# Beware the complex number

## Main Question or Discussion Point

$$4 = \sqrt {4*4} = \sqrt {4*4*i^4} =\sqrt {i^2*4 *4*i^2}$$

$$= i\sqrt{4}*i\sqrt{4} =2i*2i =-4$$

this is wrong but which setp
=)

You don't even have to go through that much work.

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

That should be:
$$4 = |\sqrt {4*4}| = |\sqrt {4*4*i^4}| = |\sqrt {i^2*4 *4*i^2}|$$
$$= |i\sqrt{4}*i\sqrt{4}| = |2i*2i| = |-4|$$
Or just:
$$4 = |\sqrt{16}| = |-4|$$

hehe oh? i was told that it sippose to be wrong
in somewhere that i forgot.

Well, it's wrong in that you're taking the wrong sign in front of the square root, but that's about it. Squares and square roots tend to generate extra solutions that are not necessarily correct. This is one such case.