Verified Answer :

wizard123

\(z^5 = -32 \\ \\ r^5 (cos \theta +i sin \theta)^5 = -32 \\ \\ r^5 (cos (5 \theta) +i sin(5\theta)) = -32 \\ \\ 2^5 (cos (\pi+n2\pi) +i sin (\pi+n2\pi) ) =-32, n = 0,1,2,3,4 \\ \\ \theta = \frac{\pi}{5}+n \frac{2\pi}{5} \\ \\ \theta_1 - \theta_0 = \frac{3\pi}{5} - \frac{\pi}{5} = \frac{2\pi}{5}\)

Cricetus

The angle between the consecutive roots will be "72°".

Given:

  • \(x^5+32 = 0\)

then,

  • \(z^5 =-32\)

Let,

→ \(z = r(Cos \Theta+ iSin \Theta)\)

Now,

→ \(r^5(Cos \Theta + iSin \Theta)^5 = -32\)

  \(r^5 (Cos (50)+i Sin (50)) = -32\)

  \(2^5(Cos(\pi +n 2 \pi))+ i Sin \Theta(\pi+ n2\pi) = -32\)

Where, n = 0, 1, 2, 3, 4

→ \(\Theta = \frac{\pi}{5}+ n \frac{2 \pi}{5}\)

  • \(\Theta_0 = \frac{\pi}{5}\)
  • \(\Theta_1 = \frac{\pi}{5} + \frac{2 \pi}{5}\)

The consecutive difference will be:

= \(\Theta_1 - \Theta_0\)

= \(\frac{2 \pi}{5} +\frac{\pi}{5} -\frac{\pi}{5}\)

= \(\frac{2 \pi}{5}\)

= \(\frac{2}{5}\times 180^{\circ}\)

= \(72^{\circ}\)

Thus the above approach is correct.

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