Can someone explain to me how I find the real and the imaginary part of $e^{\theta i}$?
I'm learning complex numbers but I don't quite understand how $e$ is intertwined in all this.
$\endgroup$ 15 Answers
$\begingroup$$$z=e^{i\theta}=\cos\theta+i\sin\theta$$ $$\mathrm{Re} z=\cos\theta,\mathrm{Im} z=\sin\theta$$
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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!!!
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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.
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