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

Regards

- Thread starter DivGradCurl
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- #1

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

Regards

- #2

Curious3141

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[tex]i = e^{i\frac{\pi}{2}}[/tex]

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

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Daniel.

- #4

Curious3141

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

Daniel.

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Daniel.

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

- #6

Curious3141

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[tex]i^{\frac{4}{3}} = e^{i(\frac{\pi}{2} + 2k\pi)(\frac{4}{3})}[/tex]

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

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

- #7

ehild

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You can write if you add "or 1, orthiago_j said:

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

Regards

[tex]-\frac{1}{2}-i\frac{\sqrt{3}}{2}[/tex]",

as

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

1 has got n different n-th roots,

[tex]\sqrt [n] {1} = \cos(\frac{2\pi}{n}k) + i \sin (\frac{2\pi}{n}k) [/tex]

k=0, ...n-1.

- #8

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

[tex]h= 1 = \mbox{cis } 0[/tex]

[tex]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}[/tex]

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