Understanding Complex Numbers: Find the Answer

[kay]

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.

 

[micromass]

https://www.physicsforums.com/showthread.php?t=637214

 

[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

 

[kay]

https://www.physicsforums.com/showthread.php?t=637214

i am really not familiar with Euler's constant that much, and complex calculus, but thanks. :)

 

[kay]

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. :|

 

[electronicsguy]

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.

 

[HakimPhilo]

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 $a,b\lt0$.