sqrt(i) but ASMR

The principal square root of i in 60 seconds. #shorts #ASMR #Fast

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bprp #fast

[bprpfast]

Adults don’t need to rationalize the denominators!

[xywing]

Bprp confirming that beard gives big brain calculations at a faster rate confirmed

[damienwilson]

Also if you look at i as e^i(pi/2) then when it is in the square root you multiply the exponent by 1/2 getting e^i(pi/4) which in the complex world associates to cos(pi/4)+isin(pi/4) which is equal to sqrt(2)/2 + sqrt(2)/2*i

[ricardoguzman]

This is the easiest method---

Multiply inside square root:
√(i × 2/2)=
√(2i)/√2=
√((1+i)^2)/√2=
(1+i)/√2
done.

[angeluomo]

How about this? sqrt(i)= sqrt(e^(i*pi/2)=e^(i*pi/4) = cos(pi/4) + i*sin(pi/4) = 1/sqrt(2)+i*1/sqrt(2).

[anuragsinha]

Or you could go for the Euler form... That's more like 2 steps

[Myrus_MBG]

Let’s do it even faster!!!

In polar coordinates, i has a magnitude of 1 and an angle of 90°.

The definition of complex multiplication is that when we do x*y we add their angles and multiply their magnitudes.

So if we have x*x = i (where x is thus sqrt(i) ), then we know (x’s angle) + (x’s angle) = i’s angle = 90° from the sentence above, so x’s angle is 45°. We also know fro, the sentence above that (x’s magnitude) * (x’s magnitude) = i’s magnitude = 1, so x’s magnitude = 1. We know x’s polar coordinates!

With m is magnitude and θ is angle, it is known we can use trig to convert polar to complex: mcos(θ) + imsin(θ), which when we do that for x gives us our answer.

Note: We also have x’s angles = 225° because 225 + 225 = 450 = 360 + 90, and since 360° = 0° then (360 + 90)° = 90°

[charlesreischer]

Leaving the radicals in the denominator physically hurt me

[vanadium]

How do you accomplish the third step? I don't understand how the two square roots add to the previous square root.

[ankitbasera]

I wasn't able to hear clearly so I went to my neighbour auntie she explained me what you were saying

[Tulipflowerak]

We can not have a negative number iside a root !!! The only information we have is i² = -1 but we cannot assume that i = √-1, that's not making sense in math

[pollydolly]

i think easier solution is by using de moivres formula, for any z complex number we can write z=r(cos(x+2pik)+isin(x+2pik)) and for any n ∈ C, z^n = for non integer real values of n : k = 0, 1, 2, 3, 4...n-1, thus if we write i = cos(pi/2 + 2pik) + isin(pi/2+2pik) and take the root on both sides we get, sqrt(i) = cos(pi/4+pik) + isin(pi/4+pik) where k = 0, 1 so we get two solutions sqrt(i) = cos(pi/4) + isin(pi/4) and cos(5pi/4)+isin(5pi/4)

[thecarman]

Amazing how that beard muffles his voice.

[megarotom]

Alt approach: i is pi/2 radians, half of that is pi/4 radians, so the ratio is 1:1 (trig), and sqrt(|i|=1)=1, so a^2+b^2=1^2, so 2a^2=1 (because a=b), so a=b=sqrt(1/2), so sqrt(i)=sqrt(1/2)+isqrt(1/2)

[philippebasier]

How do you get ligne 4 : √(1/2) + √(-1/2) ? Thanks

[guardiank]

I've completed this and i found 1=0

[lukaavni]

Do you have a video explaining how you got from step 3 to step 4? If not do you think you could make one?

[RGC_animation]

I don't understand why it's right but it gives the right answer so ok.

[yingkleung]

Why not use polar form for i, which is 1 angle pi/2? Taking the sqrt if i equals 1 angle pi/4 which gives cos pi/4 + i.sin pi/4. Hence the answer.

[AznJsn]

Is it still ASMR if he smacks the marker tip onto the board?