how to solve sin(x)=i? 

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

That's what she said!

HAHA, that was witty though. XD

_top comment_

when did he say this?

That's what Lilith said ( ͡° ͜ʖ ͡°)

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

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.

Same , he used the perfect technique to teach that concept

@@simpletn r/whoosh

BAD NOTATION XDD

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

cos(z) = ±√2

Muriatik yea lol

F

Hahahah I think they pay the wrong attention

Thank you!!

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.

: )

You need a high iq to get the E

@@gelatinaworld but im a silly man with a small

@@infernocaptures8739, ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-bZivCw3bB6w.html

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!

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)

LOLLLL

Excatly

Escatly

Persicely

with complex integers!

You should watch Micheal Penn on RU-vid, for number theory problems.

: ) #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.

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.

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))

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.

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

Постоянно этим восхищаюсь!

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.

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

RealComplex Wait there isn’t?!

Wow very cool!!!! I just worked out. Thanks!!!

Yup!

I said without using l'hobital's rule because it is the easiest way i want it the hard way 😂😂

tfw implied graph cut

No, you can do that, just exactly in the same way you can write +/- when evaluating square roots. It is multi-valued function. If you want to be strict about it, the arcsin is a map from C to P(C), and the element of P(C) by which it is valued is unique, so it is a function.

it's me It is a function. What you have failed to understand is that arcsin is not map from C to C, but rather a map from C to P(C). The output in P(C) is unique, though it is a set.

No it’s a thermal detonator

Oh you mentioned him xd

Yup! Alright, thanks for watching!