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Find the complex cube roots of 8(cos(4pi/5)+isin(4pi/5))

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atrinity | (Level 1) Honors

Posted May 20, 2013 at 8:53 PM via web

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Find the complex cube roots of 8(cos(4pi/5)+isin(4pi/5))

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mathsworkmusic | (Level 1) Educator

Posted May 20, 2013 at 9:14 PM (Answer #1)

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We have the complex number

`z = 8(cos((4pi)/5) + isin((4pi)/5))`

First convert z into polar coordinates (argument,modulus) on the complex plane.

` `We can see that the argument of z,  `arg(z) = (4pi)/5`, and that the modulus of z, `|z| = 8`.

Using de Moivre's Theorem to obtain the argument of `z_c = root(3)(z)` we note that

`arg(z_c) = (4pi)/5(1/3) = (4pi)/(15)`

Again using de Moivre's Theorem to obtain the modulus of `z_c` we note that

`|z_c| = root(3)(|z|) = root(3)(8) = 2`

The cube roots `z_c` of `z` are then all complex numbers with argument `(4pi)/15` and modulus 2. These take the form

`z_c = 2(cos((4pi)/15 +2npi) + isin((4pi)/15 + 2npi))` , `n = 0,1,2,...`

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