How to find the principal square root of a complex number

The principal value of the complex number 5+12i

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bprp #fast

[TheMathManProfundities]

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.

[chinjunyuan]

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

[KamustaJohn]

bro can make solving world hunger look easy 💀

[brainloading]

Bro casually committing multiple war crimes at once

[pi_xi]

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

[nanamacapagal]

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)

[TeamGCS]

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)

[JUSTREGULARSCREAMINGAAHH]

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

[matthiaspfau]

Generally best answered in polar coordinates

[__-bmej]

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

[georgesmyrnis]

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.

[mohammadshamsnur]

If you take a complex number z=a+ib and asked to find the square root of z you can write a as

[grassytramtracks]

This is where modulus-argument form shines

[Nalogpi]

5+12i = (a+bi)^2
5+12i = a^2 - b^2 + 2abi
a^2 - b^2 = 5 [Equate real part]
2ab = 12 [ Equate imaginary part ]
ab = 6

Now you can either factor 6 as 3*2 and put them into the equation or you can solve it

Then the answer is equal to a+bi

[vampire_catgirl]

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

[NotSoChillBozo]

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

[muhammadahmadabbas]

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

[bababrainstormer]

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)

[itzz_minecraftplayer]

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

[inspiredforsure]

Just use the formula ±(√Izl+x/2 ± i√lzl-x/2)

√5+12i ->

lzl = √5^2+12^2
= 13

Applying the formula we get :
±(√13+5/2 ± i√13-5/2)
±(√9 ± i√4)
±(3 ± i2)

In the equation i had positive sign so:

±(3 + i2) - answer