natural log of -2

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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[@blackpenredpen]

watch next: sqrt(i+sqrt(i+sqrt(i+...))) kzbin.info/www/bejne/ana9g4uNjJtjbcU

[@darknutgaming5510]

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

[@yacinemazouz1795]

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

[@adityaekbote8498]

@@yacinemazouz1795 bruh

[@Alians0108]

@@yacinemazouz1795 yup 0 = 2*pi*i*n 0 = n

[@abdullahamir119]

@@yacinemazouz1795 first of all e^2*pi*i ≠ 1

[@neoxus30]

@@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)

[@abhavishwakarma5035]

Anything is possible when you have complex numbers as your friend.

[@Sam_on_YouTube]

What's i + infinity?

[@angelmendez-rivera351]

Sam Undefined. Infinity is not a number

[@Tomaplen]

@@Sam_on_KZbin i + infinity diverges if written as r*e^(i*theta) so i suppose it just diverges?

[@paulchapman8023]

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.

[@angelmendez-rivera351]

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.

[@jimmorrison6177]

πm = Peyam

[@blackpenredpen]

Yup

[@brooksgunn5235]

@@blackpenredpen Outstanding move!

[@atrumluminarium]

*_ALL RIGHT THANKS FOR WATCHING!_*

[@yrcmurthy8323]

@@atrumluminarium lol

[@anilkumar-ic5ni]

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 )

[@VladTepesh409]

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.

[@kaanetsu1623]

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 😆

[@jannovotny4797]

@@kaanetsu1623 in this case, ln(2) is a. The entire (π+2πm) expression equals b.

[@s4m1rza]

@@jannovotny4797 lmao u shut him down 😬😂

[@s4m1rza]

@@kaanetsu1623 lol ur wrong bro, plus u don’t even know how to spell ‘multiply’ 😱🤣

[@savengo9089]

You don't need even to multiply with π, it is obvious

[@Jo_aJoaninha]

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.

[@ebrahimalfardan8823]

I prefer the kardashian form, it is rounder anf fuller.

[@rorygo456]

thx for the laugh

[@qo7052]

LOOOOOL

[@resonatedvirtue8746]

Bruh

[@Mica_No]

Agree

[@yacinemazouz1795]

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

[@beiyi6933]

Every blackpenredpen videos start with ok! 👍😂 Even with an intro now 😂

[@kqp1998gyy]

...and continues with a 'heeem'

[@t.a.muanza5906]

... and don't forget "anyway!"...

[@madanmohan877]

...and... here is the deal.....

[@i_am_anxious0247]

ln(-2) ln(2)+ln(-1) ln(2)+i(2n+1)(pi) Where n is an integer

[@GabeRundlett]

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).

[@PiEndsWith0]

hey, brother

[@angelmendez-rivera351]

Gabe Rundlett This is such a trivial fact it should not need to be mentioned in the video.

[@nidhiagrawal3354]

@@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. ☺️

[@horowirtz9415]

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)

[@lastwhisper4057]

You are absolutely not

[@LUKAS-bb4jc]

@@lastwhisper4057 what ?

[@geordan6740]

@@LUKAS-bb4jc How do you know about complex but havent learned logarithms lol. Nvm using them together

[@a_cats]

@@geordan6740 probably youtube, they watched this one so it's reasonable to assume they watched more similar videos before

[@yacinemazouz1795]

Your teacher was right

[@wikopl]

It would be interesting to show a graph of ln(x) function when x is negative.

[@Jeppy29]

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

[@Rudxain]

@@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

[@Jeppy29]

@@Rudxain wow now I need to delve into this, Complex numbers are awesome

[@victorfunnyman]

@@Rudxain please graph it 🙏🙏🙏

[@Z7youtube]

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?

[@idontknowwhathandle2use]

3:55 "I will come here for this pie" now your speaking my language!

[@klemmyphoenix5314]

Technically -2 could also be written as 2/-1 and will give us 2 solutions: +-ln(-1) + ln(2)

[@rarex2]

the expression has the "m" number from Z, so it considers all the possible negative values for the logarithm (sorry for my english tho)

[@ricky哥]

That's what I think when I saw his first step

[@ayeayecaptain6249]

How do log and e cancel each other ? Isn't it like natural log of e is equal to 1?

[@blitz7925]

@@ayeayecaptain6249 the exponent in any logarithm can be brought outside as the coefficient of the log. i.e. ln(a^x) = x*ln(a)

[@ayeayecaptain6249]

@@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.

[@mohan153doshi]

BPRP - simply the best channel for learning math on YT. There may be infinitely many channels for math but BPRP is the principal channel.

[@trwn87]

Don't forget about 3blue1brown! But also, 😂😂😂.

[@anton4488]

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😅

[@lookichtafer]

ln(a*b) not equal to ln(a) + ln(b) if we include complex solutions.

[@fedorsergeev8986]

Complete agreement

[@thecottonballtheater]

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! :)

[@pedrosantos1480]

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!

[@martinmerkez2907]

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

[@tingtong6224]

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

[@pedrosantos1480]

@@tingtong6224 I'm from Portugal tho

[@pedrosantos1480]

Seeing this comment again is funny, cause now I'm learning complex analysis at uni

[@michlop452]

@@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)

[@noedeverchere2833]

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].

[@kepler4192]

the congruence and mod always give me a headache

[@louisvictor3473]

@@kepler4192 Why?

[@kepler4192]

@@louisvictor3473 it’s just not my cup of tea

[@louisvictor3473]

@@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

[@kepler4192]

@@louisvictor3473 cheers to you as well

[@OLApplin]

4:07 shoutout to Payam :D !

[@blackpenredpen]

Yup!!!!!!!

[@shahmatsimplex4144]

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.

[@austinlincoln3414]

lol

[@qingyangzhang887]

Euler's formula just popped up! Beautiful ! Keep up the good work!

[@keypo790]

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.

[@bon12121]

I'm glad you boxed it for us

[@kr1593]

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?

[@namanmehta4108]

Well explained as always!! Time for some integrals though in the next video.

[@xxfazenoscoper360doesnosco7]

Oh yeah yeah

[@ABHISHEKSPACE-uu6mq]

Ya.ya.ya

[@st.t6885]

That's the best polar representation I've ever heard. Short and concise.

[@chaikarsenti4553]

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) .

[@marer125]

Cool,but 2π represents an angle that is the same as 0.

[@chaikarsenti4553]

@@marer125 its not that the error, we dont use ln(expx) with x is imaginary number

[@CalculatedRiskAK]

As soon as I saw ln(-1), I knew Euler's Identity would be here. e^(i*pi) = -1

[@cpotisch]

Ok

[@yihangzhu4247]

So ln(2) = ln(2) + i(2n)(pi), where n is integer. What does this imply?

[@TheMathManProfundities]

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).

[@emiliajojo5703]

Really liked this video, no unnecessary stuff, well done!!!

[@VladTepesh409]

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.

[@istvanszennai5209]

‘Kardashian form’ 🤣 I suppose you meant Cartesian form.

[@angelmendez-rivera351]

István Szennai No, he did not. It is a reference to another recent video he released

[@sonpham3438]

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

[@sidkemp4672]

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.

[@cordacord]

>Kardashian form, still not final

[@lukaboljevicboljevic]

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!

[@adamwright4634]

Couldn’t it also be 2ln(i) + ln(2) ?

[@andreimiga8101]

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).

[@happygood18]

This is awesome. I've learned a lot from this single video. Thanks❤❤

[@oui2001]

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

[@trwn87]

Did your English and Math improve 2 years later?

[@_DD_15]

Can you please do some videos about quaternions ? That would be cool!

[@angeldude101]

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.

[@lucarl1021]

3:52 It's for Dr Peyam

[@toddbiesel4288]

So ln(1) = ln(-1)+ln(-1), meaning ln(1) has an infinite number of values, all equaling 2pi(im).

[@Invalid571]

Please solve this differential equation: y' + y(sqrt(x))sinx = 0

[@michlop452]

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...

[@graysonjones5425]

Who needs sleep

[@RyanLewis-Johnson-wq6xs]

Ln(-2)=Ln(2)+pi i

[@williejohnson5172]

Please clarification ln(1)=0 ln 1= 1= the exponent of e Are these correct?

[@DancingRain]

You factored out the π and made Peyam disappear! Quickly, redistribute the π! We need πm back!

[@savengo9089]

As what I know, you can't use Ln in C (imaginary domain)

[@marcushendriksen8415]

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.

[@andreimiga8101]

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).

[@mokouf3]

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 ?

[@angelmendez-rivera351]

Mokou Fujiwara Yes

[@Vordikk]

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

[@bobingstern4448]

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

[@Gyrorealm]

Nice, another amazing video from our bilingual maths teacher

[@happygimp0]

There are also infinite amount of solutions for ln(2), not just ln(-2).

[@Coyote.Starrk_1]

the content is so helpful and entertaining , keep it up

[@georgecop9538]

let z = ln(-1). e^z = e^ln(-1) => e^z = -1 => e^z + 1 = 0 (transcendental). Anyways, we can use the W-function.

[@ilikecucumbers4223]

its so funny when you say isnt it when its dont we but i love when you do it

[@yashvijayvargiya2533]

Where I do mistake ln(-1)=(1/2)×2ln(-1)=(1/2)×ln(-1^2)=(1/2)ln(1)=0 Therefore, ln(-1)=0

[@xCorvus7x]

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.

[@husklyman]

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

[@angelmendez-rivera351]

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.

[@xCorvus7x]

@@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?

[@yacinemazouz1795]

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

[@lucy_derg]

wrong. e^0 = 1, 1=1. any angle greater than or equal to 2*pi is treated as less than 2*pi.

[@coleozaeta6344]

4:40 There is no punchline. It’s not a joke...”

[@user-ft2vp5yw6p]

Can you make a tangent equal 3 Like: tan(x) = 3 (and yes, I got the idea from the sin(z) = 2 video)

[@blackpenredpen]

pablo fontenla miota yes, the answer is just arctan(3) in the real world. : )

[@user-ft2vp5yw6p]

@@blackpenredpen Oh, interesting... Thanks! ^^ Keep up the good work!

[@Edwin-wn3ss]

tan(x) ‘s co-domain is from -ve infinity to +ve infinity, so no complex analysis needed

[@nathanisbored]

well the range of tanx is all real numbers, so tanx = 3 is already defined tanx = sinx/cosx so you just need cosx to be one-third of sinx. this can easily happen on the unit circle in the first quadrant or the third quadrant. so x is roughly somewhere between 0 and pi/2 or between pi and 3pi/2. basically x=arctan(3)+pi*m

[@user-ft2vp5yw6p]

@@nathanisbored oh wow, thanks!

[@zeeshan008x52]

You have become my fav

[@blackpenredpen]

Zeeshan Sayed Mohammed thank you!!

[@matias12381]

Eres increible, espero encontrar canales en español como el tuyo :C pero creo que si no lo encuentro, tendré que hacerlo yo mismo jajaja

[@GSHAPIROY]

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.

[@alex44119]

The formula also works if X is greater than 0

[@ThanksGodsYouAlive]

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?

[@GooogleGoglee]

I want a pie now...

[@andeheri]

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

[@s4m1rza]

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.

[@rezqbarakat2362]

something go wrong ... ln(-1) = Pi * i Multiply both sides by 2 2ln(-1) = 2 * PI * i ln(-1)^2 = 2 * PI * i but ln(-1)^2 = ln(1) = 0 then 2 * PI * i = 0 ?!!!!!!

[@fellipetoffolo4226]

I like your channel thanks for your channel existing !!! 👏 :-) i AM from brazil

[@lukebradley3193]

Man, I was hoping he was going to do something funky with alternating harmonic series and branches of complex ln.

[@adritobiswas1982]

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

[@WagesOfDestruction]

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

[@Anmol_Sinha]

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?

[@SartajKhan-jg3nz]

Dr Peyam will answer why u put the 'm' 🙃 AND WHEN IS PI FACTOREO COMING???

[@qutquttang8161]

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

[@NotBroihon]

ln(-1*-1) = 2*ln(|-1|) = 0

[@ymj5161]

what about including the note of the natural log of all complex number? it will make your video look more complete

[@guythat779]

Also this tells us ln(x) = ln(x) +2pi(i) Does that mean that 2pi(i)=0?

[@jeanpierrem7535]

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?

[@angelmendez-rivera351]

True Stories No, because there is no such a thing as an infinitesimally small function unless you work with hyperreal numbers.

[@UniMaestro]

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?

[@russellchido]

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.

[@Fokalopoka]

its wrong

[@russellchido]

@@Fokalopoka elaborate

[@Fokalopoka]

math.stackexchange.com/questions/683204/logarithm-rules-for-complex-numbers

[@Nikioko]

Logs of non-positive numbers are undefined.

[@trappedcosmos]

Wrong

[@chonkeboi]

Not true in the complex plane

[@JP-xm3qf]

Wow this was mind blowing, thank you!

[@aneeshnag1789]

Another amazing video. But can you help solve an integral? The integral is indefinite. Integral of f(x).dx where, f(x)= fourth root of [tanx+1*fourth root of {tanx+1* fourth root of (tanx+1...upto infinity)}]???

[@davisoncavalcante1450]

OMG! My mind is diferent now, because that explanation o.o

[@johannesmoerland5438]

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]

[@user-gt1cn2tf7s]

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.

[@lucy_derg]

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.

[@Gaark]

ARGH!! I went and checked the door cos I thought it was my doorbell lol

[@leonblattmann1118]

then the ln of a complex number z is: ln|z|+i(arg(z) +2m*pi), isn't it?

[@eidoln9206]

Does anyone know a name for this formula? Thank you for the amazing video! Love the energy! 😍

[@danieljulian4676]

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!

[@abdellh8079]

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 .?

[@YOUSIFPOTATOYT0]

Can you find way to solve 2^x=2x without graph

[@Apollorion]

I think BPRP can. Most probably he'd start his algebra by taking the natural logarithm on both sides of the equation. BTW.. while scanning where would be the boundaries for solutions, I noticed something something very nice occurs for x=1. Actually, a^x=a*x will always give a solution for x=1, because ln(1)=0. And because a^x derivative for x=1 is a*ln(a) is the line y=a*x only tangent at x=1 if a=e, and since 2 is not e, 2x needs to cross 2^x elsewhere (since 2>1), too. Maybe the Lambert W-function might help us out. edit: I checked at wikipedia & it turns out in my previous sentence I guessed right. x=-W(-ln(2)/2)/ln(2)

[@h4c_18]

In that case, x=1 and x=2 are the solutions, but the general case is a bit harder (requires the W(x))

[@angelmendez-rivera351]

a^x = ax => e^[x ln(a)] = ax => 1 = ax·e^[-x ln(a)] => -ln(a)/a = -ln(a)x·e^[-x ln(a)] => W[-ln(a)/a] = -x ln(a) => x = -W[-ln(a)/a]/ln(a), and this gives all the solutions to the equation for any a.

[@perveilov]

The question honestly look innocent enough, wth is this

[@xaxuser5033]

I don't know but i m wondering why i like so much this "i will do it for u guuuuys "

[@pranjaldas1762]

It was very helpful for me Thanks