B Complex Numbers in a Simple Example that I am Very Confused

1. Jan 23, 2017

Arman777

There a simple math example that I am confused $(\sqrt {-4})^2$
Theres two ways to think
1-$\sqrt {-4}=2i$ so $(2i)^2=4i^2$ which its $-4$
2-$\sqrt {-4}$.$\sqrt {-4}$=$\sqrt {-4.-4}=\sqrt{16} =4$

I think second one is wrong but I couldnt prove how, but I think its cause $\sqrt {-4}$ is not "reel" number so we cannot take them into one square root

Last edited by a moderator: Jan 23, 2017
2. Jan 23, 2017

Staff: Mentor

3. Jan 23, 2017

Logical Dog

Last edited: Jan 23, 2017
4. Jan 23, 2017

Staff: Mentor

It's "real" number, not "reel" number.

Formula 2 is incorrect. $\sqrt a \sqrt b = \sqrt{ab}$ only if both a and b are nonnegative.

5. Jan 23, 2017

Arman777

Thx a lot.My typo

6. Jan 24, 2017

Svein

Well - in the complex domain things are not the same as they are in the real domain. First, you have $-4=4e^{\pi i}$ so you might think that $\sqrt{-4}=2e^{\frac{\pi}{2} i} = 2i$. But you also have $-4=4e^{3\pi i}$ and therefore $\sqrt{-4}=2e^{\frac{3\pi}{2} i}=-2i$. Squaring either of the roots brings you back to $-4$.

Last edited: Jan 24, 2017