imaginary part

**Random Variable** · October 22nd 2011, 10:43 AM

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?

**SammyS** · October 22nd 2011, 11:53 AM

Originally Posted by Random Variable

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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