Square root of i

How to find the square root of i, as a complex number.

[mdashrafulislam]

I completely did it myself. And exactly same with you. I am astonished.

[snehadas]

Thanks for explaining so well. I am so grateful to you.

[overlordprincekhan]

Bro, I have some questions.


At first, why should I even declare that root of i is a complex number? we don't any idea what can be in there.


secondly, even if we have declared that 'i' as a complex number, algebraically, I would get a real number as a result of root of i.


(i)^(1/2)=a+bi
=> i = a^2 + 2abi -b^2
=>i-2abi=a^2 -b^2
=> i(1-2ab) = a^2 -b^2
=> i = (a^2 -b^2 )/(1-2ab)


So, why didn't that? I would be pleased if you answer my questions

[keezhang]

z=sqrt(w) is a function there must have a unique value, called the principal square root.. For example sqrt(1)=1 not -1, and sqrt(-1)=i not -i. If you write sqrt(w)=r(cosθ+isinθ) then you must choose θ such that -π/2<θ≤π/2

[FernandoGonzalez-soql]

Astonished that this is solvable with "simple" algebra!

[ezzatabdo]

Excellent and nice solution, thanks sir.

[byroodsemimiya]

i am grade 10 and my teacher put that as our long test. is it normal? i have never encountered this ever.

[jimmonroe]

Thank you! Though I think the solution is just the positive root as the square root operation only returns positive roots. Also, if you look at the negative root on the complex plane, -(sqrt(2)/2+isqrt(2)/2) is equivalent to i^(3/2)

[keeteo]

You have shown that i has two square roots, however sqrt(i) stands for THE principal square root of i, which is unique, namely 1/sqrt(2)+i/sqrt(2).

[timjack]

How do we know it's a complex number though?

[UnbreakablePickaxe]

you can write i in eular form as e^(i*(pi/2+2npi)) and divide the power by 2, you get e^(i*(pi/4+npi)). This has 2 solutions when n is any two consecutive integers

[MSKofAlexandria]

If you apply Eulers
√x = x1/2
so we can write i = (eiπ/2)1/2 = eiπ/4
so the fourth root of -1 is (√2/2) + (√2/2)i
i = √-1
then √i = 4th root of -1 which is (√2/2) + (√2/2)i
2 divided by 2 equals 1, it can be simplified to (√1) + (√1)i
the sqrt of 1 is 1, so it can be simplified further to (1)+(1)i
(1)+(1)i equals 1+i
if we apply a = a
√i = 1+i

[DavidM-duxo]

Dang. I’ve been taking so many Calc courses recently I went into Euler’s formula and took the square root of e^(i*pi/2), and then turned that back into a+bi form. Guess your way would’ve been a lot faster 😂.

[いーあー-hm]

I have a question.
⭕️√4=2
❌√4=±2
Why √i=±(1/√2+ i/√2) ?

[kchromaticpiano]

Thanks! Though my way is a bit different, there based on the some concept of a+bi. Thanks again!

[Mateusz-Maciejewski]

What you've done has one error. You can't write i=sqrt(-1) and sqrt(i)=(-1)^(1/4). As you see, there are two square roots of i and this is the same with any nonzero number. So it'd be justifiable to write sqrt(-1)=±i. Moreover, there are four 4-th roots of (-1) so you can't say that sqrt(i)=(-1)^(1/4) since left hand-side has two solutions, whereas right-hand side has four.

[anonymouscheesepie]

sqrt(i) = a + bi
i = a^2 + 2abi - b^2
i = a^2 - b^2 + 2abi
Re(i) = 0 and Im(i) = 1

Set real and imaginary components equal

a^2 - b^2 = 0
2ab = 1

a^2 = b^2

a = +-b

2(+-b)b = 1

+-2b^2 = 1

b^2 = +-1/2

b = sqrt(1/2) (b must be real)

a = +-b
a = +-sqrt(1/2)

So sqrt(i) = +-sqrt(1/2) + sqrt(1/2)i

[syedasayyada]

####Why we not took "i" value in equation, we have three "i"s in the same equation?
If we take some value for some thing we hv to put that for all "i"s.

[SKREFI]

I atempted this myself first:

sqrt(i) = x
so x * x = i
right?

i = i^5 =>
x = i^(5/2) = i^5 / i^2 = -i wich is false
OR
x = i^(2.5) = i^2 * sqrt(i) and this
is kind of recursive and somewhere I am doing a mistake probably and can't find where

[meifray]

±0.5=<+0.5, -0.5>
=><<+1/√2, +i/√2>, <-1/√2, -i/√2>>
you have shown the real number part, but what if the complex number part of answer?