Find the cube roots of 8(cos 216° + i sin 216°).

[tex]z^3=8(\cos216^\circ+i\sin216^\circ)[/tex]
[tex]z^3=2^3(\cos(6^3)^\circ+i\sin(6^3)^\circ)[/tex]
[tex]\implies z=8^{1/3}\left(\cos\left(\dfrac{216+360k}3\right)^\circ+i\sin\left(\dfrac{216+360k}3\right)^\circ\right)[/tex]

where [tex]k=0,1,2[/tex]. So the third roots are

[tex]z=\begin{cases}2(\cos72^\circ+i\sin72^\circ)\\2(\cos192^\circ+i\sin192^\circ)\\2(\cos312^\circ+i\sin312^\circ)\end{cases}[/tex]

Answer Link

psm22415 psm22415 · Answer 2 · 2022-02-22T10:11:32+01:00

The cube roots of the equation are;

[tex]\rm z=2(cos72+i \ sin 72)\\\\z=2(cos192+i \ sin 192)\\\\z=2(cos312+i \ sin 312)[/tex]

Cube roots

The Cube of a cube root of a number or the cube root of a cube of a number is the number itself.

The cube roots of 8(cos 216° + I sin 216°).

To find the cube roots of the equation following all the steps given below.

The cube roots are;

[tex]\rm z^3=8(cos 216 + i \sin 216)\\\\z^3=2^3(cos 6^3 + i \sin 6^3)\\\\ z = 8^{\frac{1}{3}}\left (cos\left (cos\dfrac{216+360k}{3} \right )\right)+ i \ sin \left (\dfrac{216+360k}{3} \right )\right)\\\\[/tex]

Where the value of k is 0, 1, 2, 3...

The cubes roots are;

[tex]\rm z=2(cos72+i \ sin 72)\\\\z=2(cos192+i \ sin 192)\\\\z=2(cos312+i \ sin 312)[/tex]

Hence, the cube roots of the equation are;

[tex]\rm z=2(cos72+i \ sin 72)\\\\z=2(cos192+i \ sin 192)\\\\z=2(cos312+i \ sin 312)[/tex]

To know more about roots click the link given below.

https://brainly.com/question/118600