Calculating sin of square root of i
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- เผยแพร่เมื่อ 8 ก.พ. 2025
- 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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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.
lovely result and working-through
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)
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)?
@@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)
I'll send this to people who think imaginary numbers are not real
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.
Wonderful Peyam…
I love to be multiplied with Peyam!
As an engineer I would say they are equivalent in the limit i -› 0
excellent
Doctor 👏👏👏
I feel like we should use demoivres theorem to evaluate this
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)
f(g(x)) = g(f(x))
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 ?
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}
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)
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)
@@pritivarshney2128 exp(ln(a))=a is a direct application of the fact that ln is the inverse of exp
@@pritivarshney2128 this makes sense
Thx for this example i got it finally :)
Dr Peyam!! Show BPRP who's boss!
Great!!
If you use addition rule from sin(a+b) from the beginning you'll get the answer no?
How so?
@@drpeyam sin(a+ib) = sin(a) cos(ib) + cos(a) sin(ib)...
I know the addition rule but how does that give you the answer?
Cool thing! Keep it up! :)
Sin(x) is odd function for real arguments, but if argument is complex?
Also odd if x is complex
@@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.
Absolutely beautiful. Who cares of its use?
In the last step of sin (√i), i think you missed to write i.
the dirac delta ween and usub to get f(u) was hilarious
Thanks so much 🤣🤣🤣
*Locus of Z, if arg(Z) =θ is a ray excluding origin*
Could anyone please tell me 🥺 why we exclude origin!!!!?????
What is the angle of the origin?
@@drpeyam yeah it's undefined right!!! Thankyou SIR !!! 😘😘😘
I'm just 11th.. from a while this became my biggest doubt!!!
Thankyou sir ❤️❤️❤️❤️❤️❤️❤️❤️❤️
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..
?
what country are you from?
Austria, Iran, US, and France
@@drpeyam huh?
@@drpeyam i mean where were you born?
US
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.
It depends if you mean principal square root or general square root. Unfortunately they both have the same notation sqrt(i)
@@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!
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.
Dr. Peyam: "Take one horrible expression and simplify it into another horrible expression."
0:20 I didn't understand. 😞
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.
Wow
This is 2ez😂😂😂