# Under root -i

1. Nov 28, 2006

### mohamen

wat is under root - i ???

please anser this

2. Nov 28, 2006

### Hurkyl

Staff Emeritus
Do you mean $\sqrt{-i}$? You can figure it out yourself! If $a + bi = \sqrt{-i}$, then what does the definition of square root tell you?

3. Nov 28, 2006

### dextercioby

HINT:

$$-i=e^{-\frac{i\pi}{2}+2k\pi} , \ k\in\mathbb{Z}$$

Daniel.

4. Nov 28, 2006

### Gib Z

Hey guys, dont post here often, i will more from now on. Anyway, dexter got that from the identity e^(ix)= cos x + i sin x.

5. Nov 29, 2006

### mohamen

will it be solved by the de mouvies theorm..........i dont think so...

6. Nov 29, 2006

### dextercioby

You mean "de Moivre". Yes, it will, since that theorem is a trivial consequence of the fact that

$$\left(e^{ix}\right)^{n}=e^{inx}$$

Daniel.

7. Nov 29, 2006

### HallsofIvy

Staff Emeritus
I personally like Hurkyl's suggestion best but WHY don't you think deMoivre's formula will work? It can be used to find any root of any complex number.

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