Question

Discussion : Square root of imaginary number.

Answer 1

I was thinking that we have square root of a negative number but we don't have any for :
\(\large{\sqrt{-i}}\)

Right?

Answer 2

Do we have any?

Answer 3

Or any simplified form for that?

Answer 4

Ok! Let us first take it as : \(\large{\pm \sqrt{i}}\) .

Answer 5

As I think we can do this by using euler's formula.

Answer 6

you're on the right track.

Answer 7

Hmn . Let us take \(X^2 = i\) :
X = \(\sqrt{i}\)

Surely X will be a complex number. So I have : X = x + iy .

Answer 8

you can use DeMoivre's formula, with a modified interpretation

Answer 9

So I have : \((x+iy)^2 = i\)

Expanding this : \(x^2 + i^2y^2 +2xiy = i\)
\(x^2-y^2 + 2xiy = i \)

that means 0 = \(x^2 - y^2\) and \(2xy = 1 \)

From equation \(x^2-y^2 = 0\) I have : \(x^2 = y^2\)
\(x= \pm y\)

So we have : \(2xy=1\) \(\implies \) \(2(-y)(y) = 1\) . From this I get :

\(-2y^2 = 1\)

Hmn so that I get \(y^2 = \frac{-1}{2}\)

But that is \(not\) possible. so x= y is the case I have now.

Answer 10

continue, @mathslover . you're still on the right track.

Answer 11

So I get x = y = \(\large{-\frac{1}{\sqrt{2}}}\)

What shall be the next step now ?

Answer 12

if x=y, then 2xy=1 implies \[x=\pm \frac{1}{\sqrt 2}\]
so your X=x+yi becomes...

Answer 13

your \[X=x+iy\] becomes
\[X=x+xi=x(1+i)\]
or \[X=\pm\frac{1}{\sqrt 2}(1+i)\]

Answer 14

are you sure the \(x,y\) are of the same sign?

Answer 15

With Euler's formula, or the "e" notation, this is much easier to do.
\(\sqrt{-i}=(-i)^{1/2}=\left(e^{-\frac{\pi}{2}i}\right)^{1/2}=e^{-\frac{\pi}{4}i}\).
This corresponds to a point on the unit circle, the argument is \(-\pi/4\) (-45 degrees) and the modulus is still 1, so if you want to write it in normal x+iy notation, you use the fact that there is now a 45-45-90 triangle involved, so it is:

\(\dfrac{1}{2}\sqrt{2}-\dfrac{1}{2}\sqrt{2}i\)