Breaking the rules of math π = 0. Where is the mistake?

Very few people will be able to identify the wrong step. We start with Euler's identity and derive several equations until we get to π = 0. Complex exponentiation is tricky.

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[mikefochtman]

I always found it fascinating that anytime you have an exponent with i in it, things get 'weird' really fast. Beginners often try to factor things and move 'i' around just like any other variable. But of course you've just shown that isn't always the case. Wonderful explanation of how 'i' is NOT always 'just another variable'. Thanks for this.

[amoswittenbergsmusings]

Presh did not explain *why* this careful handling of complex exponentiation is necessary. Complex exponentiation is basically trigonometric: e^iϑ = cos ϑ + i sin ϑ. Therefore we have e^iϑ = e^(iϑ + 2ikπ). This means that the inverse function log z is multivalued.

The faulty proof really boils down to:
e^2ikπ = e^0 ==> 2ikπ = 0 ==> π = 0.

[Jacob.Peyser]

Most of the confusion in the algebra of complex numbers comes down to the branches of the logarithm. If one is simply consistent with a branch cut of the natural logarithm (such as the principle branch), everything works out without discrepancy. Here, I'll make one critique of the video. In the video, it was expressed that ln(exp(2(pi)i+1))=1. This, however, is only true given that it was specified that ln(z) is the principle value of the natural logarithm (i.e. Log(z)). Without context, one could've easily and 'correctly' written that ln(exp(2(pi)+1))=2(pi)i+1, which was indirectly done initially. In general, the natural logarithm is not a function and can be expressed as ln(z)=Log(z)+2(pi)ki, where "k" is some arbitrary integer. So if anything, this video only adds to the confusion surrounding this number system, further mystifying it. Admittedly, I doubt this comment will clear anyone of confusion, but then again, I'm not a math educator.

[therealsachin]

A simpler one. At 1:00, we have e^(2i x pi) = 1. Now taking natural log on both sides gives us 2i x pi = 0, which means pi = 0.

[mike.]

This whole thing boils down to the fact that some complex functions (like the power function) are not actually functions unless you include branch cuts. This creates a lot of need for careful watching as you go. For example, when you took the natural log of both sides at the end of your demonstration, you really should have included the coterminal answers also. It's still already incorrect, but that's besides the point.

[DrEbrahimian]

Nice one!

I would be careful in defining z^w as e^(w ln z), which is not quite accurate, as ln z is not well-defined for non real values of z. In other words, ln z is not a function from C to C. It is a multi-valued function. For ln z to be well-defined you need to use an extended notion of the complex numbers called a Riemann surface.

Defining ln z as a function would cause all sorts of unintended consequences. For example it cannot be continuous. To see that note that ln 1=0. If you approach 1 from the fourth quadrant of the complex plane the natural log approaches 2 i pi, which is different from zero.

[him]

You could have just gone from e^{2pi i}=e^0, “so” 2pi i=0 so pi=0. This makes the same fundamental mistake.

[f-th]

From the first few seconds of the video, I know where this is going and the answer is: the exponential function is not injective over C, so you can't cancel it.
exp(x) = exp(y) implies x-y is an integer multiple of pi, so not necessarily equal.

[trnfncb]

I think the origin of all trouble lies in the simple fact that the polar representation of a number is not unique. Two numbers in different sheets may be the same but you cannot equate their nat logs (like you would instead equate two real exponents of the same real base) because they in fact differ by a multiple of 2pi.

[lesnyk]

The fallacy occurs at the 4:00 mark, when concluding that -4(pi)^2 = 0 implies that pi=0. The correct conclusion is that -4 = 0. You're welcome.

[kevinmartin]

Around 7:00 it is also important to point out that ln(e^z)≠z in the general case for complex z. The logarithm of a complex number in the general sense has an infinite number of values (just as the inverse tangent function has for real arguments), but like arctan, we define it to produce its principal value whose imaginary part is in the range (-π π] (or perhaps [0 2π), I can't recall) by removing multiples of 2π. Either way ln(e^(2iπ+1)) is 1, not 2iπ+1.

[ZacharyVogt]

I've always loved this channel, for years. I always watch the vids, but I also pause the vid early, and solve the problem for myself, then restart the video and get confirmation of my solution. This is my new favorite video, because I completely fell into the trap. I was plain WRONG and needed the video to correct me; and I really like that.

[nicholassmith]

The mistake is @1:40 e^((2ipi+1)*(2ipi+1)) does not equal e. it equals e^((1 + 2 i π)^2). the rules of exponents are slightly different when dealing with e.

[raj.sharma]

0:52 One can get it directly by this:
e^(iπ) = -1
Square both sides
⇒ e^(2iπ) = 1
⇒ e^(2iπ) = e^0
⇒ 2iπ = 0
Square both sides
⇒ -4 π^2 = 0
⇒ π^2 = 0
⇒ π = 0

[mathmachine]

The mistake was in the assumption that (a^b)^c always equals a^(bc). This statement holds true for real numbers, but in the case of complex numbers, the actual formula is a little messier:

(a^b)^c = a^(bc) * exp(τci*round(-Im(bln(a))/τ))

Where Im denotes the imaginary part, and round denotes the function which rounds numbers to the nearest integer (rounding up if it's a tie). I'm also using τ to denote 2π, not out of preference but to avoid using a confusing number of parentheses.

When c is an integer, this is just the same as a^(bc). That's also the case when a and b are real (unless a is negative or zero).

[amritsagarkar]

1:07 in the last equation at this time stamp
Exp raised to some power yields same exp means the power must be 1
So 2ipi+1=1
So either pi = 0 or i = 0

[TomFarrell-pz]

Reminds me of a favorite Tom Swifty: "Taking the natural log of -1 is as easy as pi, " Euler imagined.

[deerho]

I learned something about complex exponents -- nice way to motivate that.

[ShawnPitman]

4:08 your error is that you assumed one constant was equal to zero, and not the other… Everyone knows the value of pi therefore the value of -4 must have been zero

[dlp]

It was a very important point that you mentioned it well
The truth is, I had not paid attention to this point until now.

Thanks dear Prof.❤