Proving Euler's Identity FAST

Euler's identity is very useful for dealing with complex numbers. Let's prove it in less than two minutes!

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[ronak]

In a few years: Proving Fermat's last theorem fast.





In few decades: Proving Riemann Hypothesis fast.

[michaelgeorgievski]

This is so good it almost looks like cheating

[emperorpingusmathchannel]

You can pretty much copy and paste this method for most identities eg:
sin^2(x)+cos^2(x)=1


f(x)= sin^2(x)+cos^2(x)-1
f'(x)=0
f(some arbitrary real number)=0


Boom proof.


What is ironic is that it required more steps than the original heuristic proof using a triangle.

[BluePi]

I would have never thought to do it this way. Great work!

[chirayu_jain]

You did it from derivative. Never thought of that 🤔🤔🤔. Nice video

[EpicMathTime]

I've never seen it done this way.

[w.nickel]

Nice method. It can be used to prove many other identities, e.g. sin^2 + cos^2 = 1.

[xxxprawn]

Fantastic video, I’ve been wondering about that proof for soooo long, and I have to admit the way you presented it using calculus is genius, thanks!

[Tengdbuddy]

A wild Euler appears. Mu uses speed boost. It's super effective.

[jfcrow]

Bravo I wish my math teacher knew that when I was in college 30 years ago.

[brendanchamberlain]

Can you explain how you knew that e^ix and cos(x)+isin(x) are differentiable? Wouldn't having an imaginary number in there mess things up? Also, is there a proof of the rule that shows (e^u)' = ze^u, u=zx, where Im(z) is nonzero? I have only seen this derivative rule proven for real numbers. I haven't taken complex analysis or anything, so I apologize if the answer is obvious. Thank you!

[EricTai]

Easy to understand and very fast proof

[michaelholm]

One issue in your video: When you say that we don't really care what the denominator is. If the denominator also turns out to be zero, you're in trouble and your approach is broken.

[Invalid]

One thing is bothering me with this proof:
We showed that y(t) = c everything is good up to this point.
Then we chose t = 0 and got Euler formula as we wanted but what if we chose another value?
E.g. let t = π/2 or t = π
If t = π/2: c = ... = -i*(e^(iπ/2))
If t = π: c = ... = -e^(iπ)
So why do we prefer t = 0 over any other value?
🤔

Have I missed something?

[VibingMath]

I have a feeling that you do this intentionally for other future videos 😆

[zakariaelghaouti]

Very fast😂😂😂😂 I liked this video. Thanks

[AsadullahJamalFaizi]

But I can't ever thought that It can be solve from derivative AWESOME mu prime math

[willnewman]

What is you definiton of the function e^t?

[swapnilghosal]

A clever one!!! I must appreciate your efforts but not satisfied by the fact that you can differentiate complex functions, as per I know calculus is all about real analysis. I apologize, if I say something wrong. Thank you.

[MarkMcDaniel]

Instructor gives you kudos for a correct answer, but only gives you 3/5 points for not formatting the exact way they like it.