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I have seen a visual demonstration of $e^{i\pi}+1=0$ that goes like this:

  • Consider a particle that after $t$ seconds has the position $(\Re(e^{it}),\Im(e^{it}))$ in the complex plane. Since $\frac{d}{dt}(e^{it})=ie^{it}$, the velocity of the particle must be at right angles to its displacement at all times.
  • If your velocity is always at right angles to your displacement, then the path you take must be circular. (I tried proving this by solving the differential equation $\frac{dy}{dx}=-\frac{x}{y}$.)
  • Given that your starting position is $(1,0)$, the radius of the circle is $1$. This means that the magnitude of your displacement is $1$, and since $|e^{it}|=1$, $|ie^{it}|=1$, i.e. you are moving with unit speed counterclockwise around the circle.
  • After $\pi$ seconds, you must have travelled $\pi$ units counterclockwise around the circle, meaning that your position is $(-1,0)$.
  • Hence, $e^{i \pi}+1=0$.

I think I mostly understand this explanation, but what bothers me is geometrically understanding the statement $$ \frac{d}{dt}(e^{it})=ie^{it} \, . $$ If I draw a tangent to an arbitrary point $(\Re(e^{it}),\Im(e^{it}))$, then the gradient of the tangent doesn't seem like it would be $ie^{it}$, if that even makes sense. Since both axes are being used to describe the particle's position, I am unsure about the geometric significance significance of this tangent, if any. In the 'complex world', is there an analogue to drawing a tangent as you would when plotting a real-valued function?

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Drawing a tangent to an arbitrary point does not make sense because you can have infinite lines passing through it. Thinking of $(ℜ(𝑒^{𝑖𝑡}),ℑ(𝑒^{𝑖𝑡}))$ as a position vector makes it easier: When you differentiate this vector with respect to time, you will get the velocity vector. And since, $$\frac{𝑑}{𝑑𝑡}(𝑒^{𝑖𝑡})=𝑖.𝑒^{𝑖𝑡},$$ it is conclusive that the velocity vector should be perpendicular to the position vector.* Remember that we differentiate with respect to time, but we have two position axes. In a normal position-time graph, the derivative of the position function (w.r.t time) gives the velocity function, which is the same as saying the slope of the tangent of the position curve gives the velocity curve. The slope here is defined as: $$\frac{\delta x}{\delta t}$$

You can see easily how this slope equals the velocity through the units. Since we have two position axes, it would not make sense to say that the slope of the tangent to the position vector gives us velocity- distance divided by distance does not give the units of velocity. The velocity vector here, is just another vector which happens to be the time derivative of the position vector. Considered this way, the tangent vector is just the derivative. The fact that it is tangential to the original vector tells us that for any value of $t$, the path traced by the particle has to be circular- which is a significant result, at least for proving the equation you have mentioned.

If you think of tangent vectors to curves as complex numbers you can recast conformality in terms of complex numbers. See:1. Conformal mapping is extremely important in complex analysis, as well as in many areas of physics and engineering. 2

*Both the position co-ordinates are multiplied by $i$, which means the final vector can be written as $i.(original\_vector)$. Multiplication by $i$ gives a 90° counterclockwise rotation about $0$.

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