# Thread: Complex Numbers

1. ## Complex Numbers

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

Question:

For a real number $\theta$, define

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

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

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

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

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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 $\theta$, define

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

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

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

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

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

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

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

Idea:
This is simple. Write the LHS using the definition, then substitute the sine and cosine equations. Then factorize and show it's the exponential product!

3. Terrific stuff, Isomorphism. Thank you so much!