Why does it make sense to describe complex numbers through powers of $e$?

Question

I'm studying in-depth complex numbers but I keep wondering why it makes sense to describe complex numbers through powers of $e$.

Answer 1

Because of the convenience of logarithms and antilogarithms.

Logarithms were invented to turn multiplications and divisions into additions and subtractions, and to implement arbitrary powers.

This feature carries to the complex numbers, and you can write

$$e^{\log w+\log z}=wz.$$

This creates a close connection to the polar form of the complex numbers, i.e. as vectors characterized by a magnitude ($|z|=\sqrt{x^2+y^2}$) and an argument ($\angle z=\arctan y/x$) instead of Cartesian coordinates.

Indeed, the product of two complex number is a complex number having for magnitude the product of the magnitudes, and for argument the sum of the arguments. Then if you form a complex number with the real part equal to the logarithm for the magnitude and the imaginary part equal to the argument, you define a complex logarithm:

$$\log wz=\log|wz|+i\angle wz=\log|w|+\log|z|+i\angle w+i\angle z=\log w+\log z.$$

Conversely, the antilogarithm (i.e. the exponential) is the number such that the logarithm of its module is the real part of the given number, and its argument is the imaginary part:

$$\log\rho=x,\\\theta=y$$ which you can invert with

$$e^{x+iy}=\rho(\cos\theta+i\sin\theta).$$

There is much more to be said on the topic. For now, remember that the exponential/logarithm create interesting connections between Cartesian/polar coordinates and a symbiosis with the trigonometric functions, in such a way that many computations are simplified.

Answer 2

Well when you are working with complex numbers in the polar form, expressing them in exponential forms allows easier multiplication of such complex numbers. Thats one reason.

Answer 3

An exponentiated complex number $a+ib;a,b \in \mathbb{R}$ has the form $e^{a+ib} = e^a e^{ib}$.

With Taylor series expansion of $e^{ib}$ it holds

$e^{ib} = 1+ib-\frac{1}{2!}b^2-\frac{i}{3!}b^3+\frac{1}{4!}b^4+\dots$.

If you order it by real and imaginary part you will obtain elementary trigonometric functions. Therefore exponentials of complex numbers are a very good and elegant tool to handle trigonometric entities.

Answer 4

Suppose that we have the following function:

$$f(x)=\frac{\cos(x)+i\sin(x)}{e^{ix}}$$

Differentiating once, assuming $e^{ix}$ can be reasonably defined for differentiation,

$$f'(x)=0$$

Thus, it follows that $f(x)$ is constant. Since it is trivial that $f(0)=1$, then $f(x)=1$ for all $x$, or

$$e^{ix}=\cos(x)+i\sin(x)$$

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One may also alternatively note that $y=e^x$ is analytic, and thus, by Taylor's theorem,

$$e^x=\sum_{k=0}^\infty\frac{x^k}{k!}$$

Similarly, sine and cosine are analytic, with the following Taylor expansions:

$$\cos(x)=\sum_{k=0}^\infty\frac{(-1)^kx^{2k}}{(2k)!};\quad\sin(x)=\sum_{k=0}^\infty\frac{(-1)^kx^{2k+1}}{(2k+1)!}$$

Simply by adding, Euler noticed that

$$e^{ix}=\cos(x)+i\sin(x)$$