natural log of -2
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- Опубликовано: 10 фев 2025
- Can we have a negative number inside of a logarithm? The answer is yes but the answer is not real!
We will see how log(negative) gives us imaginary numbers. This is quite similar to sqrt(negative) gives us imaginary numbers as well.
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watch next: sqrt(i+sqrt(i+sqrt(i+...))) ruclips.net/video/4EZRXWW607c/видео.html
That was the shortest, most brilliant and succinct description of e^pi*i=-1 I’ve ever seen, well done. No one ever just converted Cartesian to polar form. It was so quick, thank you for posting
Let do some magic;
1 = e ^2*pi*i
ln(1) = ln(e ^ 2*pi*i)
0 = 2 * pi * I
So
i = 0 or pi = 0 or 2 = 0
@@yacinemazouz1795 bruh
@@yacinemazouz1795 yup
0 = 2*pi*i*n
0 = n
@@yacinemazouz1795 first of all
e^2*pi*i ≠ 1
@@abdullahamir119 It is a full revolution, so e^(2πi) is 1)
On a more mathematical form, e^(πi) = -1, right? And e^(2πi) is the square of e^(πi) because (a^k)²=a^2k, that's to say, the square of -1, which as you know, is 1)
Anything is possible when you have complex numbers as your friend.
What's i + infinity?
Sam Undefined. Infinity is not a number
@@Sam_on_RUclips i + infinity diverges if written as r*e^(i*theta) so i suppose it just diverges?
Is 1/0 possible? I thought it was simultaneously every point on the circumference of a circle with infinite radius on the complex number plane, and such a circle was impossible.
Paul Chapman Such a circle is possible on the Riemann sphere. This is important in complex analysis when doing calculus. 1/0 is defined as you would imagine it in this topology. However, in order to do arithmetic, you need to equip the Riemann sphere with axioms of a wheel instead of axioms of a field. Hence, by defining 1/0, you no longer have a field.
It's fun how this gives out an exact formula for any negative value:
ln(-x) = (2m + 1)i*pi + ln(x) {m is an integer}
Love it.
An interesting note, if you take a look at the solution;
(2m+1)iπ+ln(2) can be written as -> ln(2) + (π+2πm)i such that it represents the a+bi format. It makes computation easier in some cases.
No it does not represent a+bi i is outside bracket so it will be ai+bi you don't even know how to multiplicate lol 😆
@@kaanetsu1623 in this case, ln(2) is a. The entire (π+2πm) expression equals b.
@@jannovotny4797 lmao u shut him down 😬😂
@@kaanetsu1623 lol ur wrong bro, plus u don’t even know how to spell ‘multiply’ 😱🤣
You don't need even to multiply with π, it is obvious
I prefer the kardashian form, it is rounder anf fuller.
thx for the laugh
LOOOOOL
Bruh
Agree
Let do some magic;
1 = e ^2*pi*i
ln(1) = ln(e ^ 2*pi*i)
0 = 2 * pi * I
So
i = 0 or pi = 0 or 2 = 0
πm = Peyam
Yup
@@blackpenredpen Outstanding move!
*_ALL RIGHT THANKS FOR WATCHING!_*
@@atrumluminarium lol
Indians have so many types of pi
Piece
Pikoda
Pie pie
Tai pie
Charpie
Teepie
Pictures
Pickle
Pilot
Pick
Pita
Pie
Pista
Pistachios
Pilana
Pilani
Pichuty glands
Pixel story
Piculets
Pie pie chalna
Pitra
Piheya
Pihu pihu
Pinterest
Pirody
Pintukley
Piyakked
Piccolo
P(i)eacock
Pinki
Pithoraghad
Pitch
Piano
Pingaksh
Pingla
Pingal desh
Pain (shift the letter in between)
Points (shift the letter in between )
Every blackpenredpen videos start with ok! 👍😂 Even with an intro now 😂
...and continues with a 'heeem'
... and don't forget "anyway!"...
...and... here is the deal.....
counter-example :
according to Euler:
exp(iπ)=-1 exp(2iπ)=1 ln(exp(2iπ))=ln(1) 2iπ=0 and its false...
The error comes from the fact that the relation ln(exp(x))=x is only valid for real numbers x. And here, you apply it with x=i(π+2πm) .
Cool,but 2π represents an angle that is the same as 0.
@@marer125 its not that the error, we dont use ln(expx) with x is imaginary number
3:55 "I will come here for this pie" now your speaking my language!
ln(-2)
ln(2)+ln(-1)
ln(2)+i(2n+1)(pi)
Where n is an integer
Since he explained this using the rules of logarithms, I feel like he should have mentioned that ln(-x) = ln(x)+ln(-1) AND = ln(x)-ln(-1) (because rules of logs say log(a/b)=log(a)-log(b)), which is implicit in saying it equals ln(x)+i*pi*(2n+1) because 2n+1 multiplied or divided by -1 is still 2n+1, preserving equivalency, however it's not entirely clear in the video. I think he should have brought this up and said ln(-1)=-ln(-1).
hey, brother
Gabe Rundlett This is such a trivial fact it should not need to be mentioned in the video.
@@GabeRundlett The video wasn't about properties of log so it isn't really needed to be mentioned.
ln(-1)
=ln(-1/1)
=ln((-1)²/(-1•1))
=ln(1/-1)
=ln(1) -ln(-1)
=0 -ln(-1)
= -ln(-1)
Btw future people in this comments might think the following so let me clarify that:
x = ln(-1)
x = -x
Before dividing by x check if x is 0 or not.
Case 1:
0 = -0
Conclusion for case 1: yes 0 is a possible solution
Case 2:
1 = -1
Conclusion for case 2: no possible solution
Therefore x = 0 or undefined.
Now let me clarify this.
You found out x is 0 by assuming it's a single solution but it's infinitely many solutions.
Suppose we found a solution then it doesn't have to be the negative of itself: it can be the negative of another solution.
☺️
I calculated by myself in class when discovering logarithms, and got this result, then I asked the teacher if it was correct and she told me I wasn't allowed to do that x)
@Last Whisper what ?
@@LUKAS-bb4jc How do you know about complex but havent learned logarithms lol. Nvm using them together
@@geordan6740 probably youtube, they watched this one so it's reasonable to assume they watched more similar videos before
Your teacher was right
Let do some magic;
1 = e ^2*pi*i
ln(1) = ln(e ^ 2*pi*i)
0 = 2 * pi * I
So
i = 0 or pi = 0 or 2 = 0
Great video!
I wanted to add that for logarithms of other bases besides Euler's Number (let's assume 10, for simplicity) you'll have to multiply the new final result of (2m+1)(i)(pi)+log(x) by log(e), (derived from the change of base formula) to account for the Euler's Identity substitution shown at 2:40 not being able to cancel with the non-natural logarithm. This should work for any base!
Have a good one! :)
Any time I want a dose of some really insane math, I watch one of your videos. You always deliver the goods and cause my brain to go into overdrive.
lol
Why is this so easy when I’m 27 and don’t really need it, and was barely understandable in school when it was actually needed😅
Technically -2 could also be written as 2/-1 and will give us 2 solutions:
+-ln(-1) + ln(2)
the expression has the "m" number from Z, so it considers all the possible negative values for the logarithm (sorry for my english tho)
That's what I think when I saw his first step
How do log and e cancel each other ? Isn't it like natural log of e is equal to 1?
@@ayeayecaptain6249 the exponent in any logarithm can be brought outside as the coefficient of the log.
i.e. ln(a^x) = x*ln(a)
@@blitz7925 I'm aware mate. Im just asking how can log and e be cancelled w each other while we can simply write loge/lne = 1 as natural log of e is 1.
You factored out the π and made Peyam disappear!
Quickly, redistribute the π! We need πm back!
Euler's formula just popped up! Beautiful ! Keep up the good work!
It would be interesting to show a graph of ln(x) function when x is negative.
ln(x) has a negative divergence for x=0 and it's not defined for negative x. It could be interesting in a Im-Re axes graph
@@Jeppy29 the interesting thing about it is that if we calculate the limit from the positive side, we get negative infinity (just as you said), but the limit from the negative side approaches i*Pi - Infinity, where "Infinity" is actually Real infinity, not Complex Infinity.
This means that if we graph the negative lim of ln(0) on the Complex plane, we get 2 parallel horizontal lines stretching out to negative Real Infinity from 2 points, 0 + Pi*i and 0 - Pi*i
@@Rudxain wow now I need to delve into this, Complex numbers are awesome
@@Rudxain please graph it 🙏🙏🙏
is it even possible to graph an imaginary function? i mean there're 2 dimensions for the input and another 2 for the output so these are 4 dimensions, how can we see it?
I'm glad you boxed it for us
This year in high school, I've learned both logarithms and complex numbers. And when I learned that you could do math with squareroots of negative numbers, I started to wonder if you could also do math with logarithms of negative numbers since in the real numbers it was considered impossible. So yeah, thanks for showing this, ot just confirmed my hypothesis!
Well you can't really, logarythm is only defined with real positive numbers, when used with the polar form like this, it's not a function anymore since it has multiple images for a same value of z
wtf! our high school didn't teach us complex numbers and logarithms. That's why college maths is difficult bec we weren't taught at hs
@@tingtong6224 I'm from Portugal tho
Seeing this comment again is funny, cause now I'm learning complex analysis at uni
@@martinmerkez2907 It's called an extension
Also if you want it to have one value you can just consider the principal value as the only one (removing the 2npi nonsense)
ln refers to the principal log and will only take one value. You cannot simply cancel ln and e when the numbers are not real, you have to use ln(z)=ln|z|+iArg(z) where Arg(z)∈(-π,π]. If you ignore this rule as you have done you can have things such as 0=ln(1)=ln(e^2πi)=2πi which is clearly wrong. The multi-value part comes from the relationship eᶻ=w⟺z=ln(w)+2kπi ∀k∈ℤ (w≠0).
If some of watchers are like me that got confused on 4:02, basically it just means that when dealing with rotation you can represent infinite value of theta(angle) for a number(scalar) and so 'm' being multiplied by '2' will always output an even number and so every value of that Integer z will represent that same arrow pointing at the left -1.
To avoid adding a new variable, here n or m, we can simply say that Theta, the angle between the segment OM and the Re(z) axis, with M the point we are studying there, is simply congruent to pi mod 2pi. We can write this as:
θ≡π [2π].
the congruence and mod always give me a headache
@@kepler4192 Why?
@@louisvictor3473 it’s just not my cup of tea
@@kepler4192 I getthat, it is definitely t almost no ones cup of tea. It is just that this is a topic often improperly explained that it always makes me assume something more in that direction instead. Anyway, cheers
@@louisvictor3473 cheers to you as well
That's the best polar representation I've ever heard. Short and concise.
to be honest i don't speak english, i suck at math and i don't know how i ended up here. but i think i like the video but i'm still looking for a use for it
Did your English and Math improve 2 years later?
ln(a*b) not equal to ln(a) + ln(b) if we include complex solutions.
Complete agreement
4:07 shoutout to Payam :D !
Yup!!!!!!!
Really liked this video, no unnecessary stuff, well done!!!
Also, an interesting note. The Complex Coordinate Plain is not to be called the, "Cartesian Coordinate Plain." It is a modified-Cartesian Coordinate Plain, called the, "Argand plane."
The Cartesian Coordinates are Real numbers of x and y coordinates on a two-dimensional plain.
The Argand Coordinates are complex numbers of Real and Imaginary numbers of Re and Im coordinates on a two-dimensional plain.
This is awesome. I've learned a lot from this single video. Thanks❤❤
The answer is also complex numbers which mean different number from R x R cartesian coordinate. If you say i (2m+1)pi = ln -1 , it can say i(2m+1)pi = i(2n+1)pi (for any m,n from Z). Still make sense?
Ln(-2)=Ln(2)+pi i
Well explained as always!! Time for some integrals though in the next video.
Oh yeah yeah
Ya.ya.ya
Good stuff. I would love to see the location of the simplest solution shown as a point on the complex plane to complete the explanation and demo.
The reason he used m instead of n is I assume he wants to write ln(2) as ln(2*e^i2npi), which would result in (2n+2m+1)ipi + ln(2). (might be a bit different I'm too lazy to check for mistakes)
Though honestly considering n and m are both integers and are both multiplied by 2, you still get the exact same solutions so might as well just remove the 2n to make computation easier...
BPRP - simply the best channel for learning math on YT. There may be infinitely many channels for math but BPRP is the principal channel.
Don't forget about 3blue1brown! But also, 😂😂😂.
Nice video! I figured this myself out when I learned that e^ipi is equal to -1, but I never actually knew how it came to that
So ln(1) = ln(-1)+ln(-1), meaning ln(1) has an infinite number of values, all equaling 2pi(im).
As soon as I saw ln(-1), I knew Euler's Identity would be here. e^(i*pi) = -1
Ok
How about ln(-2)=ln(-1*2)=ln(-1)+ln(2)=iπ+ln(2)?
Because ln(-1) means: what power can I raise e to in order to get -1? to which the answer is iπ.
the content is so helpful and entertaining , keep it up
3:52
It's for Dr Peyam
My guess before playing the video: ln(-2) = ln(2) + ln(-1) = 0.6931 + ipi
My guess after playing the video: moot point, because I got it right (if considering only the most basic solution) and because it's not a guess if I've watched it lmao
I love how simple complex numbers actually are! Every time I think about them, I'm awed by their majesty.
Please clarification
ln(1)=0
ln 1= 1= the exponent of e
Are these correct?
ln(1) = 0, which implies that e^0 = +1
I did it in 10 seconds without pen and paper
good job!
I don't understand one thing:
Why don't we do this for ln2 as well, since we are in the complex world?
ln_c(2)=ln_r(2)+2*i*pi*n, n is an integer
Where ln_c is the natural log in the complex world and ln_r is the natural log in the real world (the sexy ln2=0.69).
For the special case n = 0 (the principal value), ln_c(2)=ln_r(2).
>Kardashian form, still not final
I'm curious about this:
As we know, for any complex number z=re^iθ, ln(z) = ln(re^iθ) = ln(r) + (2nπ+θ)i
If θ = 0 i.e. z=r (r is non-negative real number), can we write ln(z) = Re(ln(z)) + (2nπ) i ?
Mokou Fujiwara Yes
let z = ln(-1). e^z = e^ln(-1) => e^z = -1 => e^z + 1 = 0 (transcendental). Anyways, we can use the W-function.
its so funny when you say isnt it when its dont we but i love when you do it
Hey BPRP, I thoroughly enjoy your videos and I watch 'em every day, and I'd like to suggest one - prove that sqrt(n) + sqrt(n+1) is an irrational number, for all natural numbers(mainly because this is my homework for Calc 1 class and I don't know how to prove it B-))
Thanks and keep it up!
There are also infinite amount of solutions for ln(2), not just ln(-2).
So ln(2) = ln(2) + i(2n)(pi), where n is integer. What does this imply?
It implies that you can rotate any arbitrary number of full rotations around the complex plane, and end back on the real number line. In other words, there's not enough information to uniquely determine which exponent you raise e to, in order to get 2. The easiest solution to find is ln(2), but there are infinitely many more solutions that rotate any integer number of times around the unit circle to get to 2 as well.
I used a different method. Since ln(-1) can be written as e^x = -1 I called x some complex value (a + bi). Then I modified Euler’s identity a little to from e^(ix) = cos(x) + isin(x) to e^(a+ix) = e^a(cos(x)) + ie^a(sin(x)) since e^(a+ix) = e^a * e^(ix). Then I solved the equation for -1 and the solution was a=0 and b=pi. Then I just used log properties to find out ln(-2), the same thing you did
Now i want to draw picture of ln(x) in 3D with X, Y and Z (imaginary) axes
So there will be multiple lines for ln(positive), repeats each 2*i*pi, and mirrored to them relatively to Y axis ln(negative) but moved by i*pi
As what I know, you can't use Ln in C (imaginary domain)
I like your channel thanks for your channel existing !!! 👏 :-) i AM from brazil
I have a video-idea for you. If you first look at ln(-a) where a>0, and subsitute it into the formula for ln(-a), where a
Because when you substitute 2pi back into the exponent form, e^ix, you get e^2ipi = 1, as 2pi is just a full rotation back to the real axis. Take ln of both sides, then u get 2ipi = 0. This is the case for any 2n*i*pi, as they always results in the final angle being 0.
I disagree: ln(z) is NOT a (single valued) function for all complex numbers or even for z0
the relation ln(exp(x))=x is only valid for real numbers x
The question honestly look innocent enough, wth is this
‘Kardashian form’ 🤣 I suppose you meant Cartesian form.
István Szennai No, he did not. It is a reference to another recent video he released
Now you have to prove that ln(a*b)=ln(a)+ln(b) is valid for negative numbers.
Let do some magic;
1 = e ^2*pi*i
ln(1) = ln(e ^ 2*pi*i)
0 = 2 * pi * I
So
i = 0 or pi = 0 or 2 = 0
wrong. e^0 = 1, 1=1. any angle greater than or equal to 2*pi is treated as less than 2*pi.
"Proof" that 0=i2pi
0=ln(1)=ln(e^i2pi)=i2pi*ln(e)
=i2pi
Can you please do some videos about quaternions ? That would be cool!
Quaternions and other hypercomplex systems make roots and logarithms even more complicated since you now have i^2 = j^2 = k^2 = -1, but i =/= j =/= k. So what's the square root of -1,? well there are uncountably many such square roots. any quaternion xi + yj + zk such that x^2 + y^2 + z^2 = 1 will have (xi + yj + zk)^2 = -1. In the complex plane, there are only two such numbers, +i and -i. Now the number of solutions to ln(-1) is the same uncountable infinity _times_ the countable infinity of the solutions for the complex plane (which is still just uncountable infinity).
So ya, roots and logarithms kind of don't work since the functions they're inverses of get even farther from being 1-1.
P.S. If quaternions seem too complex, it might help thinking of the basis of i,j, and k less like vectors and more like basis planes in 3D space, where multiplying by one of them rotates an object by 90 degrees within the corresponding plane. In 2D, there is only 1 such plane and no axes orthogonal to it, so there's only one "basis plane" within which objects can rotate, which is i.
The formula also works if X is greater than 0
Please solve this differential equation:
y' + y(sqrt(x))sinx = 0
Logs of non-positive numbers are undefined.
Wrong
Not true in the complex plane
So ln(e), if we see it as ln(-1*-1*e), is ln(-1) + ln(-1) + ln(e), that is equal to ln(e) + 2iп(2m+1)? That means that 1 = 1 + 2iп(2m+1), so 2iп(2m+1) = 0. But that's wrong.
ln(-1*-1*e) = 2*ln(-1) + ln(e) = e + ln(-1) = e + 0i = e. numbers greater than or equal to 2*pi are treated as less than 2*pi, and greater than or equal to 0.
I don't know but i m wondering why i like so much this "i will do it for u guuuuys "
what about including the note of the natural log of all complex number? it will make your video look more complete
Couldn’t it also be
2ln(i) + ln(2) ?
If you write ln(i), it is basically the same problem as writing ln(-1). You must prove that value exists (you need to write it as i×pi/2).
This is great! I already knew the Cartesian form for complex numbers! I might sometimes want to call this the "Kardashian form". Much more glamourous!
Eres increible, espero encontrar canales en español como el tuyo :C pero creo que si no lo encuentro, tendré que hacerlo yo mismo jajaja
Nice, another amazing video from our bilingual maths teacher
Let φ be an infinitely small function defined by: φ (π / 2-dθ).
This function is infinitely small it can be written in the form -kdθ
Is there a mathematical principle that justifies this form?
True Stories No, because there is no such a thing as an infinitesimally small function unless you work with hyperreal numbers.
why is the part (2m+1) necessary?
you use it to see that equations like e^(3iП) = -1 or e^(5iП) = -1 are right
but u can just take cube roots from both sides of e^(3iП) = -1 and get the right equation e^(iП) = -1
so the answer is just iП + ln2, isn't it?
Man, I was hoping he was going to do something funky with alternating harmonic series and branches of complex ln.
Ive just learned about complex numbers and that r*e^i*theta form. Its good to finally understand what I couldnt a few years ago. 😂
Would it be legal to do the following:
ln(-1)
(2/2)ln(-1)
ln((-1)^2)/2
ln(1)/2
0/2
0
In which case all negative log are zero.
its wrong
@@Fokalopoka elaborate
math.stackexchange.com/questions/683204/logarithm-rules-for-complex-numbers
There is another solve, but it give only half solves to ln(-2).
ln(-2)=
ln(-1*2)=
ln((i^2)*2)=
2ln(i)+ln(2)=
2ln(e^(i(pi/2+2piM)))+ln(2)=
2i(pi/2+2piM+ln(2)=
(4M+1)i*pi+ln(2).
Where i did error? or, if we go by another way, we will not get full answer?
6:02 Was that 王力宏 Wang Lee-hung's 唯一 playing towards the end?
Who needs sleep
I have a doubt (I am in school at 10th so I am not aware of complex numbers).. if we have log(-1), then multiply by 2, we get 2log(-1) = log(-1²) = log(1) = 0. So 2log(-1)=0, then log(-1) = 0. What is wrong with this approach?
Where I do mistake
ln(-1)=(1/2)×2ln(-1)=(1/2)×ln(-1^2)=(1/2)ln(1)=0
Therefore, ln(-1)=0
You are toying with a multi-valued operation.
Note that 1/2 * log(1) = log(1^(1/2))
= log(sqrt(1)) which is either log(1) or log(-1), since both 1 and -1 are roots of 1.
Thus your mistake should be in the third or fourth step where you get rid of the negative sign, while still claiming equality.
Yash Vijay
We are not saying "where do I mistake". we say "where did I mistake".
This is your mistake, also, 0 can be equal to 2mπ while m is integer
Omer No, 0 = 2πim is not generally true. The correct statement is that ln(1) = 2πim. m = 0 is not an appropriate branch choice for the equation, because the branch cut happens precisely on the negative integers.
@@angelmendez-rivera351
Could you elaborate on your last sentence?
Why is m=0 an inappropriate branch choice; can't I choose any branch, including the principal?
And why should branch cuts happen only at negative integers?
ARGH!! I went and checked the door cos I thought it was my doorbell lol
consider 0= ln(1)= ln(-1x-1)= 2ln(-1) = 2(2m+1)i Pi
so, is it a clue for sollving the non-trivial zero of The Riemann zeta function
ln(-1*-1) = 2*ln(|-1|) = 0
I'm just wondering why these questions have an infinite number of solutions, but sometimes when you deal with complex equations they don't, considering any answer could be spun to have an additional 2π rotation thus making the solutions infinite. That is, essentially you're just adding e^2niπ to every solution, but e^2niπ is essentially just 0.
bloodyadaku No, e^(2πni) = 1, not 0.
@@angelmendez-rivera351 My mistake. I suppose then I mean a multiplication of e^(2niπ) then? Either way, every answer could have that potential coefficient and/or constant giving an additional array of solutions.
Some days ago I though I had the value of ln(-1) that is iπ but no. YOU LITERALLY FOUND IT BEFORE ME WHEN I DID NOT EVEN KNEW WHAT IS LN
if the in(-x) = (2m+1)iPi+in(x) , therefore m = 1 or m=10 will equal the same result so 3ipi=21ipi=>18ipi=0 which is nonsense
if i want a non ln answer can i just use the "change base" property???
I want a pie now...
Hi man how's your day ? When I was in high school , I learn the complex numbers but when we study the function from R to R. Can we study the function from Z to Z .?
I don't think that there is a principal value for the log of a negative value as the main branch of the complex log is not continuous on (-infty,0]
If you add 2m*pi*i in log of negative numbers, why don't you add it in log of positive numbers?
because this video is bad at explaining.
when you use an angle greater than or equal to 2*pi, it is treated as it's less than 2*pi, greater or equal to 0 equivalent.
so when you have a positive number it becomes ln(x) = ln(x) + 0i.
the logarithm of any complex number is ln(z) = ln(r) + i*theta
for all real numbers greater than 0, theta = 0.
Wouldn't you also have to factor in the new radius if "x" isn't 1? You would then have an additional ln(x) from the ln(re^i(theta)), wouldn't you? So you'd end up with (2m+1)(pi)(i) +2ln(x).
Correct me if I'm wrong, not totally sure!
(From the final equation you gave at the end)
The radius is always one because it is always -1 times inside the ln. Which translate to the sum of ln(-1) + ln(n). And we only transform the - 1 to complex number
From logarithm rule you can separate product inside logarithm to addition outside. So ln(-x) is always equal to ln(-1) + ln(x). Hope this helps.
I don't think it's advisable to explain the complex logarithm this way. Sure, it has multiple branches, but just adding 2m doesn't really help matters because you didn't exactly specify what m is. If you just say that m is an integer, it gives off the impression that you can use any m, but that would of course be wrong because (2m+1)ipi+ln(2) = (2n+1)ipi+ln(2) already implies m=n. While I'm at it, I might also add that it may have been a good idea to mention that ln(wz) doesn't need to be equal to ln(w)+ln(z) for complex w and z. The same goes for ln(e^z) and z. I can't see why it would be a good idea to ignore these properties entirely.
Arteyyy You must be on acid. No, no one would think you can choose any m because the direct implication of saying "m is an integer" is that the logarithm is a multi-valued relation on the complex numbers. i(2n + 1)π = i(2m + 1)π implying m = n does not invalidate anything he said here. In fact, this inference is trivial and basically requires no explanation. And ln(zw) = ln(z) + ln(w) for any appropriate branch choice, as well as ln(e^z) = z. If you evaluate the logarithm as multivalued or set valued, then the above identities hold true trivially. Your comment goes on about literally nothing.
A better criticism would have talked about principal values and defining branch cuts, but this technically is unnecessary anyway as long as we never try to work with a single-valued logarithm, which in this case, we are not required to.
No, I'm not. There's no reason to be rude. I just think it's imprecise to say that m is an integer, so I would have appreciated some more information. For example, how can one determine which m is to be used? The ln I know, which allows for complex inputs, is a function and not just a relation. And of course, it's the principal branch of the complex logarithm. Maybe you got taught it differently, but I'm not sure all viewers immediately understood that ln is supposed to be a multi-valued relation. I also obviously didn't mean that (2m+1)ipi = i(2n+1)ipi doesn't imply m = n, if I understand you correctly. My point was that he just used ln(wz) = ln(w) + ln(z) and ln(e^z) = z without explaining that you can't always do that. But maybe the source of confusion is just that we aren't talking about the same ln.
4:40 There is no punchline. It’s not a joke...”
then the ln of a complex number z is: ln|z|+i(arg(z) +2m*pi), isn't it?