Question

Find the fifth roots of 32(cos 280° + i sin 280°).

Answer 1

Answer on Question #53944– Math – Complex Analysis

Question

Find the fifth roots of $32(\cos 280{}^\circ + \mathrm{i} \sin 280{}^\circ)$.

Solution

Definition:

Assume that the complex number $z$ is presented in trigonometric form

$$z = \rho (\cos \varphi + \mathrm{i} \sin \varphi),$$

then the $n$-th roots of $z$ are defined by the formula

$$\sqrt[n]{z} = \sqrt[n]{\rho} \left( \cos \frac{\varphi + 2\pi k}{n} + \mathrm{i} \sin \frac{\varphi + 2\pi k}{n} \right),$$

where $k = 0,1,2,\dots,n-1$.

In the given problem the complex number $z$ has the form

$$z = 32 \left( \cos 280{}^\circ + \mathrm{i} \sin 280{}^\circ \right).$$

For convenience, let's rewrite the polar angle $\varphi$ in radians:

$$280{}^\circ = 280{}^\circ \cdot \frac{\pi}{180{}^\circ} = \frac{14}{9} \pi.$$

Hence, for (3) we obtain

$$z = 32 \left( \cos \left( \frac{14}{9} \pi \right) + \mathrm{i} \sin \left( \frac{14}{9} \pi \right) \right).$$

Now using (2) we find the fifth roots of (5):

$$\sqrt[5]{z} = \sqrt[5]{32} \left( \cos \frac{\frac{14}{9} \pi + 2\pi k}{5} + \mathrm{i} \sin \frac{\frac{14}{9} \pi + 2\pi k}{5} \right), \quad k = 0,1,2,3,4;$$

for $k = 0$: $z_1 = 2 \left( \cos \frac{\frac{14}{9} \pi + 0}{5} + \mathrm{i} \sin \frac{\frac{14}{9} \pi + 0}{5} \right) = 2 \left( \cos \frac{14}{45} \pi + \mathrm{i} \sin \frac{14}{45} \pi \right)$

for $k = 1$: $z_2 = 2 \left( \cos \frac{\frac{14}{9} \pi + 2\pi}{5} + \mathrm{i} \sin \frac{\frac{14}{9} \pi + 2\pi}{5} \right) = 2 \left( \cos \frac{32\pi}{45} + \mathrm{i} \sin \frac{32\pi}{45} \right)$

for $k = 2$: $z_3 = 2 \left( \cos \frac{\frac{14}{9} \pi + 4\pi}{5} + \mathrm{i} \sin \frac{\frac{14}{9} \pi + 4\pi}{5} \right) = 2 \left( \cos \frac{10\pi}{9} + \mathrm{i} \sin \frac{10\pi}{9} \right)$

$$\begin{array}{l} \text{for } k = 3: \ z = 2 \left( \cos \frac{ \frac{14}{9} \pi + 6 \pi }{5} + \text{isin } \frac{ \frac{14}{9} \pi + 6 \pi }{5} \right) = 2 \left( \cos \frac{68\pi}{45} + \text{isin } \frac{68\pi}{45} \right), \\ \text{for } k = 4: \ z_5 = 2 \left( \cos \frac{ \frac{14}{9} \pi + 8 \pi }{5} + \text{isin } \frac{ \frac{14}{9} \pi + 8 \pi }{5} \right) = 2 \left( \cos \frac{86\pi}{45} + \text{isin } \frac{86\pi}{45} \right). \end{array}$$

Therefore, the fifth roots of $z = 32 \left( \cos \left( \frac{14}{9} \pi \right) + \text{isin} \left( \frac{14}{9} \pi \right) \right)$ are

\left\{ \begin{array}{c} \boxed{ \begin{array}{r} z_1 = 2 \left( \cos \frac{14}{45} \pi + \text{isin } \frac{14}{45} \pi \right), \ z = 2 \left( \cos \frac{32\pi}{45} + \text{isin } \frac{32\pi}{45} \right), \\ z_3 = 2 \left( \cos \frac{10\pi}{9} + \text{isin } \frac{10\pi}{9} \right), \ z_4 = 2 \left( \cos \frac{68\pi}{45} + \text{isin } \frac{68\pi}{45} \right), \\ z_5 = 2 \left( \cos \frac{86\pi}{45} + \text{isin } \frac{86\pi}{45} \right). \end{array} } \right.

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