1. ## Complex Numbers

Hey guys, I'm totally stumped on this question. Any help will be appreciated.

Question:

For a real number $\displaystyle \theta$, define

$\displaystyle e^{i \theta} = \cos \theta + i \sin \theta$

Show that the identity $\displaystyle e^{i(\theta + k)} = e^{i \theta}e^{i k}$ is equivalent to the trigonometric rules:

$\displaystyle \cos (\theta + k) = \cos \theta \cos k - \sin \theta \sin k$

$\displaystyle \sin (\theta + k) = \sin \theta \cos k + \cos \theta \sin k$

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2. Originally Posted by WWTL@WHL
Hey guys, I'm totally stumped on this question. Any help will be appreciated.

Question:

For a real number $\displaystyle \theta$, define

$\displaystyle e^{i \theta} = \cos \theta + i \sin \theta$

Show that the identity $\displaystyle e^{i(\theta + k)} = e^{i \theta}e^{i k}$ is equivalent to the trigonometric rules:

$\displaystyle \cos (\theta + k) = \cos \theta \cos k - \sin \theta \sin k$

$\displaystyle \sin (\theta + k) = \sin \theta \cos k + \cos \theta \sin k$

First assume $\displaystyle e^{i(\theta + k)} = e^{i \theta}e^{i k}$ and prove

$\displaystyle \cos (\theta + k) = \cos \theta \cos k - \sin \theta \sin k$
$\displaystyle \sin (\theta + k) = \sin \theta \cos k + \cos \theta \sin k$

Idea:
$\displaystyle e^{i \theta} = \cos \theta + i \sin \theta$
And by given data we have,
$\displaystyle e^{i(\theta + k)} = e^{i \theta}e^{i k}$
$\displaystyle e^{i \theta + i k} = \cos (\theta + k) + i \sin (\theta + k)$
Now,
$\displaystyle e^{i \theta}e^{i k}=(\cos \theta + i \sin \theta)(\cos k + i \sin k )$
Multiply term by term and don't forget to use $\displaystyle i^2 = -1$
Then equate Real and Imaginary pats on both sides to get what you want.
Now assume
$\displaystyle \cos (\theta + k) = \cos \theta \cos k - \sin \theta \sin k$
$\displaystyle \sin (\theta + k) = \sin \theta \cos k + \cos \theta \sin k$
$\displaystyle e^{i(\theta + k)} = e^{i \theta}e^{i k}$