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expscv
Mar31-04, 06:53 PM
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
=)

cookiemonster
Mar31-04, 06:58 PM
You don't even have to go through that much work.

4 = \sqrt{16} = -4

cookiemonster

Chen
Mar31-04, 07:08 PM
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|

expscv
Mar31-04, 08:19 PM
hehe oh? i was told that it sippose to be wrong
in somewhere that i forgot.

cookiemonster
Mar31-04, 10:36 PM
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.

cookiemonster

expscv
Mar31-04, 10:50 PM
yep i agree, i remmber now haha