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How to prove Euler's formula: $\\exp(i t)=\\cos(t)+i\\sin(t)$ ?

Hi, I've been curious for quite a long time whether it is actually possible to have an intuitive understanding of euler's apparently magical formula: $$e^{ \pm i\theta } = \cos \theta \pm i\sin \theta$$

I've obviously seen the taylor series/differential equation based proofs, and perhaps I'm just going to have to accept that it's not possible to have an intuition on what it means to raise a number to an imaginary power. I obviously realise that the formula implies that an exponential with a variable imaginary part can be visualised as a complex function going around in a unit circle about the origin of the complex plane. But WHY is this? And why is e so special that it moves at just a fast enough rate so that the argument of the exponential is equal to the arc length of the path made by the locus (i.e. the angle in radians we've moved around the circle)? Is there any way anyone out there 'understand' this?



2 Answers 2


If I recall from reading Analysis of the Infinite (very nice book, at least Volume $1$ is), Euler got it from looking at $$\left(1+\frac{i}{\infty}\right)^{\infty}$$ whose expansion is easy to find using the Binomial Theorem with exponent $\infty$.

There is a nice supposed quote from Euler, which can be paraphrased as "Sometimes my pencil is smarter than I am." He freely accepted the results of his calculations. But of course he was Euler.

  • $\begingroup$ You also get it just using the Taylor expansion of $\exp(i\theta)$. $\endgroup$ May 7, 2011 at 5:52
  • $\begingroup$ It's interesting thinking where he got that expression from... I can't quite see what it represents in terms of e and i... $\endgroup$
    – tom
    May 7, 2011 at 6:07
  • $\begingroup$ @tom: it represents the result you get from applying Euler's method to the system of two first-order ODEs describing a particle moving with unit velocity on a circle. See the question listed as a duplicate, in particular the .gif in my answer which illustrates this visually. $\endgroup$ May 7, 2011 at 6:21
  • $\begingroup$ @tom: He got it after working with the same expression with a real number $x$ in place of $i$. In that case there is a clear intuition. Presumably the fooling with $i$ was the pencil playing. $\endgroup$ May 7, 2011 at 6:51
  • $\begingroup$ haha, yep, it just hit me, was rushing on to share the enlightenment. And thanks for the link Qiachu! Looking forward to digesting it... $\endgroup$
    – tom
    May 7, 2011 at 8:22

There's a book length treatment of this question: Where Mathematics Comes From by George Lakoff and Rafael E. Núñez.

  • $\begingroup$ Thanks, I might check it out $\endgroup$
    – tom
    May 7, 2011 at 5:54

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