What is the square root of i^2

[murshid_islam]

what does \sqrt{i^2} equal to? is \sqrt{i^2} = i or \sqrt{i^2} = \pm i?

 

[meee]

im guessing its plus&minus

 

[matt grime]

The square root is usually taken to be the principal branch of the function. I.e. the root whose argument lies in [0,pi), so you would conventionally obtain i.

In general the n'th root of a complex number is taken to be the principal root unless otherwise stated: the root with argument in the interval [0,2pi/n)

 

[Data]

I gave you the usual definition of the square root function in your other thread. You can use that to figure out what it is.

 

[murshid_islam]

I gave you the usual definition of the square root function in your other thread. You can use that to figure out what it is.

you said, \sqrt{z} = z^{1/2} = \sqrt{r}e^{i\theta/2}.

So let's try out this definition. If z=-1, then we write z = e^{i\pi}, and we get \sqrt{z} = e^{i\pi / 2} = i. As you might expect.

but if z = x + iy, then r = \sqrt{x^2 + y^2} and \theta = \tan^{-1}\left( y \over x \right)

if z = -1 = -1 + 0.i, then x = -1, y = 0.

r = \sqrt{\left(-1\right)^2} = \sqrt{1} = 1

\theta = \tan^{-1}\left( {0} \over {-1} \right) = 0

\sqrt{z} = \sqrt{-1} = \sqrt{r}e^{i\frac{\theta}{2}} = \sqrt{1}e^{0} = 1

but where did i go wrong?

 

[matt grime]

Because arctan only returns values in the region -pi/2 to pi/2, so it is wrong to say that theta is arctan(y/x) alone. You must also use your common sense to see that the argument is actually pi, not zero. It is correct to say that tan(theta)=y/x, but that is *not* the same as theta =arctan(y/x) at all.

 

[murshid_islam]

yeah, i can see that \theta = \pi from the position of z = -1 in the argand plane. so,

\sqrt{z} = \sqrt{-1} = \sqrt{r}e^{i\frac{\theta}{2}} = \sqrt{1}e^{i\frac{\pi}{2}} = 1.(\cos{\frac{\pi}{2}} + i.\sin{\frac{\pi}{2}}) = i

am i right?

 

[Robokapp]

what does \sqrt{i^2} equal to? is \sqrt{i^2} = i or \sqrt{i^2} = \pm i?

This is my assumption:

\sqrt{i^{2}} = \sqrt{-1} = i

i^{2} is equal to -1 and square root of -1 is defined as i.

 

[Data]

you said, \sqrt{z} = z^{1/2} = \sqrt{r}e^{i\theta/2}.but if z = x + iy, then r = \sqrt{x^2 + y^2} and \theta = \tan^{-1}\left( y \over x \right)

That expression for r works for all z, but the one for \theta only works for x>0, assuming you're using the usual branch of arctan. In general finding the polar representation of a complex number involves solving

\sin{\theta} = \frac{y}{r}
and
\cos{\theta} = \frac{x}{r}

simultaneously (taking \arctan (y/x) makes the possible solutions obvious though!).