# Imaginary nos.

Folks, I was just wondering why I can write:

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

Regards

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Curious3141
Homework Helper
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.

dextercioby
Homework Helper
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.

Curious3141
Homework Helper
dextercioby said:
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.
True, I know exponentiation is multivalued, but since he just wanted the principal value, that's what I gave him.

dextercioby
Homework Helper
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.

Curious3141
Homework Helper
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.

ehild
Homework Helper
thiago_j said:
Folks, I was just wondering why I can write:

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

Regards
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.

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

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

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

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