# What is the square root of i^2

1. Oct 4, 2006

### murshid_islam

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

Last edited: Oct 4, 2006
2. Oct 4, 2006

### meee

im guessing its plus&minus

3. Oct 4, 2006

### 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)

4. Oct 4, 2006

### 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.

5. Oct 5, 2006

### murshid_islam

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 $$\theta = \tan^{-1}\left( y \over x \right)$$

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?

Last edited: Oct 5, 2006
6. Oct 5, 2006

### 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.

7. Oct 5, 2006

### 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?

8. Oct 7, 2006

### Robokapp

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.

9. Oct 8, 2006

### Data

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!).