how to solve sin(x)=i?
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- čas přidán 2. 12. 2018
- Learn how to solve this complex impossible-looking trig equation sin(x)=i. Of course, we need to use Euler's formula and the complex definition of sine.
sin(sin(z))=1 • Math for fun, sin(sin(...
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sin(z)=2, czcams.com/video/3C_XD_cCeeI/video.html
sin(sin(z))=1 czcams.com/video/8dp35ZhUr_o/video.html
i don't need to be on the bottom if i don't want to
That's what she said!
HAHA, that was witty though. XD
_top comment_
when did he say this?
That's what Lilith said ( ͡° ͜ʖ ͡°)
I solved this very easily:
sin(z)=i
z=arcsin(i)
Easy isn't it?
The point is to figure out what arcsin(i) is through algebraic manipulation.
Did you doing in your head... without prompting‽‽
CrosisBH what ._.
Future mathematician right here ☝️
@@crosisbh1451 he is horribly UnFun in the explanation process
8:00 "pi is an integer".
I am personally offended.
But he wrote only n is an integer so was not wrong
π ∈ ℤ
well pi is an integer isnt it?
Sarcastic Name
I feel I have to make a public apology now Bc of that. : )
We are young
Heartache to heartache
We stand
No promises
No demands
Pi is an integer
Whoo
you are the only person in this world that makes math look fun. well done
VJZ thanks!!!
Definitely not the only one
@@branthebrave who else?
@@VJZ-YT Definitely a lot o teachers out there at any level that do that. Math always looks fun to me, so that doesn't really matter. You're probably asking for other youtube channels, so really any of the popular ones sometimes do like numberphile, 3blue, mathologer, standupmaths, but it depends what you call "fun," like maybe you mean really interesting. Think twice does that.
I get impressed when you use the euler's identity by using as well the logaritms rules good video man
Same , he used the perfect technique to teach that concept
sin inverse both sides -> z= sin^-1 (i)
ez
@@simpletn r/whoosh
BAD NOTATION XDD
One answer is the negative imaginary natural log of the silver ratio.
*how. Cool. Is. Thaat.*
Clemente Wacquez : )))))
Maths does weird things with seemingly unrelated areas sometimes.
has to do with the quadratic equation that appeared
Silver ratio what's that
I've heard of golden ratio only
alan smithee *math
Easy solution:
sin(z) = i, so cos(z) = sqrt(2) by sin^2 + cos^2 = 1.
Therefore, e^(iz) = cos z + i sin z = sqrt(2) + i(i) = sqrt(2) - 1,
and z = -i ln(sqrt2 - 1) as desired.
cos(z) = ±√2
i discovered you from the sin(z) = 2 video. i remember a lot of people were fighting in the comments because you said "conplex axis"
Muriatik yea lol
F
Hahahah I think they pay the wrong attention
"You got to do more work to please people" ...
If you want a single formula you could rewrite the final answer as arcsin(i)=-i*ln(sqrt(2)+(-1)^(k+1))+k*pi for integer values of k.
Now do the formula Sin(z)=a+bi
Z is just sin^-1(a+ ib)
@@shre6619 And what is the inverse sin of a complex number?
@Seife Zwei you take the formulas of sin-1(I) and switch i with a+bi
@@thanoskalamaris3671 That's not how math works
You can write any real number in a+bi form.
4:10 you always make really interesting videos, and some .real. good puns
You can also do by saying that e^iz=cos(z)+isin(z)
Since cos^2+sin^2=1
Cos^2+(i)^2=1
Cos^2-1=1
Cos^2=2
Cos(x)=+-sqrt(2)
Using that you have that
e^iz=+-sqrt(2)-1
The rest is the same as the video
:)
I am missing "Blackpenredpen Yay!"
None of your videos are possible without e.... XD
: )
You need a high iq to get the E
@@gelatinaworld but im a silly man with a small
@@infernocaptures8739, czcams.com/video/bZivCw3bB6w/video.html
For logarithm to be bijective you need to specify which theta you are taking and which half line you are removing also
Love this dudes passion for math!
I love your way of making maths fun , hoping to be the same person like you .
Thank you!!
i think a clever way to do it without knowing the complex form of sin would be by
sin(z)=i
sin²(z)=-1
1-cos²(z)=-1
cos²(z)=2
cos²(z)=plus or minus sqroot 2
then by reemplazing in euler's formula
e^iz=cos(z)+isin(z)
e^iz=plus or minus sqroot 2 +i²
i really enjoyed the video
thanks for being an awesome teacheeeer
" _i_ don't have to be on the bottom if _i_ don't want to"
Brilliant pun and very useful.
I just stumbled along this problem the other day and I solved it. I came on here to check my answers. Very cool problem indeed.
This is awesome i love the content you make ❤️
seriously i have fun by this video teaching style,This video is amazing
I am absolutely passionated about complex numbers... It is just completely different from anything I have learned for 19 years of my life, sometimes is just crazy. Yet it is so useful that we use these numbers in electricity, quantum mechanics, Riemann's hypothesis which is the biggest mystery in Maths. Just amazing. Thank you for your work sir.
I get fun to look your math problems. Thnx sir🙂🙂
Szerintem ez totál elméleti, talán csak elméleti matek szempontból érdekes, de sok apró lépés eszembe jutott. Nagyon jól tanítasz, minden részletet bemutatsz gratulálok !
Thank you so much
Excellent presentation ! Vow !!
8:04 "You write down where n is Z so people know you are cool" LOL
This video was great :D
Wow its very useful for me...tanku for that
I love this "This like that"
this guy is my spirit animal
Would you mind doing a video on Fermat's Little Theorem?
this is ez for me now, but it wouldnt be if it werent for ur vids
thank u so much
I Love Math. Blackpen Redpen YAY
10:15 You can't call that z=arcsin (i), because arcsin is a function so it can't have infinitely many values, it gives a "principled" angle, while the set of all solutions to sin z = i include arcsin + 2pi*n and pi - arcsin + 2pi*n. The same way how the square root function gives you the principal root, and the set of roots are +/- the square root function.
I want number theory videos 😭😭
Excatly
Escatly
Persicely
with complex integers!
You should watch Micheal Penn on CZcams, for number theory problems.
Math is so beautiful and helpful in our life.❤️❤️❤️
"We have to do more work to appease people."
Man, I fucking felt that. Great video.
The two general solutions look very similar. Is there a correspondence on the complex plane?
Bprp: "Okay, thanks for watching"
*almost dies laughing
LOLLLL
Hello RPBP i really like your videos and I need some help. I was doing some math for fun the other day and tried to solve the integral of the xth root of x (or xˆ(1/x)) dx. How should I tackle this problem?
Nice way of teaching..
8:05 pi is an integer? Almost as good as three is smaller than two^^
Also, Sin(z)=2 was cooler, imho.
Lol!!! Yup I still remember that one too!! Btw, the time mark should be 8:00
@@blackpenredpen Oh. I BPRP'uped the time stamp.
Do you remember which video that was?
AndDiracisHisProphet log_2(3) vs log_3(5)
@@blackpenredpen such a trauma, that you still remember^^
AndDiracisHisProphet
lol!! So did you!
I have question:
e^(iz)=+-sqrt(2)-1
=cos(z)+isin(z)
But sin(z)=i
=cos(z)+ii
=cos(s)-1
Cos(z)-1=+-sqrt(2)-1
Cos(z)=+-sqrt(2)
Arccos(cos(z))=arccos(+-sqrt(2))
Z=arccos(+-sqrt(2))
It is correct?
Yeah but you still have to do more work for the arccos, cuz it's domain is only [-1;1] and sqrt(2)>1/-sqrt(2)
Best part of the video " i don't need to be at the bottom if i don't want to"
Can you generalize it for sin(?)=a+bi
Yeah, ill give it a try
@@Tom-qz8xw I've worked it down to a formula where you can input a and b and have it pop out the answer. But I must have made an algebra mistake somewhere because its slightly wrong. For the case in the video it gives me a value where I take the sin and it gives 1.5*i rather than i. So I'll do the derivation again and fix it.
@@shoopinc Done yet?
if you still need it. It's sin(z)=pi/2-i*ln(z±sqrt(z^2-1))
“i don’t like to be on the bottom” hahaha I see what you did there
"2 screw"
Gotta love subtitles.
The important lesson from this, is :
"Keep people in peace by not to do logarithm of negative numbers. Just don't do that and even touch it" (Prof. Steve) 😆😆😆😆😆👍👍👍👍👍
: )
#SteveIsMyStageName
Please don't be mad Professor ☺☺. I was joking. He he he he...
Cause math is fun, isn't it Professor ??? 👍😊👍
@@af8811
Oh no, I wasn't mad at all.
I just wanted to put that harsh tag whenever people comment "steve" : )
@@blackpenredpen :') i thought it's your real name, Professor.
vous etes formidable!
Actually it is interesting , great job , keep going ,
Are you sure you can just invert sin without specifying the domain and range for this question? The formal definition for the inverse of sin is for every x in [-π/2, π/2] , y in [-1,1] , arcsin(y) = x if and only if y = sin(x) though.
for the second answer wouldnt the pi and the 2*pi*m term add together giving us pi*(1+2m)? or in other words pi*q where q is an odd number? by the way love your videos
My professor told me that it is possible to solve for z in the equation |z|=-1
Wolfram Alpha couldn't do it, can you make a video about it, if it is possible?
Fact Sheet I think your professor is wrong. In the real number set, there’s obviously no number that fulfills this. In the complex world, all complex numbers have positive magnitude. In the quaternion world, all quaternions have positive magnitude. Et cetera, I think. But I’m just a sophomore in high school, so what do I know?
It's impossible because it's undefined just like 1/x when x is 0.
I mean probably not as the distance definition but as square root of x to the power of 2
And yes Wolfram Alpha will say there is no solution but it also says that for x^(1/3)=-2, while clearly - 8 is the solution.
depends on how norm is defined. is this the usual norm?
Okay, so for the real numbers |x| is just defined as
|x| = { x if x ≥ 0
{-x if x < 0
i.e. throw away the minus if there is one.
For other numbers like the: Complex, Quaternions, Octonions, etc
The ||z|| function is the distance from 0 to z.
As such, you won't find an example where ||z|| < 0, in any of these sets of numbers.
Distance functions map to values ≥0 as part of their definition.
But supposing ...
You had your own set of numbers "😂",
Then you define a function f : 😂 → ℝ , where f(z) = -2 for some z∈😂
Everyone would think you're an ass if you wrote "Let |z| = f(z) when z∈😂"
By all means you could... it's your mathematics...
But... my recommendation would be for you to extend |z| with a new function "😊"
And just put:
😊(z) = { |z| if z∈ ℝⁿ
{ f(z) if z∈😂
That way everyone will be happy with your notation.
thank's for watching...... 😁😘
1:45 Ёкарный, я думал он по-русски сейчас зашпарит!
Yep but note that (1+√2)=(-1+√2)^(-1) (reader exercise)
So ln(1+√2)=-ln(-1+√2), and therefore we can wtite it as:
z=nπ +((-1)^n)ln(-1+√2)
Is that a mic ?
No it’s a thermal detonator
I would like it more if you had taken sin(z) = i/2. -1+sqrt(2) and 1+sqrt(2) are not as nice as the golden ratio in your answer.
wouldn't the full answer just by -i*ln(1+sqrt(2)) + pi*n since both answers are the same just offeset by pi and they both have the 2pi factor?
If we plugged in the first solution into the z of the second solution, wouldn’t that make pi*(2m+1) = 0?
Thx i ♡ maths
You missed a golden opportunity to mention that "e to the i pi equals -1" is the famous Euler's identity.
pretty sure he thinks most his viewers know that tho lol
All of his veteran viewers know that.
He also missed a golden opportunity to mention that 1 + 1 = 2, which is probably the most famous equation of all!
Хорошо что язык математики и без переводчика понятен)
Постоянно этим восхищаюсь!
I learned complex numbers for high school, still though, didn't see any of this e number and all, is it only for calculus at a university? I'm from Brazil so I don't know if it's just cause here we don't have this for HS curriculum
Does anyone know of an app I could get for an android phone that plots complex number equations on graphs?
please the integral of e^cosx
When are you going to teach Linear Algebra!?
Well well
Make a video showing that there is no z such that tan(z) = i
RealComplex
Wait there isn’t?!
Wow very cool!!!! I just worked out. Thanks!!!
Good video :)
We have -1+sqrt(2) = 1/(1+sqrt(2)) therefore using the fact it is a quotient: -i*ln(-1+sqrt(2)) = (-i)*ln(1) - (-i)*ln(1+sqrt(2)) = i*ln(1+sqrt(2)).
Can we now write down a shorter solution in only one expression ?
I don't think so, but we could have found the second solution easily, as if z is a solution to sin(z)=v, then (π-z) is also a solution…
z = (-1)^n * i * ln(√2 + 1) + πn
cases : n = 2m and n = 2m -1
Finally, bc is because!
By writing 2π, it is aproximating aproximated tau or 6.28
4:10 when she wants to be on top for once
when he said "this guy" at around 5:00 he had the most perfect american accent
Sir can you define log 0
There may be a problem in this case. According to Wikipedia
(complex logarithm)
the property Log (z1 z2) = Log (z1) + Log (z2) is not generally valid when there is a negative number.
What do you think about it?
Before watching the video, I tried this problem, and when I got to that step, I used the polar coordinate definition of the complex number (z=a+bi, z= r*cos[theta] + i*r*sin[theta]), then applied euler's formula (z=r*e^i[theta]), then took the natural log (ln(z)=ln(r) + i*[theta]).
In this case r is |-1-sqrt(2)| and since this is a negative real number, on the complex plane the angle theta would be pi. So you end up with i*z=ln(1+sqrt(2))+i*pi, and then it's the same steps as the video.
This doesnt require the product property of logarithms and I got the same answer, so I think we are good here. BPRP shows exactly what I'm describing in the sin(z)=2 video.
7:59
"you have to denote that pi is an integer"
wait what
Can you juste right -iln(-1+V2) + n pi ?
So the sin inverse of a complex number is also multivalued ?
Hey BPRP, I have an Oxford maths interview in a week. Are there any cool maths tricks I should know about which would blow the interviewer away?
Hmmm, I am not sure about math tricks, especially they should be the finest math people in the world. If I have to do it myself, I will definitely show them how to do math with blackpen and redpen in one hand. Best luck to you!!!! Keep me updated. I would love to hear how it goes!
I have a computer science interview next week as well!
Prove there are no boring positive integers.
0
1 is not boring because it is the identity for multiplication
2 is not boring because it is the smallest prime
3 is not boring because it is the closest prime to the previous
4 is not boring because it is the smallest composite
3 5 and 7 are not boring because they form the smallest value series of primes in arithmetic progression
6 is not boring because it is the first number with distinct prime factors
This proof IS becoming boring so can we generalise it?
Reductio ad absurdum
If any numbers were boring one of them would be smaller than all the other boring numbers, and therefore would be interesting simply for that.
Therefore the smallest boring number is NOT boring: which is absurd.
Therefore there are no boring positive integers. QED
Please do a video about 1/(2+3/(4+5/(6+7/(8+9/(10+11/(12+...))
gotta do calc 2 again.
Yup!
Hi - I am probably wrong, but in your final answers you could start with (Z+2.pi.n) on the left hand side, so wouldn't it transfer across as -2.pi.n on the right hand side? I suppose it might depend on the odd and even symmetry of the sin() and cos() functions?
Yes, you are right notation wise. But it doesn't make any difference since n (and m) can be any integer (negative and positive). So the solution is still correct.
When u did ln-1 shouldnt u have also put a plus 2pin so in the answer u should have at the end smth like 2pi(m+n)?
Alternative solution:
Let z = a+ib
sin(a+ib) = sin(a).cosh(b)+cos(a).(i.sinh(b)) = i
Equating real and imaginary parts gives
a=0 and sinh(b)=1
So final answer is: a=0, b=0.88137
8:00 know π is an integer.
(quoted verbatim, but out of context 😂)
8:00 "π is an integer"
But that's the hypotenuse of the sine equation
So good^_^
If 1-root2 is rationalised then it becomes one answer provided +2npi
Right??
I do this way
sin(z)=sin(a+bi)=sin(a)*cosh(b)+i*sinh(b)*cos(a)=i
sin(a)*cosh(b)=0 and sinh(b)*cos(a)=1
4:41
it's pretty easy to prove that
for every x where x is real and x
The thanks for watching was good :)
I follow all the steps and everything's fine but I still can't understand how a function can be both periodic and unbounded at the same time. please help so I can make sense of this complex world..