How exactly does Euler notation work for complex numbers?

Question

When textbooks say $e^{i\theta} = \cos\theta + i\sin\theta$, is the $e$ actually Euler's number, 2.71828..., or is this purely a symbolic equivalent notation where the $e$ does not represent this number? If it is the actual number, why is it used rather than another, more intuitive number? Does $\theta$ have to be measured in radians or can it be measured in either radians or degrees? Does the formula work when an angle is larger than 2$\pi$ and goes around the circle over and over? If so, why do the larger values of $\theta$ in the exponent not increase the distance the point is from the origin?

Answer 1

$e$ is the same $e$ you see in natural logarithm bases and whilst $\theta$ should ideally be between $0$ and $2\pi$ for basic examples, you can choose larger or smaller values but radians must be used.

There are a few nice proofs of this formula to be found by doing a quick search online.

The larger (or smaller) values of $\theta$ do not affect the distance from the origin because both $\sin{\theta}$ and $\cos{\theta}$ are oscillating (periodic) functions so the $x$ and $y$ or real and complex parts will always remain in the range $[-1,1]$.

More specifically, the identity $\sin^{2}{\theta}+\cos^{2}{\theta}=1$ holds true for all values of $\theta$ so the point will always be a distance of 1 from the origin unless a dilation factor $r$ is applied to make $r\times e^{i\theta}=r(\cos{\theta}+i\sin{\theta})$ in which case the point will always be at a distance of $r$ from the origin.

Answer 2

Yes, $e$ is the same value of $e$ that is called Euler's number.

There isn't any "intuitive" reason I can think of why another number would work in this equation. What kind of number could be the "intuitive" one to put in this equation?

In fact, the logarithm with base $e$ is called the natural logarithm because the powers of $e$ have particularly nice properties. You might even say that $e$ is the "natural" base for exponentiation, since exponentiation and logarithms are so closely related. In that sense, $e$ is the most "intuitive" number to find raised to a power in an equation, if you had to make a wild guess as to what number to raise to a power.

Yes, you need to assume $\theta$ is in radians when using the formula $e^{i\theta} = \cos\theta + i\sin\theta$, including the assumption that both of the trig functions expect their input values to be in radians. That's because $e^{i\pi} = \cos\pi+ i\sin\pi = -1$ whereas $e^{i180} \neq -1$.

No, you don't get any farther from the origin if you use values of $\theta$ greater than $2\pi$, because, as you say, the result "goes around the circle over and over" and no matter how far you go around the circle you are still somewhere on the circle and all point on the circle are the same distance from the origin.

Answer 3

To directly addresses the last question:

If so, why do the larger values of θ in the exponent not increase the distance the point is from the origin?

Because that's how exponentiation works with complex arguments, when you change the complex part of the number (and as $\theta$ is multiplied by $i$ that is what happens here, for $\theta\in\mathbb R$) you just go around, to get further from (or closer to) the origin you have to change the real part of argument.

Answer 4

More likely, this 'Euler's identity' is just a notation in complex analysis. I personally would not put this in real analysis branch. In complex world, you should forget most of the stuff you have learnt in real case. The $e^z$ here is defined similar to the real's definition, using Euler's limit definition, but with complex limit (at least when Im learning complex analysis)$$\lim_{n\to+\infty}\left(1+\dfrac{z}{n}\right)^n=e^z$$ And the complex trigonometric function is defined as $$\cos z=\dfrac{e^{iz}+e^{-iz}}{2}\quad\sin z=\dfrac{e^{iz}-e^{-iz}}{2i}$$ One preliminary result is that they coincide with the real definition, so we are happy with that. You can try to express $e^z$ with the use of $\cos z$ and $\sin z$, which you should get the following results $$\cos z+i\sin z=e^{iz}$$ This $z$ here can be arbitrary complex number, which does not restrict to $(0,2\pi)$.