# Complex numbers help

by kay
Tags: complex, numbers
 P: 12 We know that i^3 is -i . But I am getting confused, because I thought that i can be written as √(-1) and i^3 = √(-1) × √(-1) × √(-1) = √(-1 × -1 × -1) = √( (-1)^2 × -1) = √(1× -1) = √(-1) = i ( and not -i ). Please help. Sorry I couldn't use superscript because I was using my phone.
 Mentor P: 18,036
 P: 1 i definitely is not \sqrt{-1}. If you like (abuse of notation) \sqrt{-1} = \pm i Using (this not correct notation) \sqrt{-1}^3 = \pm i. Much better is of course i^3 = (i*i)*i = -1*i = -i
P: 12
Complex numbers help

i am really not familiar with Euler's constant that much, and complex calculus, but thanks. :)
P: 12
 Quote by dieterk i definitely is not \sqrt{-1}. If you like (abuse of notation) \sqrt{-1} = \pm i Using (this not correct notation) \sqrt{-1}^3 = \pm i. Much better is of course i^3 = (i*i)*i = -1*i = -i
I didn't understand anything. :|
P: 7
 Quote by kay i am really not familiar with Euler's constant that much, and complex calculus, but thanks. :)

The link given by micromass has everything you need to know and you don't need to know Euler's Formula to understand what he meant. I suggest read (not skim) the link provided by micromass.
 P: 61 When you got to this point: $$\sqrt{-1}\cdot\sqrt{-1}\cdot\sqrt{-1}=\sqrt{(-1)\cdot(-1)\cdot(-1)},$$ you made a mistake since $\sqrt{a}\sqrt{b}=\sqrt{ab}$ isn't true when $a,b\lt0$.

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