Anyone know if this number here is anything interesting?

[u/homestar_galloper]

I was messing around with complex numbers in desmos, and I got this graph. Anyone know if this number here is anything important? Like anything derived from pi or e? To give you a clear definition it's the largest imaginary part a complex number can have such that |cos(c)| = 1. Looks to be equal to about 0.881

[u/Magicman432]

I looked up what you have in the screenshot and apparently 0.88137 = arcsinh(1) = ln(1+sqrt(2)).

[u/homestar_galloper → u/Magicman432]

Oh, that's very interesting. Thank you for the answer.

[u/Low_Step_9035 → u/homestar_galloper]

Thank you for showing me desmos, I’ll be playing around with it for sure.

[u/ESHKUN → u/Low_Step_9035]

r/desmos

[u/Cobsou]

As someone noted, it is arcsinh(1). Here is the proof:

Let's look at the equation |cos(x+iy)| = 1. Let's rearrange the cosine: we know, that cos(x+iy) = cos(x) cos(iy) - sin(x) sin(iy). Then, as cos(iy) = cosh(y), sin(iy) = i sinh(y), we get, cos(x) cosh(y) - i sin(x) sinh(y). Then |cos(x+iy)|² = cos² (x) cosh² (y) + sin² (x) sinh² (y). Using the trigonometric identity and it's hyperbolic analog, we get: cos² (x) cosh² (y) + sin² (x) sinh² (y) = cos² (x) cosh² (y) + (1-cos² (x)) sinh² (y) = cos² (x) (cosh² (y) - sinh² (y))+ sinh² (y) = cos² (x) + sinh² (y). And we want cos² (x) + sinh² (y) = 1, or, equivalently, sinh² (y) = sin² (x). Taking the root and keeping only the positive (since they are symmetric), we get sinh(y) = sin(x). Since sinh is bijective, it is equivalent to y = arcsinh(sin(x)) -- that's one of the branches of this graph, the other is y = -arcsin(sin(x)). And we are interested in the maximum of this graph, so we need to calculate the derivative: y' = cos(x)/(1+sin² (x))^1/2. Equating it to zero, we get that the extremums are at points where cos is 0, so sin is ±1. Then, the maximum value of the function is arcsinh(1). Q.E.D.

BTW, the cos itself of the point pi/2 + i arcsinh(1) is -i

[u/Forklad2]

Using the exponential definition of cosine, you want complex solutions to abs((e^iz + e^-iz)/2) = 1, or by multiplying the 2 over you want abs(e^iz + e^-iz) = 2. Write z=x+iy.

Rewriting: e^i(x+iy) + e^-i(x+iy) = e^ix e^-y + e^-ix e^y = (cos x +isin x)e^-y + (cos x - isin x)e^y = cos x(e^y + e^-y) - isin x(e^y - e^-y).

Note that e^y and e^-y are real numbers, and that e^y + e^-y has a minimum value of 2 whereas e^y - e^-y can be smaller, as low as 0.

By changing x, we are changing the weight we give each of these two numbers, so intuitively if we want to maximize y while keeping the value the same, we should give the most weight to the smaller function, e^y - e^-y. In particular this means picking x such that cos x = 0, so x can be any of the odd multiples of pi/2, which is the x values you see the maximum occurring at.

Now that cos x = 0 we are just considering the term e^y - e^-y. Its absolute value is just its value because it’s a positive real number. So you’re looking for solutions to e^y - e^-y = 2, or (e^y - e^-y)/2=1.

By definition this means sinh(y)=1, so the value is arcsinh(1) as the other commenter mentioned.

[u/Frangifer]

Not really: it's just arcsinh1 = ㏑(1+√2) .

If we call x & y the real & imaginary parts, respectively, your condition amounts to

(cos(x)cosh(y))² + (sin(x)sinh(y))² = 1 ,

& rearranging that into a plot of y versus x yields

y = ±arcsinh(sin(x)) ...

... so the maximum & minimum values of y are

±arcsinh1 .

It probably appears as a constant in some mathematical scenario or other (it looks familiar: I think I've seen it somewhere in that sort of connection), so it's 'interesting' in that degree ... but what it definitely isn't is any kind of fundamental mathematical constant such as π or e or Euler-Mascheroni 𝜸 , etc etc.

[u/Airrows]

No