How to find the square root of a complex number

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#math #algebra #mathbasics

[محمدالنجفي-ظه]

In iraq we solve it this way
Let X+Yi =Sqr(5+12i)
Then square both sides we get
X²-Y²+2XYi=5+12i
So we spilt the imaginary part and real part and make equations:
Real part =Real part ;
X²-Y²=5
.
.
Imaginary part= imaginary part;
2XY=12
XY=6
We take either x or y it's up to you.

.
.
Put (2) in (1)
We get ;
(36/Y²) -Y²=5
Multiple the equation (Y²)
We get :
36-Y⁴=5Y² ..*(-1)
-36+Y⁴= -5Y²
Y⁴+5Y²-36=0
(Y²+9)(Y²-4)=0
EITHER Y=±2.
.
OR Y²+9=0 ( DON'T BELONG TO REAL NUMBERS SO WE GET RID OF IT)
SO WE GET Y=±2
THEN X =±3
SO OUR TOW SOLUTIONS
WE GET
C(1) =3+2i
C(2)=-3-2i
.

[manudude]

I just do it as (a+bi)^2 which gives a^2-b^2+2abi, set a^2-b^2=5 and 2ab=12, no messing around

[seanwilkinson]

Things like this are so cool to me. It's been almost twenty years since I had to do high level math, and listening to mathematicians get excited about weird and obscure math stuff is my late-night ASMR. 🤯👍👍

[GillesF]

I prefer this (universal) way to get the answer:

√(5 + 12i)

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|| z = 5 + 12i
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|| module: |z| = √(5² + 12²) = √(25 + 144) = √169 = 13
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|| argument: θ = arctan(12/5) = 1.176 rad
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|| z = 13·e^(1.176i)
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|| √z = √[13·e^(1.176i)]
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|| √z = √(13)·√(e^(1.176i))
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|| √z = √(13)·(e^(1.176i))^(1/2)
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|| √z = √(13)·(e^((1/2)·1.176i))
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|| √z = √(13)·(e^0.588i)
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|| √z = √(13)·(cos(0.588) + i·sin(0.588))
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|| √z = √(13)·cos(0.588) + i·√(13)·sin(0.588)
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|| √z = 3 + 2i
||

/// final result:

■ √(5 + 12i) = 3 + 2i

🙂 AND THANK YOU FOR ALL YOUR VERY INTERESTING VIDEOS!!!

[mathmachine]

The formula for √(a+bi) = √((√(a²+b²)+a)/2) + sgn(b)i*√((√(a²+b²)-a)/2)

Alternatively, you can just compute one component, and the other component is b/(2*(that component)).

√(5+12i) = √((13+5)/2) + i*√((13-5)/2) = √(9)+i*√(4) = 3+2i.

[TheMathManProfundities]

√(a+b+2√ab) isn't necessarily √a+√b It depends on the values of a and b. For example, a=b=-1 gives √(a+b+2√ab)=0 but √a+√b=2i. The issue here is the assumption that √ab is always equal to √a√b. This assumption is incorrect.

[TurtleGod]

Thanks for this, we did imaginary numbers for the first time in my algebra class and im sure we might have something similar to this.

[guillermolp]

I have a question
If I have y=√x, where y and x belong to the field of complex numbers. Is it necessary to specify if the symbol "√" refers to the principal root? Is there a convention?
For example: y=√4. If "√" is the principal square root, then y takes a single value. But if "√" refers to finding all the roots, then y takes two values. I give this example, because if I am working in the field of real numbers, there is a convention, and y only takes one value which is 2.
Thanks for the answers

[عليعليعلي-ظخج]

18-6i/1+9x18=1-2yi/1-yi+2y2 pleas solve this equation this from my teacher

[JakubS]

(x+yi)^2=5+12i, and then expand and make simultaneous equations from the real and imaginary parts

[Jominer]

interesting, i thought you’d chuck it in cis form and use de moivres theorem. or you could said (a+bi)^2=√5+12i and use simultaneous equations

[yannickweber]

Hello, thanks for the video!
If i may, i have a stupid question: With a+b=5 and ab=-36, how do you efficiently find that a=9 and b=-4 ? i can't remember how to solve this without just trying out the different possibilities.

[rogierownage]

Call the unknown 'z'
z = sqrt(5+12i)
Let z = a+bi (where a and b are real) and then square both sides
a^2 -b^2 + 2abi = 5+12i
The real parts must be equal, and the imaginary parts must be equal. Therefore:
a^2 -b^2 = 5
2ab = 12

Isolate a in the second equation
a = 6/b
Fill in this a in the first equation
36/b^2 - b^2 = 5
Multiply everything by b^2, put everything on the left, and divide by -1 so the b^4 becomes positive (makes it easier)
b^4 +5b^2 - 36 = 0
Let u=b^2
u^2 +5u -36 = 0
(u + 9)(u - 4) = 0
b^2 = -9 or b^2 = 4
b is real so the first answer is invalid
b=+-2

Fill this b into the other equation. This eliminates the +- because b^2 is 4 either way.
a^2 - 4 = 5
a^2 = 9
a = +- 3

This gives you 4 answers, but only 2 are valid. You can check this manually but i just used a calculator.

This gets you:
z = 3+2i or z = -3-2i

[khansahab]

You could also just do DeMoivre’s Theorem

[glitchy]

A nice formula:
sqrt(z) = sqrt(abs(z)) * sgn((sgn(z)+1))

[TheMathManProfundities]

Let z=5+12i. z=13e^{iArg(z)} where Arg(z)∈(-π, π] so √z=√(13)e^{iArg(z)/2}.

[RealKangarooFlu]

How does sqrt(a+b+(2 * sqrt(a+b))) reduce to sqrt(a) + sqrt(b)??

[RealQinnMalloryu]

{5x+5x ➖ }+{12xi+12xi ➖ }={10x^2+24x^2i^2}=34x^4i^4 2^17x^4i^4 2^17^1x^2^2i^2^2 1^1^1x^1^1i^1^2 i^1^2 (x ➖ 2x+1i)

[Limited_Light]

(sqrt(a)+sqrt(b))^2=a + b + 2 * sqrt(a * b) = a + b + sqrt(4 * a * b)

a + b = 5.

12 i = sqrt(-144) = sqrt(4 * -36).
a * b = -36

b = 5 - a

a(5 - a) = -36
a^2 - 5a - 36 = 0
(a - 9)(a + 4) = 0

a = 9 or a = -4.

sqrt(5 + 12i) = sqrt(9) + sqrt(--4) = 3 + 2i

[gp-htug]

What if the sqrt is not solvable as shown? Thanks