Complex Analysis: sin(x)=pi

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[Tomaplen]

I study physics so sin(x) is approximately x so x=pi

[simohayha]

Would x = arcsin(pi) be a legal solution 😂

[FernandoVinny]

I think the quadratic formula teached in Deutschland is different from the rest of the galaxy

[lookat-thestars]

It's 04:43 am and I just opened my phone and found your video. Now I can't go back to sleep lol-
Thank you so much!!♡

[MichaelRothwell]

Solution to sin(z) = a (a > 1) using compound angle formula and hyperbolic functions
Let z = x + yi (x, y real)
Then sin(z) = sin (x + yi) = sin x cos(yi) + cos x sin(yi) [This is true for real z, so will be true for complex z also]
Now cos(x) = (exp(xi) + exp(-xi))/2, so cos(yi) = (exp(-x) + exp(x))/2 = (exp(x) + exp(-x))/2 = cosh(x) (by definition of cosh(x))
And sin(x) = (exp(xi) - exp(-xi))/2i, so sin(yi) = (exp(-x) - exp(x))/2i = i(exp(x) - exp(-x))/2 = isinh(x) (by definition of sinh(x))
Hence sin(z) = sin(x)cosh(y) + i cos(x)sinh(y).
For sin(z) = a (a > 1), imaginary part cos(x)sinh(y) = 0, which happens if cos(x) = 0 or sinh(y) = 0
In the second case, sinh(y) = 0, we have y = 0 [sinh(0) = 0 and sinh'(x) = cosh(x) ≥ 1, so sinh(x) is a strictly increasing function], so z is real, and this doesn't solve sin(z) = a (a > 1), so is impossible.
So we have cos(x) = 0, and hence sin(x) = ±1.
But sin(x)cosh(y) = sin(z) = a > 0, and cosh(y) > 0, so sin(x) > 0
So sin(x) = 1
So cosh(y) = a and x = π/2 + 2nπ for some integer n
As cosh(x) is an even function, with cosh(0) = 1, continuous, strictly increasing for x > 0, and tending to ∞ as x tends to ∞ [easily verified from the definition of cosh(x)], there will be exactly two real solutions of cosh(y) = a (a > 1), which will be opposites of each other. Call the positive solution arccosh(a).
So y = ±arccosh(a)
Hence z = (π/2 + 2nπ) ± i arccosh(a)
I think this solution makes it clear as to why the real and imaginary parts take the form they do.
We can also evaluate x = arccosh(a) in terms of natural logs:
We have cosh^2(x) - sinh^2(x) = 1 (can be easily verified using the above definitions of cosh(x) and sinh(x))
As cosh(x) = a, a^2 - sinh^2(x) = 1, so sinh^2(x) = a^2 - 1, sinh(x) = ±√(a^2 - 1)
Hence e^x = cosh(x) + sinh(x) [easily verified from the definitions] = a ± √(a^2 - 1)
So x = arccosh(a) = ln[a ± √(a^2 - 1)]
So we can rewrite the solution as z = (π/2 + 2nπ) ± i ln[a ± √(a^2 - 1)]
Again, I think this solution makes it clear as to why the imaginary part takes this particular form.

[ActionJaxonH]

Shortcut once you get to x = -i * ln[i(pi +/- sqrt (pi^2 - 1))] is to separate log by addition (product rule for logs), substitute pi/2 for ln(i) with multiples if desired, then distribute the -i into the parenthesis. Done and done. I solved the problem halfway through the video and thought I had made a wrong turn, only to find out you got the exact same answer at the end :)

[ikaros]

bprp brought me here. quite a good channel

[DrQuatsch]

The nice thing about these solutions is that you can also pull the +- in the ln outside of the ln, because ln(a - sqrt(a^2 -1)) = -ln(a + sqrt(a^2 - 1)) when a>=1.
My favourite version of these, is when you set the cos(x) = -i/2, which gives us the golden ratio somewhere in the answer :D.

[Zonnymaka]

Well done!
Yet as usual, you overcomplicate matters :)
x= -i*ln(i*BLAH)= -i*ln(i)-ln(BLAH)
Hence all we are interested about is -i*ln(i) and you might want to show that or link the answer to another video (ln(i)=i*(pi/2+2n*pi)). That's it!

[HarrySarge]

"...square root of
𝔱𝔥𝔯𝔬𝔞𝔱 𝔰𝔮𝔲𝔦𝔯𝔱
2 pi i over..."

[HighlandHellboy]

I have to thank you so much because I had a question in worksheet that needed to be handed in on when sin(2*theta)=pi and your video helped so much!

[douglasstrother]

Hier ist fintage Papa Flammy fideo!
Your englisch ist gut, und die zerman aksent is not zu bad!

You kids, (I'm 55+.) are fortuntate to have guys like Papa Flammy and other young folks who publish informative videos on mathematics, physics, etc. to cover some very sophisticated techniques and concepts. It's impressive and very generous!

Soak it up! ... And get your work done.

[alberteinstein]

It’s so trippy seeing these videos with actual full-on math content instead of math memes and constant swearing 😂

[willsonbasyal]

I don't understand why you have just 15K subscribers. You're worth a million! But then again, not everyone is smart enough. Thanks a lot for all the work, man. These videos are helping a lot!!

[immersionmusic]

This video is a good introduction to complicated numbers

[eric]

we know by the fundamental Theorem of Engineering, thatsin(x)=x, thus x must be pi. Q.E.D.

[Metalhammer]

man i know ib is the same as bi but my inner monk get´s aggression when i see z=a+ib instead of z=a+bi. don´t ask me why. i´m not even the kind of person that has every sock always in the very same place (looking around my room i´m about the total opposite^^) guess it´s just how we learned things, even if they are nonsensical (like the definition of a derivative it will always be lim h=0 (f(x+h)-f(x)/x+h-h) despite a blind man seeing the x´canceling in the bottom. i learned it that way in school, canceling the x´was always the second step and as insane and impractical as it is, i´ll write it like that first^^) ^^ but great video.

[doodelay]

"So this was the absolute worst way to evaluate pi because we are using pi to evaluate pi" haha

[juanpiedrahita-garcia]

Holy shit😅. I solved this in five easy steps. But only if you solve the residues correctly

[MegaBreckenridge]

Fantastic content. Really enjoying this series