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I'm trying to understand $$e^{it}=\cos(t)+i\sin(t)$$ This comes from the definitions $$\cos(t)=\frac12(e^{it}+e^{-it}) \quad\text{and}\quad \sin(t)=\frac1{2i}(e^{it}-e^{-it})$$ and those are consistent with the power series definitions of $\cos$ and $\sin$, which are their Taylor series $$\sum_{k=0}^{\infty}{\frac{(-1)^k z^{2k}}{(2k)!}} \quad\text{and}\quad \sum_{k=0}^{\infty}{\frac {(-1)^k z^{2k+1}} {(2k+1)!}}$$ respectively.

I see all those things. They make sense and are consistent with each other, but I can't believe that $e$ and cosine and sine are so related coincidentally.

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marked as duplicate by Aloizio Macedo Sep 17 at 1:36

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  • $\begingroup$ See this answer by whuber - it may be helpful. math.stackexchange.com/a/3593/399334 $\endgroup$ – John Doe Dec 29 '18 at 19:07
  • $\begingroup$ Thinking in polar coordinates helps—multiplying a number by $i$ rotates its position by $90°$, ie $\frac{\pi}{4}$, $\cos$ and $\sin$ calculate the co-ordinates, multiplying by a fractional power of $i$ rotates by the equivalent fraction of $\frac{\pi}{4}$ and so on. Euler's formula expresses the same thing in imaginary powers of $e$. $\endgroup$ – timtfj Dec 29 '18 at 20:41
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    $\begingroup$ Related (duplicate?): Simple proof of Euler Identity $\exp i\theta = \cos\theta+i\sin\theta$. Also, this possible duplicate has this answer, with a nice visual demonstration of the result. There are more instances of this question floating around Math.SE. Try searching for variations of "euler identity proof"; if no existing answers satisfy you, try to convey what it is about them that you find lacking, so that people don't waste time repeating them. $\endgroup$ – Blue Dec 29 '18 at 20:52
  • $\begingroup$ There is an excellent 3 blue 1 brown video related to this. It cleared everything up to me, in an intuitive sense. I would consider it a must watch for anyone who wants an awesome explanation for Euler’s formula. youtube.com/watch?v=mvmuCPvRoWQ $\endgroup$ – D.R. Dec 29 '18 at 21:16
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If you're after an intuitive understanding, I think this page at the Better Explained website might help. It's very visual and imagines exponentiation to an imaginary power in terms of growth which happens "sideways". (I was going to attempt an explanation based on polar coordinates and so on, but I think the page I've linked does it better.)

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Another way to make it feel less coincidental? Differential equations. The function $y=e^x$ satisfies $y'=y$, while $y=\sin x$ and $y=\cos x$ satisfy $y''=-y$. These are fundamental; we could even define the functions based on these equations. Now, the two are obviously different equations - but we don't have to stop there.

Consider the equation $y''=y$. That has both $e^x$ and $e^{-x}$ as solutions, along with linear combinations such as $\sinh x=\frac{e^x-e^{-x}}{2}$ and $\cosh x=\frac{e^x+e^{-x}}{2}$. Now that's looking a lot more like the equation for $\sin$ and $\cos$ - and we can get even closer. What happens with the equation $y''=ay$ for some constant $a$? Well, if $a>0$, we get $y=e^{\sqrt{a}\cdot x}$ and $y=e^{-\sqrt{a}\cdot x}$ as solutions. We can even apply a substitution $t=\sqrt{a}\cdot x$ to transform one to the other. Similarly, for negative $a$, we get $y=\sin(\sqrt{-a}\cdot x)$ and $y=\cos(\sqrt{-a}\cdot x)$ as solutions.

So now, we have a differential equation $y''=ay$, and the form of the solutions changes when the parameter $a$ changes sign. Outside of that change, we have a substitution that can change one case of the equation to another. What happens if we cross the divide, and use the substitution to go from positive to negative $a$? Start with $y''=y$ and it's solutions $y=e^x,y=e^{-x}$. Apply the substitution $y=\sqrt{-1}\cdot x = ix$. The new equation is $y''=-y$, with transformed solutions $y=e^{-it}, y=e^{it}$ - or, at least, those should be solutions to the new equation. But we already know that $y=\cos t$ and $y=\sin t$ are solutions; for things to work out, the new solutions have to be linear combinations of the old and vice versa. Work out the details using values and derivatives at zero, and we get the formulas in the question statement.

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  • $\begingroup$ Damped vibrations seem relevant here (eg a meter needle settling). Mixture of exponential decay and sinusoidal oscillation. $\endgroup$ – timtfj Jan 3 at 23:08
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The Euler formula $e^{iz}=\cos(z)+i\sin(z)$, $z \in \mathbb{C}$, can be shown by using the exponential series along with the series you wrote for sin and cos:

We have $e^{iz}=\sum_{k=0}^{\infty}{\frac{(iz)^{k}}{k!}} = \sum_{k=0}^{\infty}{\frac{(iz)^{2k}}{(2k)!}}+\sum_{k=0}^{\infty}{\frac {(iz)^{2k+1}} {(2k+1)!}} = \cos(z)+i\sin(z)$ using the fact that for convergent $\sum{a_k}$ and $\sum{b_k}$ we have that $\sum{a_k+b_k}$ converges to $\sum{a_k}+ \sum{b_k}$ and the fact that $i^{2k}=(-1)^k$ as well as $i^{2k+1} = i (-1)^k$.

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    $\begingroup$ I don't see how this answers the question. OP says they are aware of the power series proof, and all the maths makes sense logically. The question was about how to gain intuition about the result. $\endgroup$ – John Doe Dec 29 '18 at 19:27
  • $\begingroup$ From my understanding he only was aware of the power series definition of sin and cos as well as that $\sin(t)=\frac{e^{it}-e^{-it}}{2i}$ and $\cos(t)=\frac{e^{it}+e^{-it}}{2}$ hold. $\endgroup$ – Riquelme Dec 29 '18 at 19:31
  • $\begingroup$ Hmm... I feel like the wording that the equations are "consistent" with the power series for $\sin$ and $\cos$ means he is also aware of the power series for $e^t$. In any case, the final sentence and the question tag suggest maybe OP wanted a different angle. $\endgroup$ – John Doe Dec 29 '18 at 19:40

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