Math for fun, sin(z)=2

We know the range of sin(x) is between -1 and 1, inclusively, but that's just with real numbers x. What if our input for the sine function is a complex number? In fact, we can derive the complex definition of sine from the Euler's formula and we can write sin(z) in terms of complex exponential (e^(iz)-e^(-iz))/(2i) and we will be able to solve sin(z)=2.
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This is my equation of the year in 2017.
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-ln(2+-sqrt(3)), • small problem that i o...
Euler's formula: • Euler's Formula (but i...
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*Sorry I forgot the square root. |z| =sqrt(a^2+b^2)
**Also, I should have written the horizontal axis as "Re" and the vertical axis as "Im"
***The last time I did complex analysis was back in 2012

[@blackpenredpen]

Get an Euler's Identity e^(iπ)+1=0 t-shirt: amzn.to/427Seae

[@proton..]

booooo

[@danobro]

I am SO getting one!

[@MoreCharactersThanNeeded]

0/5 The channel is called blackpenredpen but he only used white during the entire video.

[@blackpenredpen]

hmm... sorry... lol~

[@LevatekGaming]

Jonas Lecerof yes, but look at the colour of his top O_o

[@thatoneguy9582]

Jonas Lecerof and it’s not even a pen it’s a chalk

[@mdjameel8494]

Black board red shirt

[@edrishansari2286]

ঈট

[@Fummy007]

Title didn't lie. the math was indeed fun.

[@blackpenredpen]

thanks!!!

[@elymajdou5668]

- said no one ever*

[@arnavanand8037]

@@elymajdou5668 shut up and leave

[@mrinmoysen3860]

@@arnavanand8037 he left ages ago

[@rohitashwadey4510]

Indeed it was..

[@TheMatthew6565]

if sin(z)=2, z is obviously arcsin(2)...

[@blackpenredpen]

: )

[@Tomaplen]

thats like saying, z = " insert correct answer"

[@edwardaa8692]

lol the fun part is arcsin‘s domain is -1， 1

[@tweeweekes5309]

Edward AA yes , in the real numbers there is no answer (ie. undefined). Need to be in the complex number domain to have an answer and, as shown, it’s tricky.

[@shayanmoosavi9139]

Inverse sine domain is [-1,1] :) but nice one. You almost got me there😂😂

[@kiseryota6960]

Watching 3b1b is like taking a lecture from a perfect professor and watching black pen red pen is like doing math with your smart friend.

[@gabestewart-guido8299]

Good I thought was the only one

[@brockobama257]

That’s almost a diss but my smart friend has helped me more than my greatest professors just because I do the homework with my smart friend.

[@abird9724]

I actually almost liked this better, I feel like I'm getting a little more use to seeing i and sin and all that here, like their real and workable, watching 3b1b makes me question a lot of stuff, I love them both equally

[@tictactoe101]

@@abird9724 you can be as good as black/red pen if you do lots of exercises, but that alone can't help you reach 3b1b level

[@lazarussevy2777]

In general I think, Grant is great at instilling necessary intuition for problems, but black black pen red pen does more interesting and somewhat unusual topics, ie ln(negatives, a+bi . . . .)

[@brahmandsaraswat867]

Actually for the people living in 'real' world, and with 0 'imagination', this calculation is a bit 'complex'.

[@brendawilliams8062]

✅

[@Arcticroberto9376]

Shut up, that was funny. You're a dad now

[@divinebanana8400]

Damn the amount of continuous puns is enough to change even a woman into a dad

[@GeorgeBezerra85]

I got that exact equation as a question in my first Complex Variables exam at the university. It was actually a lot of fun to calculate that on the spot and get to the same result. Complex numbers are amazing, you just need to abandon whatever constraints you took as granted for real numbers, especially with trigonometrical functions, logarithms and such.

[@PANZER7910]

Note to self: 1/i makes mathematician feel uncomfortable

[@Nylspider]

USE -i PLEASE

[@GamingWithJumbo]

or use i^-1

[@GamingWithJumbo]

or i^i^2

[@gtsiam]

@@GamingWithJumbo WHY? or maybe i^i^(π/2 - i*ln( 2+sqrt(3) ))

[@paolo6219]

Good thing I'm not a mathematician

[@ABHIGAMING-yo9my]

i have small solution for this We have • sin(z)=2 •cos(z)=root(3)i Then cos(z)+isin(z)=(2+root(3))i e^(iz)=(2+root(3))i z=[ln(i)+ln(2+roor(3))]/i

[@blackpenredpen]

This is very nice!!

[@Javier-id4lq]

Very elegant, love it! 👌🏼

[@wyattsullivan2714]

We can find the general solution to sin(z) = n, using this method, to be z = (pi/2) - ln(n +/- sqrt(n^2 - 1))i

[@blackpenredpen]

Wyatt Sullivan nice!!!

[@black1blade74]

that does look very similar to inverse hyperbolics

[@omeraydindev]

this is actually quite amazing!!!

[@shreyassarangi6106]

I was fascinated enough by this video to try doing this by myself in 12th grade :") along with the relevant solutions for arctan and arccos. Btw, since (n+sqrt(n²-1))×(n-sqrt(n²-1))=1, you can turn the second part of the RHS into +/- ln(n+sqrt(n²-1))

[@selimhassairi]

That "be careful" at the end lmao

[@blackpenredpen]

hehehehe!

[@sleepyfella]

Replying to a 4 year old comment with no reply other than bprp..... Ok

[@marmar8021]

I'm still in high school and watching this just fills me with determination

[@zyrohnmng]

If you enjoyed this, consider taking complex analysis in university. It's filled with this :)

[@viktorramstrom3744]

I'm in the second semester of 10th grade and this video was relatively easy to understand.

[@vorkutavorkutlag]

@@viktorramstrom3744 not much of a flex, he explains it really well

[@Arcticroberto9376]

Keep learning math, grab some practice books (calc is fun imo), start practicing! Mathematics is a very powerful tool to learn

[@nolancheck1465]

I like the idea that he sneaks into the math classroom after hours to do the forbidden math

[@temkin9298]

Dark magic!

[@seagullkilla8252]

did you mean r = |z| = sqrt(a^2+b^2)? at 13:05

[@newsoupvialt]

otis neill I was wondering about that too!

[@blackpenredpen]

otis neill yes

[@ffggddss]

Yes, and this has now been added to the description. Good catch, otis! And good correction, blackpenredpen!

[@MrRyanroberson1]

wait... r cant = z so easy! especially since z has a complex component... unless absolute value assumes the Pythagorean formula on the ocefficients to get a single real number

[@seagullkilla8252]

the absolute value of a complex number (z) is defined to be the "distance" (r) away from the origin of the co ordinate system of all combinations of real and imaginary numbers (R x I). if z = a + bi then a is the real component of z and b is the imaginary component; or a is how far along the real axis z is and b is how far along the imaginary axis z is. if the modulus of z (|z|) is defined to be the distance away from the origin, we can draw a right triangle out of the values a, b and the distance r; where r is the hypotenuse of the right triangle. so if we wanted to know the absolute value of z and we knew the components of it were a and b we could use the Pythagorean theorem to work out how far away from the origin it is since we know the hypotenuse squared (r^2) is the sum of the square of the other two sides then we can take the square root of both sides to find that r = sqrt(a^2 + b^2). this would lead us to the conclusion that is z = a + bi then |z| = sqrt(a^2+b^2); if we define |z| to be the distance away from the origin. i hope that helped :)

[@vallabhdeshpande8428]

I don't know why I laughed at 16:00!

[@blackpenredpen]

LOLLLL

[@rokurosora2404]

"I have another one!" that made me laugh so hard

[@epatrono33]

sameee!!

[@ufalistarotp6170]

same

[@matt-ih5nq]

Valhalla

[@sarthakmandal4082]

A very negligible error at 13:15 i think its supposed to be (a^2+b^2)^1/2. Enjoy the video alot though👍

[@TheStumbleGuysPlayer]

12:34 that r is still stuck inside of that ln

[@TheStumbleGuysPlayer]

This needs 10,000 likes

[@hashimiyazib]

You can also easily find ln(i) using Euler’s identity. I never thought of imaginary natural logarithms this way, but it’s definitely more interesting and useful than just hoping that you’re finding the ln of an easy power of -1.

[@mueezadam8438]

my new favorite math channel

[@blackpenredpen]

mueez adam thank you!!!

[@積分定数C-o5n]

Amazing! I love how range of sin(z) was extended in complex world!

[@cedricp.4941]

Nice ! Can you show us the last magic step ? 😃

[@leorio7416]

yes pls

[@selimhassairi]

Which one?

[@fmakofmako]

Cédric P. it is equivalent to saying -ln (2-sqrt (3)) = ln (2+sqrt (3)), which is equivalent to 1/(2-sqrt (3)) = 2+sqrt (3). if you multiply the left expression by (2+sqrt (3))/(2+sqrt (3)) you'll see its true.

This was fascinating and enlightening, thanks! I watched the whole 19.5 mins, although in the beginning I thought it doesn't have a solution.

[@Jesin00]

Picard's theorem says: If a non-constant function has a taylor series with an infinite radius of convergence, that function must output every complex number with at most 1 exception. e^z outputs every value except 0, and sin(z) outputs every value with no exceptions.

[@iplaysandro3473]

Very instructive video course. I love how you take your time to explain the methodology step by step and clearly.

[@arekkrolak6320]

I think if you introduce new definition of trygonometric functions you cannot assume old formulas hold true like cos -z = cos z. you need to prove it from the definition.

[@MuffinsAPlenty]

True, he's not entirely careful about a lot of the properties still holding (the video would probably be twice as long if he did!). But you can check these properties using the definitions he gave for the functions. To pull the same trick as a math professor would, those are good exercises for the viewers ;)

[@arekkrolak6320]

I agree, I watched the video again carefully and this new definition is derived from old one, so everything is fine!

[@highdino2181]

Arek Krolak you guys are virgins

[@neonet310]

@@highdino2181 minecraft

[@maikmeier5032]

@@MuffinsAPlenty : Not sure he can use the Euler equation to prove it, since that is a circular proof then. To prove it, you would need to write sin and cos as an infinite series and show the equivalency.

[@AgglomeratiProduzioni]

I was loving him, until he wrote C for the "complex axis". Then my love has been tested, and I went on loving him with a bit of bitter.

[@emilsvahn5400]

Ruben i like your profile pic. Diabolic tutor

[@hitzcritz]

I am planning on taking IB HL and I found this video. Now, I'm super pumped to be starting advanced math!

[@kiseryota6960]

Is that how complex analysis work? I haven't taken that course yet. I hope it will be fun as much as this video is.

[@MohaMMaDiN55]

8:53 You made me laugh when you said “what in the world is ln(i)”

[@marcoscandido3793]

we gunna over-look the supreme shirt?

[@blackpenredpen]

lol!

[@brianfisher1305]

For blackpenredpen, yes. Anyone else, no.

[@komeleduc]

Thanks for reminding haha

[@vikasbalani4310]

If we directly put the value of sin Z =2 in Euler's form we get cos Z= sqrt (-3) which gives us sqrt3 I. Then we can take log both sides and simplify we get the same answer. Is this method correct too?

[@NavSci]

probably yes , because another comment simplifies this by a vast amount using cosx=root of 1-sin^2x and substituting sinx=2 cosx=root3i

[@ProCoderIO]

Did you do custom subtitles or has KZbin developed awareness of complex analysis??

[@blackpenredpen]

Unfortunately I do not have the time for that. I believe sometimes some viewers contribute to it and sometimes YT automatically generates it.

[@Caesim9]

the "problem" is at 8:30 . On the complex plain, ln(z) isn't fully defined. Of course this worked, but life isn't that easy..

[@anmoljhamb8775]

AM I THE ONLY ONE WHO NOTICED THE DERIVATION OF FIBONACCI'S GENERAL TERM AT 16:43? (In the erased part I mean 😂)

[@minecrafting_il]

Finally, a video worthy of watching while I eat

[@AdityaKantKushwaha]

Do you know- The quadratic formula of -b±√(b²-4ac) x=------------- 2a was given by Indian Mathematician 'ShreeDhar Acharya' Hence this relation is also called ShreeDharAcharyan Formula

[@zetsuu1106]

No homo but he looks cute when he smiles innocently like that every now and then

[@shuyoe100]

may I ask, sin(∅) , ∅ is an angle. yet sin (z), z is a complex number, what does it mean?

[@gardening_vibes]

But we can take the "complex number" as a radian angle. That will still hold.

[@carultch]

The equation sin(theta) has an application in geometry of relating the height of a position on a circle to the angle from standard position in radians. It also has the application of right triangle trigonometry as the ratio of opposite over hypotenuse when theta is between zero and pi/2. To understand its meaning beyond real values of theta, you determine its Taylor series, and work with the pattern of terms that generates the infinite series polynomial. Given a polynomial with integer exponents and real number coefficients, we can evaluate its terms when you give a complex number as the input to this function. This is what it means for trigonometric functions and the exponential function to have complex inputs. It ultimately means hyperbolic trigonometric functions, when given an imaginary input. This is also what is ultimately behind Euler's formula, that connects trigonometry to exponential functions.

[@seagullkilla8252]

you should get a red chalk so its still black and red

[@materiasacra]

white and red? i do hope he does not get a black chalk, and so should you with your sunglasses :-)

[@rarebeeph1783]

He should get cyan chalk so it would be negative black and negative red on a negative white board

[@fourier07able]

ln(z) = ln|z| + i Arg(z), forgot "(z)" on the RHS second term, no big deal anyway. Also ln(z) is a multivalued function, but he is taking the principal value, so it's okay.

[@rolwindpinto1516]

This solution to a crazy looking problem completely blew my mind. Thanks for uploading a video on this :))

[@mariodiaz3976]

How can somebody explain such a difficult thing at first sight, in such a simple way. Thank's so much, pretty rare to find such a good explanation

[@engjayah]

At 18:10 you should explain With the identity (2+√3)(2-√3) = 1 That ln(2+√3) + ln(2-√3) = 0 and how to take out the +or- outside

[@dreamiitdep2521]

best thing i saw on internet today i had the knowledge of all the formulas you used still i got amazed

[@michaellane1316]

Great, just forwarded this to my two grand children.

[@dalenassar9152]

At about timestamp 9:30 , shouldn't the vertical axis be the "imaginary" axis, -not- the "complex" axis??

[@mike4ty4]

Yupp :)

[@esajpsasipes2822]

the +2pi*n is just the period, nice to see that sine is periodic with period 2pi even in the complex world

[@memegamer1383]

This dude just disproved 2 years of my high school math

[@weixiong1.0]

Many terrible memories of Evan's Hall are brought back in the first 7.2 seconds, all the midterms and finals that I struggled with and lectures I couldn't stay awake in.

[@weixiong1.0]

So in results, a thumb up and a sub.

[@СергейЮбилейный-т2п]

I got 3 seminars in technical university for free/ Excellent video. Thank u. I gues now i can solve at least 1 exercise on exam(actually there are 2 of them).

[@xyzabc3091]

Sir I was searching magic values of sin It came first

[@SlidellRobotics]

around 12 minutes, replace r = e^lnr then make exponents of the product into a sum is more straightforward. Alternately, note i=√-1 and e^iπ=-1, so i=e^iπ/2.

[@deepdutta4671]

I am a student of class 12 from india and I have understood all the things he has explained very clearly ❤️ , He is a very good teacher

[@AlessioQ]

nice one!!!! never tought about sin and cos with complex number! that's good

[@stapler942]

The beginning felt like an FP game where you're a math professor. -- Found CHALK. --

[@reckigaming8686]

In 13:06 why the value of r was (a^2 + b^2) rather than √(a^2 + b^2)????, Btw you are epic

[@hasanrakib391]

🧠🧠🧠🧠🧠🧠🧠Ohhhhh My GOD............What a amazing relation between trigonmetery and complex number....It's really AMAZING.....❤❤❤❤❤❤❤

[@benderamp]

very nice. could not stop until the end

[@tunneloflight]

Now take the next step. Plot the entire sin function of s in complex space. Trust. You want to do this. The result is beautiful. And - it provides a huge insight into the interrelationship of hyperbolic and regular trigonometry. Next, do the same for sinh(z). Finally, repeat all of this for the other trig functions.

[@kvthe2nd903]

Dam i feel like a 3rd grader again

[@eswyatt]

This is the best math video I've ever seen

[@bernard3690]

Greetings and Kind Regards Thank You for the videos I always learn something Please consider editing and shortening the videos by for example in this case merely point out one can use the quadratic formula to solve the problem in the next step by utilizing substitution and not require us to watch you write out the entire formula which we all know Thank You Kindly

[@NeverSnows]

I understoon 90% of that and i came to the conclusion that: SORCERY.

[@localtitans4166]

1 month ago

[@AdmiralJota]

Was it everything after the even/odd function bit? Because after that is when I started being able to follow along for the rest of it.

[@JayTemple]

It's funny the things that grab one's attention. I was distracted each time you uses parentheses to write "sin (z)" and "ln (z)."

[@casperyc]

On a complex plane, it's acutally pretty colorful. With[{a = 2.5}, ComplexPlot[Sin[z], {z, -a - a*I, a + a*I}, PlotLegends -> Automatic]]

[@pranavsharma203]

Just assuming and jumping...

[@karolakkolo123]

It's not assuming and jumping. It is a factual way to do it. And if you ever take complex analysis in college, this will be only like the first week of class

[@rektyboi5356]

Wtf I thoughtZ=sin^-1(2)

[@rahimdina6980]

That wouldn't work on a calculator as it would have to be between 0 and 1

[@ammyvl1]

You can take the cosmologist approach and round 2 to 1 then just say that the answer is 1 (since we don't have enough precision to know what pi is)

[@saswatsarangi6669]

Why did you not say straight hyperbolic function? the way you pronounce is decent but that 4:45 time I had to turn on the captions. BTW I've gone through some of your videos earlier because KZbin, these are nice

[@pritigajjar4105]

Blackpenredpen:* calculates arcsin of 2 Me:wait that's illegal

[@RiccardoTommasini-kz1zw]

9:54 small imprecision... In the graph z=a+bi, the vertical axis is not "complex", but "imaginary" 13:42 r is not a²+b², but the square root of it

[@umermehmood342]

You are great sir This video saves a lot of my time You have tought every single step extraordinarily.

[@RexxSchneider]

At 3:00 - you know you don't need to multiply by -1 and add, there's a thing called subtraction. *** and the last time I did complex analysis was in 1970.

[@ShadowCooper77]

For ln(i) you can also do e^(ipi) = -1, e^(ipi/2) = i, ln(e^(ipi/2)) = ln(i), ln(i) = ipi/2

[@superlinux]

Hello. I hope you mark this comment with a heart sign as it's very important. what you have done in this video is trying to answer the question : when or what would make any fraction give you the result that's greater than one when in reality it must not exceed one. please solve it one more time by replacing the number 2 with an unknown variable: sin(z) = C. thanks. if you do it like that, in my opinion you have formulated a physical law.

[@kennethgee2004]

no as the range for sine is [-1,1]. This is of course assuming a unit circle. so maybe if we are using something other than a unit circle in the complex plane otherwise no.

[@paulchapman8023]

First thing to note is that 2 is not in the real range of sinx, meaning that if there is any number that satisfies the equation sinx=2, it must be imaginary or complex. That means it can be written as a+bi, where b=/=0. So what you’re really solving is: Sin(a+bi)=2+0i. Knowledge of the hyperbolic functions can help solve this equation too: Specifically, cos(bi) = cosh b, and sin(bi) = i * sinh b. It’s good to see another way to solve this, though. (Btw, arcosh (2) = ln (2 + sqrt(3)).)

[@paulchapman8023]

sin(a+bi)=2+0i sina * cos(bi) + sin(bi) * cosa = 2+0i sina * coshb + i * sinhb * cosa = 2 + 0i Next, isolate the real and imaginary elements of the above equation: sina * coshb = 2 sinhb * cosa = 0 Since sinhb can only equal 0 where b equals 0, and since b cannot equal 0 (because that would make a+bi real, and we know that a+bi is not real), sinhb cannot equal 0. This proves that division by sinhb is a valid operation: cosa = 0/sinhb = 0 a = arccos(0) = pi/2 sina * coshb = 2 sin(pi/2) * coshb = 2 1 * coshb = 2 coshb = 2 b = arcosh(2) a+bi=pi/2 + i*arcosh(2)

[@rahulchoudharybhardwaj4818]

When you wrote r=|z|= a^2+b^2 Won't it be √(a^2+b^2) ???? Or I'm just confusing things up Please reply

[@gabrielfernandes5545]

Short equation, giant answer: how perfect! This really got me with that math-loving silly smile in my face!

[@rogerdudra178]

Greetings from the BIG SKY. Good math lesson, thanks.

[@NonTwinBrothers]

This is a certified hood classic.

[@greatestgrasshopper9210]

Everytime somebody shows me stuff like this, I just want some software that could graph a 3rd dimension with axis x, xi, and y. Another example of an imaginary graph I want to see is something like (1/x+1), or conic sections

[@potterworld2584]

The math was really fun. But I think it would have been easier to find the value of ln(i) through euler's identity. e^(iπ)=-1 e^(iπ)=i^2 ln(e^(iπ))=ln(i^2) iπ=2ln(i) ln(i)=iπ/2

[@Omar-mn1eb]

Awesome video! This taught me so much!

[@ravenmusic932]

I can't believe I watch a video about math on saturday morning >.

[@zemoxian]

Neo: This can't be. Morpheus: Be what?...Be real?

[@jucom756]

This video made me realize that the real part and imaginary part function can be written algebraically

[@BleachWizz]

Boy this video changed a lot This video means: The image lof the sin (probably cos as well) function is not [-1,1] I'll need to understand more log functions, I'd personally would like so see a plot of log z. The sum of inverses should receive more atention

[@alwaysgonnagiveyouup4332]

This guy is king of integrals:D

[@cpotisch]

I think it's easier to just completely derive the complex arcsine formula as -i ln(ix +/- sqrt(1 - x^2)) = -i ln(i (x +/- i sqrt(x^2-1)) = pi/2 - i ln(x +/- sqrt(x^2-1)).

[@dw5chaosfan]

I was gonna comment something about branch cuts until the end of the video XD this is actually a really nice complex analysis homework problem, I enjoyed solving it last semester (also glad that I found this video late wasn't tempted to copy this back then)

[@shawnlee1]

like actually, i think this man is some sort of genius

[@KnowingEyes]

I think he just solved world hunger.

[@knutritter461]

Nice one! ;-) Just one mistake about your pythagoras as it is: |z|^2 = a^2 + b^2

[@bobzarnke1706]

The first part of the solution can be simplified by noting that cos²(z) = 1-sin²(z) = -3, making cos(z) = ±√3ɩ. So, exp(ɩz) = cos(z) + ɩ sin(z) = ±√3ɩ + 2ɩ = (2±√3)ɩ.

[@astyanax8913]

Wait. You are trying to define sin for a complex variable, but then assume that cos is even and sin is odd for complex variables, without even having the definition yet. What am I missing here?

[@aashsyed1277]

nothing wrong??????????????????????????????

[@user-lp1bu3gv2f]

Amazing it was Sir Well I'm from INDIA currently I'm a student of 11th grade & I'm Preparing IIT-JEE EXAMS (2022) Your Video really helps me a lot ❤ Thankyou so much ♾

[@binderchannel9454]

Interesting but what does it mean? What does a complex angle mean? We use a real angle and real number as straight line to describe a complex number. What is a complex angle to describe?

[@junwooabbei487]

Theirs a problem in the vetagors .... |z|=√a^2+b^2 Also you r awesome ,thank you for sharing the information ❤️

[@captainteach007]

Awesome maths. Everything you put down made sense but I would never have been able to work that out on my own.

[@wikopl]

When at the beginning you plugged to Euler’s formula that minus, and concluded based on even/odd cos/sin and negative Z that it is cosZ - i sinZ. Couldn’t you plug that minus to ‘i’ to achieve that straight away? And yet another, perhaps dumb question. How to verify, ie calculate sin (a+ib) for a given a and b?