# Need a quick help with a simple identity

1. May 26, 2006

### hooker27

Hi
Where is the error is this 'identity'?:

$$(-e^i)^{\frac{1}{2}} = (-1)^{\frac{1}{2}}*e^{\frac{i}{2}}$$

My calculator says that the right side is minus one times the left but I can't see the mistake I'v made. Help me please, thanks.

2. May 26, 2006

### HallsofIvy

Staff Emeritus
What happened to the "i" on the left? I assume that was a typo and you mean $(-e^i)^{\frac{1}{2}} = (-1)^{\frac{1}{2}}*e^{\frac{i}{2}}$

You have to be careful about apply "laws of exponents" to complex numbers- we all remember the "proof" that 1= -1:
$$1= ((-1)(-1))^\frac{1}{2}= (-1)^\frac{1}{2}(-1)^\frac{1}{2}= (i)(i)= -1$$

3. May 26, 2006

### hooker27

- I am not sure which 'i' (and 'typo') you are reffering to, your 'indentity' is, as far as I can say, identical to mine.

- So make this clear for me: when can I use the fact that $$(A*B)^x = A^x*B^x$$ when A,B are complex and x real? (I have little knowledge of complex analysis)

I need to get $$(-e^i)^x = something * e^{ix}$$ but I am not sure what the 'something' should be, obviously it is not $$(-1)^x$$, could you please help me?

Thanks, H.

4. May 26, 2006

### HallsofIvy

Staff Emeritus
Okay, I didn't see that you had "i/2" rather than "1/2". Maybe I need to have my eyes checked!

Actually it is but (-1)x, like most complex valued functions, is multi-valued. You need to state which value you are using.

In your particular case, $(-1)^\frac{1}{2}$ has two values: i and -i.

Last edited: May 26, 2006