Finding Real and Imaginary Part of Exponential Function

$$z=e^{i\theta}=\cos\theta+i\sin\theta$$ $$\mathrm{Re} z=\cos\theta,\mathrm{Im} z=\sin\theta$$

If your definition of $e^{i\theta}$ is the power series, use $i^{2k} = (-1)^k$ to $i^{2k+1}=i(-1)^k$ and split the summands into real and imaginary part. Have a close look at them and you will notice that you just wrote down the power series of cosine and sine.

If you expand $e^{i\theta}$ in a power series, you see that

$e^{i\theta} = \sum_0^\infty \dfrac{(i\theta)^n}{n!} = 1 + i\theta - \dfrac{\theta^2}{2!} - i\dfrac{\theta^3}{3!} + \dfrac{\theta^4}{4!} + . . . , \tag{1}$

and a careful inspection of the real and imaginary terms show that (1) is in fact equivalent to

$e^{i\theta} = \cos \theta + i \sin \theta.\tag{2}$

And there it is!

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

You seem to not know what the complex exponential function is. I'm sure you know that any complex number $z$ can be written as $z=cos\theta + isin\theta$.

What would you get if you were to take the derivative of this function?

$$\frac{d}{d\theta}(\cos\theta+i\sin\theta) = -\sin\theta + i\cos\theta$$

Looking at this, you know it is possible to write as $i(\cos\theta + i\sin\theta)$, correct?

Now, knowledge of elementary calculus will tell you that the only solution to $\frac{dz}{d\theta}=iz$ is $z=e^{i\theta}$

So, we can define the complex exponential as being:

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

So, the polar form of any complex number z can be written as:

$$z=re^{i\theta}$$

Using this new definition, you should be able to answer your question. Comment if you need any further help.

For $A e^{i\theta}$, where $i=\sqrt{-1}$, and $A,\theta \in \mathbb{R}$, the real part is given by $\operatorname{Re}(Ae^{i\theta}) = A \cdot \cos \theta$ and the imagniary part by $\operatorname{Im}(Ae^{i\theta}) = A \cdot \sin \theta$.

You can derive the relation $e^{i\theta} = \cos \theta + i \cdot \sin \theta$ using Taylor series expansions.