# Imaginary nos.

1. Feb 17, 2005

Folks, I was just wondering why I can write:

$$i^{4/3}=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$$

Regards

2. Feb 17, 2005

### Curious3141

Do you know the identity $$e^{i\theta} = \cos\theta + i\sin\theta$$ ?

$$i = e^{i\frac{\pi}{2}}$$

Use that, and the basic rules for exponentiation to get the result. Then express it in Cartesian form.

3. Feb 17, 2005

### dextercioby

Sure,complex exponentiation is multivalued,meaning that "i" can be written in an infinte ways using various arguments of the exponential...But worry about that less and use the formula which (though incomplete,hence inaccurate) was already given...

Daniel.

4. Feb 17, 2005

### Curious3141

True, I know exponentiation is multivalued, but since he just wanted the principal value, that's what I gave him.

5. Feb 17, 2005

### dextercioby

Sorry to be nit-picking,but nowhere in his post is that "he just wanted the principal value" noticeable...

Daniel.

P.S.It doesn't matter,it's good if the OP got the simple part,at least.

6. Feb 17, 2005

### Curious3141

OK...

$$i^{\frac{4}{3}} = e^{i(\frac{\pi}{2} + 2k\pi)(\frac{4}{3})}$$

The possible values in Cartesian coords are what was given, $$-\frac{1}{2} -i\frac{\sqrt{3}}{2}$$ and 1.

I know this. The point is, he just wanted help to "see" the first answer.

7. Feb 17, 2005

### ehild

You can write if you add "or 1, or

$$-\frac{1}{2}-i\frac{\sqrt{3}}{2}$$",

as

$$i^{4/3}=(i^4)^{1/3}=\sqrt[3]{1}$$.

1 has got n different n-th roots,

$$\sqrt [n] {1} = \cos(\frac{2\pi}{n}k) + i \sin (\frac{2\pi}{n}k)$$
k=0, ...n-1.

8. Feb 17, 2005

Thank you for all the help! I think I finally get what you're saying...

$$z=i\, ^{\frac{4}{3}}=\left( i^4 \right)^{1/3} = \sqrt[3]{1}$$

$$h= 1 = \mbox{cis } 0$$

$$z=h_w ^3 = \mbox{ cis } \left( \frac{0+2k\pi}{3} \right) = \cos \left( \frac{2k\pi}{3} \right) +\mbox{ } i\mbox{ } \sin \left( \frac{2k\pi}{3} \right) = -\frac{1}{2} +\mbox{ } i \mbox{ } \frac{\sqrt{3}}{2}$$