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[tex]z^3=64(\cos219^\circ+i\sin219^\circ)[/tex]

Note that [tex]64=4^3[/tex]. The third roots are then given by

[tex]z=4\left(\cos\left(\dfrac{219+360k}3\right)^\circ+i\sin\left(\dfrac{219+360k}3\right)^\circ\right)[/tex]

where [tex]k=0,1,2[/tex], so that

[tex]z=\begin{cases}4(\cos73^\circ+i\sin73^\circ)\\4(\cos193^\circ+i\sin193^\circ)\\4(\cos313^\circ+i\sin313^\circ)\end{cases}[/tex]

The cube roots of the considered complex number are:-

  • z = (4(cos(73°) + i*sin(73°))
  • z = (4(cos(193°) + i*sin(193°))
  • z = (4(cos(313°) + i*sin(313°))

What is De Moivre's theorem for exponentiation of complex numbers?

Any complex number [tex]z = a + ib[/tex] can  be written in polar form as:

[tex]z = r(\cos(\theta) + i\sin(\theta))[/tex]

Raising this to nth power (n being an integer), we get:

[tex]z^n = r^n (\cos(n\theta) + i\sin(n\theta))[/tex]

For this case, the complex number given is:

[tex]64(\cos(219^\circ) + i\sin(219^\circ))[/tex]

Let z be the family of complex numbers whose cube will give that above complex number.

Then we have:

[tex]z^3 = 64(\cos(219^\circ) + i\sin(219^\circ))\\[/tex]

Converting the rigth side such that we could write it in power of 3:

[tex]z^3 = 4^3(\cos(3 \times \dfrac{219^\circ + 360^\circ k}{3}) + i\sin(3 \times \dfrac{219^\circ+360^\circ k}{3}))\\\\z^3 = (4(\cos(73^\circ + 120^\circ k) + i\sin(73^\circ+ 120^\circ k))^3[/tex]

For full round of angle, k will have to be 3. So, under that value, there are unique angles(ie for k = 0,1 and 2, there are unique values). If we take negative, then also -1 is same as +2, -2 is same as +1, so we ignored negative values of k.

Thus, for k = 0, 1 and 2, we get:

[tex]z = (4(\cos(73^\circ ) + i\sin(73^\circ))\\z = (4(\cos(73^\circ +120^\circ) + i\sin(73^\circ + 120^\circ))\\z = (4(\cos(73^\circ +240^\circ) + i\sin(73^\circ + 240^\circ))[/tex]

These are the all possible cube roots for distinct angles.

Thus, the cube roots of the considered complex number are:-

  • [tex]z = (4(\cos(73^\circ ) + i\sin(73^\circ))[/tex]
  • [tex]z = (4(\cos(193^\circ) + i\sin(193^\circ))[/tex]
  • [tex]z = (4(\cos(313^\circ) + i\sin(313^\circ))[/tex]

Learn more about cube root of complex numbers here:

https://brainly.com/question/16048106