High School Complex Numbers in a Simple Example that I am Very Confused

[Arman777]

There a simple math example that I am confused $(\sqrt {-4})^2$
there's 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 couldn't prove how, but I think its cause $\sqrt {-4}$ is not "reel" number so we cannot take them into one square root

 

[fresh_42]

I suggest to give this a read: https://www.physicsforums.com/insights/things-can-go-wrong-complex-numbers/

 

[Logical Dog]

edit: wrong suggestion

the formula for finding nth roots of a complext number is this (slightly complicated):

https://www.math.brown.edu/~pflueger/math19/1001%20Complex%20roots.pdf

 

[Mark44]

2-$\sqrt {-4}$.$\sqrt {-4}$=$\sqrt {-4.-4}=\sqrt{16} =4$

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

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.

 

[Arman777]

Thx a lot.My typo

 

[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.