0

I know about the famous identity $$ e^{ix}=\cos x + i\sin x $$ I have proved the above identity by comparing the Maclaurin series of $e^{ix}$,$\cos x$ and $\sin x$. But I want to know about other proofs of this identity. I have no idea about any such proofs. Any help would be appreciated.

Myst1cal
  • 267
  • 2
    I hope that you mean $e^{ix}$ instead of just $e^x$ – Matti P. Apr 12 '21 at 07:00
  • What do you think of this? https://en.wikipedia.org/wiki/Euler%27s_formula#Proofs – Matti P. Apr 12 '21 at 07:01
  • Yes Matti P. I just edited – Myst1cal Apr 12 '21 at 07:03
  • You can't prove that $e^{i\theta} = \cos(\theta) + i\sin(\theta)$ any more than you can prove that $(i^2) = -1.~~$ $e^{i\theta} = \cos(\theta) + i\sin(\theta)$ is an artificial contrivance that extends the exponent rule $(e^a \times e^b) = e^{(a+b)}$ rule to complex numbers, based on the easily provable idea that $[\cos(a) + i\sin(a)] \times [\cos(b) + i\sin(b)] = [\cos(a+b) + i\sin(a+b)].$ – user2661923 Apr 12 '21 at 07:05
  • @Matti P. Not so good proofs on wikipedia. – Myst1cal Apr 12 '21 at 07:05
  • @user2661923 Please explain in detail. – Myst1cal Apr 12 '21 at 07:06
  • I will post an answer, giving more details. – user2661923 Apr 12 '21 at 07:07
  • @user2661923 if you use infinite series as the definition of $e^x$ For complex $x$ then you can . – Vivaan Daga Apr 12 '21 at 07:08
  • I'd like to propose a proof shown on 3 Blue 1 Brown, which basically shows that $z(t) = e^{it}$ parameterizes a path in the complex plane where the derivative is a 90 degree rotation of the position ($z'(t) = ie^{it}$) and $z(0) = 1$, and these two conditions can only be met if the path is the unit circle. – Stephen Donovan Apr 12 '21 at 07:22
  • See also https://math.stackexchange.com/questions/72113/has-anyone-talked-themselves-into-understanding-eulers-identity-a-bit. – Hans Lundmark Apr 12 '21 at 07:28

1 Answers1

1

First, see my comment, following the query.

It was desired to facilitate solving problems involving complex numbers. Based on the Taylor series, it is know that for Real $x$

$$ e^x = \lim_{n\to\infty} \sum_{k=0}^n \frac{x^k}{k!}.\tag1$$

This formula, as is, (prior to 1700) would have been considered nonsensical, if $iy$ was substituted for $x$, where $i = \sqrt{-1}$ and $y$ is a real number. That is, how can you possibly have a real number taken to a non-real exponent. Just as $2^3$ represents $2 \times 2 \times 2$, it does make sense to have any real number (or at least any positive real number) taken to any real exponent.

However, Euler noticed that if he broke the rule that equation (1) above is limited to Real numbers, and artificially decided to also apply it to imaginary numbers, or Complex numbers, that it would facilitate solving problems. He also noticed that upon breaking the rule, that it would make sense to regard $e^{i\theta}$ as $\cos(\theta) + i\sin(\theta)$, based on the Taylor series for the sine and cosine functions.

As my comment indicated, re trig identities, Euler found that artificially extending equation (1) above, to Complex numbers, would preserve the law of exponents. This means that you could have more compact syntax, which would facilitate solving problems.

Actually, $e^{i\theta}$, where $\theta$ is real, is nothing more than syntactic sugar for $[\cos(\theta) + i\sin(\theta)].$

user2661923
  • 35,619
  • 3
  • 17
  • 39