# Beware the complex number

1. Mar 31, 2004

### expscv

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

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

this is wrong but which setp
=)

2. Mar 31, 2004

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

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

3. Mar 31, 2004

### Chen

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

4. Mar 31, 2004

### expscv

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

5. Mar 31, 2004

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.