i(\beta+1)\theta} = (1+\cos \theta + i \sin \theta)^{\alpha} \Big(i\cos (\beta+1) \theta - \sin (\beta+1) \theta \Big)

     = \Big(2 \cos^{2} \frac{\theta}{2} + i \sin \theta \Big)^{\alpha} \Big(i\cos (\beta+1) \theta - \sin (\beta+1) \theta \Big)

    Now what?
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  2. #2
    Senior Member
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    Clarksville, ARk
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    Re: imaginary part

    Quote Originally Posted by Random Variable View Post
    I'm having a hard time seeing why the following is true:

     \text{Im} \  i (1+e^{i\theta})^{\alpha} e^{i(\beta+1)\theta} = \Big(2 \cos \frac{\theta}{2} \Big)^{\alpha} \cos \Big( \big(\beta + \frac{a}{2} +1 \big) \theta \Big) \ \ \alpha, \beta \ge 0

     i (1+e^{i\theta})^{\alpha} e^{i(\beta+1)\theta} = (1+\cos \theta + i \sin \theta)^{\alpha} \Big(i\cos (\beta+1) \theta - \sin (\beta+1) \theta \Big)

     = \Big(2 \cos^{2} \frac{\theta}{2} + i \sin \theta \Big)^{\alpha} \Big(i\cos (\beta+1) \theta - \sin (\beta+1) \theta \Big)

    Now what?
    First, notice that Im(iz) = Re(z). So drop the leading i, & find the real part of the rest.

    Look at (1+e^{i\theta})^{\alpha}\,. The difficult part of this is the exponentiation.

    (1+e^{i\theta})=1+\cos(\theta)+i\sin(\theta)
    =e^{\displaystyle \ln\left(\sqrt{2(1+\cos(\theta))}\right)}e^{ \displaystyle i\tan^{-1}\left(\frac{\sin(\theta)}{1+\cos(\theta)}\right)  }

    =e^{\displaystyle \ln\left(2\cos\left(\frac{\theta}{2}\right)\right)  }e^{ \displaystyle i\tan^{-1}\left(\tan\left(\frac{\theta}{2}\right)\right)}

    =\left(2\cos\left(\frac{\theta}{2}\right)\right)e^  { \displaystyle i\left(\frac{\theta}{2}\right)}
    Take this to the α power.

    Then combine the imaginary exponent with that of the other factor.
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