I just learned this from this video √[(a+b)+(2√ab)] =(√a)+(√b)

@bprpfast

3 жыл бұрын

Yup

@fikrim8819

Жыл бұрын

a^2+2ab+b^2 = (a+b)^2

@alimaisamamiri2530

Жыл бұрын

I would like to see a proof of this,

@saiph8007

Жыл бұрын

Crab claw method

@basilvanderelst128

Жыл бұрын

​@@alimaisamamiri2530 just the proof of (x+y)²= x² + 2xy + y² but with x = sqrt(a) and y = sqrt(b)

bro can make solving world hunger look easy 💀

Bro casually committing multiple war crimes at once

@changhaizhu7854

5 ай бұрын

LOL

@sherylbegby

2 ай бұрын

Is it just me, or was this just full of shocking illegal math moves to, improbably, get the right answer? I''d love a video on how to do this correctly.

@TheMathManProfundities

2 ай бұрын

It is legit but leaves out the key steps which show how it works. However it will only work when the real and imaginary parts of the root are integers and we can easily find the two magic numbers to make it work.

@spastinakos7224

2 ай бұрын

​@@sherylbegby i mean i think he used it right. U can solve it like tbis: sqrt (a+b 2(sqrt ab) )

@spastinakos7224

2 ай бұрын

Had this in math class this year and last so pretty sure its aight

I used √(e^(iθ))=√r e^(iθ0.5). I guess I could've had it easier

@wassupbros4629

7 ай бұрын

How it feels to use quadratic formula to solve x^2=16

@raghavkumar4914

4 ай бұрын

​@@wassupbros4629😂 true

You can solve this with completing the square root. 5 + 12i = - 4 + 12i + 9 = 4i² + 12i + 9 = (2i + 3)²

@birb9425

Жыл бұрын

Yeah, but that process takes too much time when the question has numbers too complex. Who knows what numbers 5 was composed of, without having the numbers given in the video as hint. Is 5 made up of 10 and -5? 8 and -3? 2 and 3?

@parveenmaraikar1980

Жыл бұрын

@pixi9278 that's a good way

@thegoofiestgoooberr

Жыл бұрын

@@birb9425 because they have to multiply to 36, as described in the video. Just like factoring quadratics

@abohamza314

6 ай бұрын

Your solve is better

@Moss_Muzz

3 ай бұрын

@@birb9425as a Asian, actually this isn’t hard to find them

bro is whispering so he doesn't awaken the maths gods

I lost at square root 9 + square root -4

@SanePerson1

Жыл бұрын

He doesn't explain what he's doing there. First, the number inside the square root sign has a positive real and a positive imaginary part, so it's inside the upper right quadrant in the complex plane. The principal root will also be in the upper right quadrant (in polar coordinates, it will have an angle that's half of it's square). Therefore, he knows that the root will have the form a + ib, with a and b both positive. Also, what you see in the radical is (a² - b²) +i(2ab), because that's the square of (a + ib)². So he just sees by inspection that 9 - 4 = 5 (that's a² - b²) and 9 × 4 = 36. In his full video, I'll bet he explains this much more completely.

Also me : Uses classic algebric way sqrt(5 + 12i) = a + bi 5 + 12i = a² - b² + 2abi 5 = a² - b² 12 = 2ab 12 = 2ab b = 12/2a b = 6/a 5 = a² - b² 5 = a² - (6/a)² 5 = a² - 36/a² 5a² = a^4 - 36 a^4 - 5a² - 36 = 0 Subtitution : u = a² u² - 5u - 36 = 0 u² + (4u - 9u) - 36 = 0 (u + 4)(u - 9) = 0 u + 4 = 0 or u - 9 = 0 a² + 4 = 0 or a² - 9 = 0 a² = -4 or a² = 9 a = ±sqrt(-4) or a = ±sqrt(9) a = ±2i or a = ±3 a can't be imaginary because it's the real part of the solution, so a = ±3 b = 6/a b = 6/3 or b = -6/3 b = 2 or b = -2 So the solution is ±(3 + 2i)

@Syndicalism

Жыл бұрын

But your way isn't finding the principal root

@Shreyas_Jaiswal

Жыл бұрын

​@@Syndicalism in complex world, there are two square roots as there is no negative or positive complex numbers.

@Name-pb8mw

Жыл бұрын

@@Shreyas_Jaiswal yea but he’s explicitly said he wants the principal square root

@moksh6395

Жыл бұрын

You chad

@kumarharsh837

Жыл бұрын

You actually bothered to type that 🔥

You take 15 minutes to go from self-explanatory step A to step B and then 1.5 seconds to go straight to step Z. Very consistent and helpful.

@Nigelfarij

5 ай бұрын

(a+b)^2 = a^2 + b^2 + 2ab

@lgndary5715

2 ай бұрын

Its cuz it's a commonly known algebraic identity, and this video is for people new to the complex world, so it makes more sense to do those steps slowly with explanation and skip over explaining the identity which most people doing complex analysis would already know

My favourite method for finding the principal square root when answers are likely to be simple is to use √(a+bi)≡t/2 + bi/t (b≠0) where t=√{2(r+a)}, r=√(a²+b²). In this case, r=13, t=6 and the answer is 3+2i.

If you take a complex number z=a+ib and asked to find the square root of z you can write a as a=[√(a+b)/2+a/2]-[√(a+b)/2-a/2]

You just made the algebra a lot easier The concept is sqrt(5 + 12i) = a + bi -> 5 = a^2 - b^2, 12 = 2 * a * b But your process of how it's done made it so much easier 5 = U - V (12/2)^2 = 36 = UV a + bi = sqrt(U) + isqrt(V)

Wouldn't it be easier to convert to polar form?

@grassytramtracks

Ай бұрын

Bingo, this is where modulus argument form shine

Thank you,sir

When you tryna do complex mathematics but your parents are asleep

You can also use polar-coordinates. Its easier to take square roots, but you have to traduce it back to the algebraic version.

Generally best answered in polar coordinates

Alternatively you can just plug it into the definition of the principal branch of the complex square root: √z = √(a±ib) = ±√((|z|+a)/2)+i√((|z|-a)/2) (± is the same sign on both sides) |5+12i| = √(5²+12²) = 13 √(5+12i) = √((13+5)/2)+i√((13-5)/2) √(5+12i) = √(9)+i√(4) √(5+12i) = 3+2i I used the positive-imaginary convention rather than the positive-real convention for my definition of the principal branch, but both give the same solution when b is positive, so it doesn't really matter for this problem.

I would just use De Moivre's Theorem

*So √(a+bi) = √[a + 2•(b/2)•√-1] = √[a + 2•√(-b/2)]. From there, we solve the system that the two numbers sum to a and they multiply to get -b/2.*

I didn't get the logic from 3rd to 4th line

Great job, i like the way you explained this

What is the method where you find the two numbers that add to 5 and multiply to 36. I missed this trick

@sahibvirk9223

Жыл бұрын

You have to use trial and error for that ,factorise 36 and check which factors add up to 9

Never heard about this identity before, thank you

Thank you so much sir🫡

check the answer :)

Where did the 2 go from step 3 to 4?

@ronndan2004

Жыл бұрын

It disappears because the whole solution is based upon the formula a^2+2ab+b^2=(a+b)^2 When he developped to the shape of (a+b)^2 then the 2ab disappears. It's easy to square root (a+b)^2 because the answer is just (a+b)

Another fun way: √(5+2.2.3.i) Since sum of squares is positive, i is surely stuck to the smaller number √(3²+2.2i.3+(2i)²) =√(3+2i)² =3+2i

Thanks for the simple trick sensei

Thank you!

There are many ways like let it =a+bi or using e I will just go for easy way: √(5+12i)=√(9+2.3.2i+4i²)=√(3+2i)²=3+2i

Wrong there is no sqrt(-1) that's not how it works. i instead is a number when squared equals -1

Not a fun of asmr(expect touhou) but this was great deffinitly wanna see fails on this one

@bprpfast

3 жыл бұрын

Just did this for you kzread.info/dash/bejne/dKecyK2JnKuxkqg.html

Was just thinking could you use de moivres theorem here?

Thank you for this trick

Right, now I remember why my major was “media communications”

Ty so much for this vid

Those last two lines remind me of the opening Matrix chase when Agent Smith jumps 50 ft from one bldg to the next and the cop says, "That's impossible." 😊

This is where modulus-argument form shines

Thank you❤

This is good idea ,when you spell equations. Will OT help complex plane?

Thanks, this video made me the square root of complex numbers very easy

Than solve this, Z=2-3i

It was one of my main issues understanding how it solve similar when I was in 12grade.

What I would do I convert 5 + 12i into it's polar coordinates, r = sqrt(5² + 12²) = 13 Theta = atan(12/5) = 1.176005207095 Then take r×e^(i×theta/2), or r(cos(theta/2) + i×sin(theta/2)) You can add pi to theta/2 to then get the other square root A lot more work

I have even more easy method just requirement is you know pythagorean triplets Let m be the triplet(hypotenuse) formed by the no.s in the radical sign Then for given 5 and 12 m= 13 Here,x= 5 and yi=12i i.e y=12 Now take √[(m+x)/2] ±(depending upon the sign between x&y) √[(m-x)/2]i √(13+5)/2 + √(13-5)/2 =±(3+2i)

Professor ,are you doing the inverse ? Are you finding the equations?

I can't see why did you split the square root with a sum?

17 square root 17 times two smz equal 24

As someone trying to learn for a test tmr, i hate this...

Or the general way: (a+bi)^k = sqrt(a²+b²)^k*(cos(k*arccos(a/sqrt(a²+b²)) + i*sin(k*arccos(a/sqrt(a²+b²))) (I derived this a while ago, please comment if I remembered something wrong.)

@markoj3512

6 ай бұрын

Jep it’s a version of the equation of De Moivre's formula

@user-ue4hn7nk4s

5 ай бұрын

z^2 = 5+12i , r = 13, theta = arctan(12/5) z = e^i(theta +2*pi*k) k is a set of integers in this case k=0,1 obviously k can be ANY integers but after k = 0 and 1 the solutions will be periodic so no point since we only want the principle root let k = 0 and done

Ty

I got u u divide nine into three of the square roots equal infinity 😂😂😂😂😂

One more step is that the modulus of the 2 numbers is 13 , so we can solve instead of guessing. a+b 5, a-b=13, -> 2b=8 -> b=4

nice from 🇮🇳 INDIA

What about that 2 behind the √-36

How did you find 2×6?

You can double-check by finding that the square length is still equal

toughest way,, simplest way is Step1:take modulus of the |a+bi| in this case |5+12i|=√5²+12²=√169=13 Step2: for real value: add the value of a which in this case is 5 with the modulus= 13+5=18 now divide the number by 2 =18/2=9 now take square root of it=√9=3 now for imaginary value do them same steps just in this case subtract instead of adding the real value a(5) with the modulus 13=13-5=8/2=4 then underroot =√4=2 now add iota with 2 i:e 2i,,,,and put the sign b/w them that sign which is present b/w original values like in this case + sign is present so it would be 3+2i

Can’t this be done faster by converting what’s in the sqrt to polar form and using DeMoivre’s Theorem for the sqrt

@carultch

9 ай бұрын

Yes. The polar form of 5 + 12*i: 13*e^(i*arctan(12/5)) Take the square root of the magnitude, and cut the Argand in half. sqrt(13)*e^(i*arctan(12/5)/2) To find the components: real: sqrt(13)*cos(arctan(12/5)/2) Imaginary: sqrt(13)*sin(arctan(12/5)/2) Use the half-angle identities to rewrite the trig terms: cos(arctan(12/5)/2) = sqrt(1 + cos(arctan(12/5)))/sqrt(2) sin(arctan(12/5)/2) = sqrt(1 - cos(arctan(12/5)))/sqrt(2) Draw a 12-high x 5-wide triangle to work out: cos(arctan(12/5)) = 5/13 Thus: cos(arctan(12/5)/2) = sqrt((1 + 5/13)/2) = sqrt(18/26) = 3/sqrt(13) sin(arctan(12/5)/2) = sqrt((1 - 5/13)/2) = sqrt(8/26) = 2/sqrt(13) Thus we get: sqrt(5 + 12*i) = 3 + 2*i

@user-ue4hn7nk4s

5 ай бұрын

@@carultch z^2 = 5+12i , r = 13, theta = arctan(12/5) z = e^i(theta +2*pi*k) k is a set of integers in this case k=0,1 obviously k can be ANY integers but after k = 0 and 1 the solutions will be periodic so no point since we only want the principle root let k = 0 and done

Omggg this is so helpful

thanks sir 🙏💯 ever

There is an easier way We call delta = x + iy that verifies delta square = square root of 5+12i And you solve this equation

Um, it looks to me like you forgot about the two in front of the nested radical for -36 when, after factorization, you arrived at 9 and -4 as the values in the separated radicals; so, shouldn't the actual solution be 3+2i^(2^(1/2))?

@jase2053

Жыл бұрын

I was wondering where the 2 went. Thanks for confirming I am indeed _not_ crazy.

So this guy can talk, that's nice to know. It had always appeared as though he's mute

Conformed mapping please

Thankyou sir for this useful trick.... But i think it works with easier complex no.s As this can be difficult to use for complex no.s like root(3+i).

5+12i=5+12-/1=6ײ+5=-36/=×=3-×(+2)

Or you can use this trick to find the roots of complex number z (z=a+ib suppose) even quicker Write the square root as ±(x+i y) X^2= (modulus of z + a)/2 Y^2= (modulus of z - a)/2 Just put the values in and you'll have the answer. For √5+12i x^2=(√(12^2 + 5^2) + 5)/2=9 x=3 (no need to account for both positive and negative 3 since we are putting ± before the root already) Similarly, y=2 So the answer is ±(3+2i)

@user-ue4hn7nk4s

5 ай бұрын

idk if u need this a faster approach is z^2 = 5+12i , r = 13, theta = arctan(12/5) z = e^i([theta +2*pi*k]/2) k is a set of integers in this case k=0,1 obviously k can be ANY integers but after k = 0 and 1 the solutions will be periodic so no point since we only want the principle root let k = 0 and done

@gaycat599

5 ай бұрын

@@user-ue4hn7nk4s oh so convert to Euler and solve basically?

@user-ue4hn7nk4s

5 ай бұрын

@@gaycat599 exponential form yes, i dont really say its Euler yes its Euler's relation but i prefer to call it exponential form of a complex number

@gaycat599

5 ай бұрын

@@user-ue4hn7nk4s is there any reason behind it?

24 smz

You are so smart and nice

That's where I got lost. When we started making up numbers to guess...I was done with math.

is there any link between the integer solution to this problem and the fact that the norm of the number is integer (13)=sqrt(5²+12²)

I didn't get it at the end

He speaks!!

Just normalize the complex vector, calculate angle and project that divided by 2 out by the square root of the length of the original vector

we can say 5 is 9-4 so we have 3²+2*3*2i+(2i)² so = 3+2i

If i is a number who can have a min and a max will be then a range of values between. So can you aprox i , take a min and max , make it a range and sample the results ? Or i is not this ?

Five plus two eaqul seven minus one square roots minus one equal infinity cero 😂😂😂😂😂 cause minus doesn't can b square root 😊

Because there's a formula that explains it. The formula is √(a±b×i) = √[(a²+b²)+a] ± √[(a²+b²)-a].

Can somebody explain step 4? How did it turn to √9 + √-4

You cannot break a-square rot in2 parts

Plz do sqr root of 4+3i

@duggirambabu7792

Жыл бұрын

*√(4 + 3i)* = √((8 + 6i) / 2) *= (3 + i) / √2* *= (3 + i)√2 / 2*

Can someone explain why you can break up the brackets like this in step 3?

@oenrn

Жыл бұрын

(√a + √b)² = a + b + 2√(ab) Take square roots: √a + √b = +- √(a + b + 2√(ab)) Since he just wants the principal root, the +- can be ditched.

Divide and multiply by root 13 and you'll get root 13 x e^i/2 . (arcos 5/13). Is that the same ans ?

@SanePerson1

Жыл бұрын

Yes.

Complex numbers. I still have nightmares after 40 years

Too confusing for me I’m taking Algebra Keystones and thankfully ik this won’t be on it but i failed the first time I took it so😅

(x+yi)² = 5+12i => x² - y² + 2xyi = 5 + 12i => {x² - y² = 5 and 2xy = 12 =>xy = 6 => y = 6/x} => (substitute) {x² - (36/x²) -5 = 0 and y = 6/x} => (multiply everything from the first equation with x² andlet x² be equal to t) {t²-5t-36 = 0 and y = 6/x} => t is: (5 +- sqrt(25+144)) / 2 = (5+-13) / 2 => t = 9 or -4 => x = +-sqrt(9) = 3 (and x = sqrt(-4) is not real) => y = 6/x => (substitute) y = -2 or 2 So sqrt(5+12i) = -(3+2i) or 3+2i but -(3+2i) is not an answer considering the initial number was positive

I was fine until "now think of two numbers"

@dannkod

Жыл бұрын

Then use the formula: √(a±√b) = √((a+c)/2) ± √((a-c)/2) where c=√(a²-b).

I'm from India🇮🇳 and u make more video from the √of complex no.

@aszpnh1538

Жыл бұрын

Why do Indians keep saying their Indian?

@user-ue4hn7nk4s

5 ай бұрын

@@aszpnh1538 they think they're so smart thats why, me personally id destroy these so called indians in any math competition

How to write for infinite root sequence.

Could sb explain how he got 9 and 4 pls?

@pepperjohnns

4 ай бұрын

He's trolling man 💀

I would've turned it into polar form, then found half the argument and the root of the modulus.

oooooo thank you mr

Wait wait wait, what the hach did just happen right there! How th did you just factor 2 complex numbers that are being added like that?

Five plus twelve 17 smz of square root of 17 smz equal not the same fractions so 😊

Infin😢 cero cause minus six minus square root of negative one it's seven 😅😅😅😂😂😂😂😂

No entiendo el tercer paso

Can someone explain, how sqrt 5+2 sqrt -36 can be sqrt 9 + sqrt -4

Sooo we are done.