Chapter 6.5, Problem 176E

a.

To determine

To find:The number of roots did not show in the given image.

Answer to Problem 176E

The number of roots did not show in the given image is 2, 3 and 4.

Explanation of Solution

Given information:

The figure is,

  

Calculation:

The graph shows a circle with 2 units, the absolute of complex number is,

  $\left|z\right|=2$

The first root is,

  $z_0=\sqrt[4]{2}\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right)$

Since,

  $\begin{array}{l}\frac{\theta +2\pi \left(0\right)}{4}=\frac{\pi }{6}\\ \theta =\frac{2\pi }{3}\end{array}$

Hence, 2, 3 and 4 root are missing.

b.

To determine

To find: The root of the given complex number.

Answer to Problem 176E

The root of the given complex numberis $\sqrt[4]{2}\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right),\sqrt[4]{2}\left(\cos \frac{7\pi }{6}+i\sin \frac{7\pi }{6}\right),\sqrt[4]{2}\left(\cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3}\right)$ .

Explanation of Solution

Given information:

The figure is,

  

Calculation:

The square root for $k=1$ .

  $\begin{array}{c}\sqrt{z}=\sqrt[4]{2}\left(\cos \frac{\frac{2\pi }{3}+2\pi \left(1\right)}{4}+i\sin \frac{\frac{2\pi }{3}+2\pi \left(1\right)}{4}\right)\\ =\sqrt[4]{2}\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)\end{array}$

The square root for $k=2$ .

  $\begin{array}{c}\sqrt{z}=\sqrt[4]{2}\left(\cos \frac{\frac{2\pi }{3}+2\pi \left(2\right)}{4}+i\sin \frac{\frac{2\pi }{3}+2\pi \left(2\right)}{4}\right)\\ =\sqrt[4]{2}\left(\cos \frac{7\pi }{6}+i\sin \frac{7\pi }{6}\right)\end{array}$

The square root for $k=3$ .

  $\begin{array}{c}\sqrt{z}=\sqrt[4]{2}\left(\cos \frac{\frac{2\pi }{3}+2\pi \left(3\right)}{4}+i\sin \frac{\frac{2\pi }{3}+2\pi \left(3\right)}{4}\right)\\ =\sqrt[4]{2}\left(\cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3}\right)\end{array}$