what is the angle between two consecutive roots in the complex plane of z^5 + 32 = 0

The angle between the consecutive roots will be "72°".
Given:
then,
Let,
→ [tex]z = r(Cos \Theta+ iSin \Theta)[/tex]
Now,
→ [tex]r^5(Cos \Theta + iSin \Theta)^5 = -32[/tex]
[tex]r^5 (Cos (50)+i Sin (50)) = -32[/tex]
[tex]2^5(Cos(\pi +n 2 \pi))+ i Sin \Theta(\pi+ n2\pi) = -32[/tex]
Where, n = 0, 1, 2, 3, 4
→ [tex]\Theta = \frac{\pi}{5}+ n \frac{2 \pi}{5}[/tex]
The consecutive difference will be:
= [tex]\Theta_1 - \Theta_0[/tex]
= [tex]\frac{2 \pi}{5} +\frac{\pi}{5} -\frac{\pi}{5}[/tex]
= [tex]\frac{2 \pi}{5}[/tex]
= [tex]\frac{2}{5}\times 180^{\circ}[/tex]
= [tex]72^{\circ}[/tex]
Thus the above approach is correct.
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