sqrt(i)

We will find both the square roots of i, i.e. sqrt(i). We will first write sqrt(i) as a complex number a+bi and then square both sides. Then we will solve for a and b by setting a system of equations! This is the algebra way to find the square root of the imaginary unit i.

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[Electric_Bagpipes]

There ya go psychologists; the root of imagination.

[jonhues]

I happened to watch this on my break from studying before my leaving cert maths exam. Square root of a complex number was on the exam and I got the right answer using this method. What a crazy lucky coincidence.

[Ambarrabma]

What I miss when I zone out for 30 seconds in class:

[nayjer]

There is a simpler way without the polar form. We know a² - b² = 0 and 2ab = 1, therefore a² = b² <=> a = b or a = - b. But because of the second equation we can cancel out a = - b as a possibility. That means a = b and if we plug this in the second equation we get: 2a² = 1 <=> a = +/ - (1/sqrt(2)). Thats it. :D

[BlackIGO]

Plot twist: He has endless layers of boards.

[PG]

Me: trying to go to sleep

YouTube: BuT wHaTs ThE sQuArE rOoT oF i???

[neobaud]

You can also convert to e^(i*π/2). Then sqrt(e^(i*π/2))=e^(i*π/4). Then convert back to get 1/sqrt(2)+i/sqrt(2)

[CannedMan]

My professor taught me something valuable when writing my master’s: Never start an with ‘as we all know’; you never know who doesn’t know, and thus risk pushing away potentially interested readers.

[TheDailyEgg]

Okay, that's easy. The real question is:

How many blackboards does this guy have?

[NotYourAverageNothing]

Wanna know what's behind my board? It's another board!

[adityellectual]

Another way can be e^(i*π/2)= i for r=1, square root both sides and it will be e^(i*π/4)=√i, which will give √i= (1+i)√2

[AniketTurkel]

Just a heads up, writing it in Euler form is way faster.

That is,
i = e^[iπ/2]
=> √i = e^[iπ/4]
= 1/√2 + i.1/√2.

Edit: forgot a - for the second root.

[stalebread]

I litterally read it as "Squ(i)rt"

[thexavier]

I, as a non native english speaker, watched your video at 2x speed. Got everything you said. Keep it up.

[kritishalli]

This showed up on my recommended, and I could feel myself getting smarter throughout the video because of your amazing teaching style. You have earned yourself a new subscriber, so thank you

[nliehugtil]

I don't know why people keep complaining about this guy's solution to the problem, and why they offer geometric proofs instead. I really like this guy's answer because it uses only the simplest arithmetic/algebra and the simplest definition of a complex number: a + bi. Also, he did it in an incredibly concise manner. (My only complaint was at the end where he could have used a² - b² = (a+b)(a-b) = 0 --> a = ±b. But, typing this out, I suddenly realise how clever he is that even *factorization* doesn't need to be used in his answer.) This answer is teachable on someone's very first lesson on complex numbers; even the average 15-year-old will comprehend it very well, and he's earned my amazement.

[a_voice_in_the_wilderness]

That was interesting to follow, I've forgotten so much math including basic algebra, this was very helpful and you did a great job of explaining it all.

[Bikeswede]

Another way to see it is that multiplying by i makes the complex number rotate around origo by 90° (pi/2). Multiplying by i^(1/2) instead rotates 45° (pi/4). So, for example, 1 × i^(3/2) = -1/sqrt(2) + i/sqrt(2) since that is where a rotation of 135° from 0° takes us.

[waxyacrobat]

6:05 when he got the answer he started moving closer to the speed of light

[chetanraikwal]

**He lifts up the black board**
Me - what the hell is thisss?