Imaginary Numbers: Solve i^(4/3) Equation

[DivGradCurl]

Folks, I was just wondering why I can write:

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

Regards

 

[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.

 

[dextercioby]

Sure,complex exponentiation is multivalued,meaning that "i" can be written in an infinite 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]

Sure,complex exponentiation is multivalued,meaning that "i" can be written in an infinite 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]

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]

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]

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.

 

[DivGradCurl]

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}