Calculating sin of square root of i

Рет қаралды 14,712

2 жыл бұрын

Calculating sin of square root i. Using complex exponentials and Euler's formula, we calculate the values of trigonometric functions of complex numbers, more specifically sine of the radical of the imaginary number. We also compare it to sqrt sin i which is much easier to compute with cosh and sinh.
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Пікірлер: 61

@SisypheanRoller 2 жыл бұрын

I love this reality where, whenever I feel down, I just watch a Dr Peyam upload and my day is better again. ❤️

@drpeyam 2 жыл бұрын

Awwwww ❤️

@txikitofandango 2 жыл бұрын

lovely result and working-through

@emanuellandeholm5657 2 жыл бұрын

Some things are just a lot less messy once you walk into the complex field forever.

@theproofessayist8441 2 жыл бұрын

OMG Dr. Peyam you mad man! I forgot if I suggested this before but this is exactly something interesting I want to see you do! Kudos and hope you're doing well in the USA! - (post edit) Personally I prefer square root of sin(i) much better than sin of the square root of i. You really need to be on the ball or love hyperbolic functions to get that right as well as that very clever -1=I^2 trick which allowed for cancellation in numerator and denominator. The last example was more elegant in having less steps - first example good to play around your head with hyperbolic functions.

@jessstuart7495 Жыл бұрын

I still remember when I first learned how you can represent trig functions with complex exponential (linear systems class) and how all the trig identities can be proved by just using simple algebra and exponent rules. I felt like my Trigonometry teacher in high-school really should have shown us this.

@aweebthatlovesmath4220 2 жыл бұрын

I'll send this to people who think imaginary numbers are not real

@nathanisbored 2 жыл бұрын

could also use the angle sum identity at 2:05 to split it up, and then use the hyperbolic-circular trig relations to pull out the i's, which i think makes it easier (havent worked it out on paper)

@theproofessayist8441 2 жыл бұрын

Hmmm I like that - that could work out too - compound angle for sine right? sin(x+y)=sin(x)cos(y)+sin(y)cos(y)?

@nathanisbored 2 жыл бұрын

@@theproofessayist8441 yeah so y here would be replaced with iy. then you can use sin(iy) = i sinh(y) cos(iy) = cosh(y) EDIT: your formula is wrong but probably just a typo. it should be: sin(x+y)=sin(x)cos(y)+sin(y)cos(x)

@aee220phmunirabad 2 жыл бұрын

Wonderful Peyam…

@Kdd160 2 жыл бұрын

I love to be multiplied with Peyam!

@mudkip_btw 2 жыл бұрын

As an engineer I would say they are equivalent in the limit i -› 0

@jesusalej1 2 жыл бұрын

Great!!

@9circlesofMATH 2 жыл бұрын

Cool thing! Keep it up! :)

@mehmetemin903 2 жыл бұрын

Doctor 👏👏👏

@vladimirzbroj6849 2 жыл бұрын

excellent

@zashra_90 2 жыл бұрын

Amazing video i like it and yeah thank you so much Mmm i have a question .. Why we always add (2πmi) in every time we use this e to the power ln a = a and (a) including [π] or even use the opposite of this rule Is it like when we add 2πk : k from z like in trigonometric functions because they are periodic functions that repeat themselves ?

@hach1koko 2 жыл бұрын

What you're saying is a bit hazy, but yes it has to do with periodicity. The real exponential is not periodic, however x->exp(ix) is, we have exp(ia)=exp(i(a+2pik) for any integer k. In addition, it is one to one on every open interval of length 2pi, so if you'd want to find all x's such that exp(ix)=C for example, if you find one x that works, the set of solutions is then exactly {x+2pik, k integer}

@pritivarshney2128 2 жыл бұрын

e^ix = cosx + isinx e^i(x + 2 pi m) = cos (x + 2 pi m) + i sin (x + 2 pi m) = cosx + isinx. So e^i(x+2pim) gives us cosx + isinx for every m (integer)

@pritivarshney2128 2 жыл бұрын

Let e^(ln a) = x Taking natural log on both sides, we get: ln a = ln x So a = x But x was e^(ln a) So a = e^(ln a)

@hach1koko 2 жыл бұрын

@@pritivarshney2128 exp(ln(a))=a is a direct application of the fact that ln is the inverse of exp

@zashra_90 2 жыл бұрын

@@pritivarshney2128 this makes sense Thx for this example i got it finally :)

@mathadventuress 2 жыл бұрын

I feel like we should use demoivres theorem to evaluate this

@JSSTyger 2 жыл бұрын

Dr Peyam!! Show BPRP who's boss!

@kumardigvijaymishra5945 2 жыл бұрын

Absolutely beautiful. Who cares of its use?

@Sqwaush 2 жыл бұрын

the dirac delta ween and usub to get f(u) was hilarious

@drpeyam 2 жыл бұрын

Thanks so much 🤣🤣🤣

@skylardeslypere9909 2 жыл бұрын

11:00 did you mean to say multiplicative? Because I don't really see the connection between commutativity and dragging the square root inside the sin(i)

@drpeyam 2 жыл бұрын

f(g(x)) = g(f(x))

@King-ve5jr 2 жыл бұрын

*Locus of Z, if arg(Z) =θ is a ray excluding origin* Could anyone please tell me 🥺 why we exclude origin!!!!?????

@drpeyam 2 жыл бұрын

What is the angle of the origin?

@King-ve5jr 2 жыл бұрын

@@drpeyam yeah it's undefined right!!! Thankyou SIR !!! 😘😘😘 I'm just 11th.. from a while this became my biggest doubt!!! Thankyou sir ❤️❤️❤️❤️❤️❤️❤️❤️❤️

@alisidheek3980 2 жыл бұрын

Wow

@midas-holysmoke7642 2 жыл бұрын

If you use addition rule from sin(a+b) from the beginning you'll get the answer no?

@drpeyam 2 жыл бұрын

How so?

@midas-holysmoke7642 2 жыл бұрын

@@drpeyam sin(a+ib) = sin(a) cos(ib) + cos(a) sin(ib)...

@drpeyam 2 жыл бұрын

I know the addition rule but how does that give you the answer?

@jarikosonen4079 2 жыл бұрын

Sin(x) is odd function for real arguments, but if argument is complex?

@drpeyam 2 жыл бұрын

Also odd if x is complex

@jarikosonen4079 2 жыл бұрын

@@drpeyam So I hope to see how it is proved. Its possible indeed, but did never see it proven. Like cases sin(1+i), sin(1-i), sin(-1+i), sin(-1-i),... etc.

@dhavalvyas77 2 жыл бұрын

In the last step of sin (√i), i think you missed to write i.

@oscaroblivion6570 2 жыл бұрын

Dr. Peyam: "Take one horrible expression and simplify it into another horrible expression."

@gbhsdhurnalfatehjangattock7023 2 жыл бұрын

Why so much lengthy calculation is used instead of using formulas of Sin (a+b) or Sin (a-b) and then using Osborn's identities..

@drpeyam 2 жыл бұрын

@SuperYoonHo 2 жыл бұрын

what country are you from?

@drpeyam 2 жыл бұрын

Austria, Iran, US, and France

@SuperYoonHo 2 жыл бұрын

@@drpeyam huh?

@SuperYoonHo 2 жыл бұрын

@@drpeyam i mean where were you born?

@drpeyam 2 жыл бұрын

@MichaelGrantPhD 2 жыл бұрын

I feel like I'm tilting at windmills here! The square root operator is single-valued. sqrt(4) is *just* 2, not 2 and -2. Likewise, sqrt(i) is a single value. Assuming we use the principal branch, it is e^(i pi/4) = (1+i)/sqrt(2). It seems like all my favorite youtubers have been abusing this notation lately.

@drpeyam 2 жыл бұрын

It depends if you mean principal square root or general square root. Unfortunately they both have the same notation sqrt(i)

@MichaelGrantPhD 2 жыл бұрын

@@drpeyam thank you for the reply! I have tried to reply twice but unsuccessful. The replies keep disappearing. But I haven't seen solid literature supporting the use of the radical for anything but the principal. Can you point me to some? Thank you!

@xinpingdonohoe3978 2 жыл бұрын

It seems to me that people take principle values when working in N, Z or R and take multivalues when in C or H. Just a thing I've noticed and it might be coincidence.

@alisidheek3980 2 жыл бұрын

0:20 I didn't understand. 😞

@theproofessayist8441 2 жыл бұрын

If you are stuck with square root of i being the same as I to the half - revise your rational exponents - otherwise don't know how to help. If it's the 2nd part take a look at Dr. Peyam's whole complex exponential and trigonometry videos. The simplest way I can explain is that I^(1/2) can be written in complex exponential/Euler's formula form and that are an infinite number of integer multiple solutions separated by period of 2pi. Everything else you need to look at carefully but I agree this particular problem is some big game algebra stuff.