Chapter 6.6, Problem 107E

(a)

To determine

To calculate:

The square roots of following complex number:

  $5\left(\cos {120}^{\circ }+i\sin {120}^{\circ }\right)$

Answer to Problem 107E

The square roots of complex number are $\sqrt{5}\left(\cos {60}^{\circ }+i\sin {60}^{\circ }\right)$ and $\sqrt{5}\left(\cos 240+i\sin 240\right)$

Explanation of Solution

Given information:

Complex number is,

  $5\left(\cos {120}^{\circ }+i\sin {120}^{\circ }\right)$

Formula used:

The formula to calculate the roots of complex number are given by,

  $n\sqrt{r}\left(\cos \frac{\theta +2\pi k}{n}+i\sin \frac{\theta +2\pi k}{n}\right)$

Where,

  $k=0,1,\mathrm{2...},n-1$

Calculation:

The given complex number is:

  $z=5\left(\cos {120}^{\circ }+i\sin {120}^{\circ }\right)$

Substituting in the above formula,

  $z=2\sqrt{5}\left(\cos \frac{120+2\pi k}{2}+i\sin \frac{120+2\pi k}{2}\right)$

Now find the square roots of complex number for $k=0,1$

For $k=0,$

  $\begin{array}{l}{z}_1=\sqrt{5}\left(\cos \frac{120+2\pi \times 0}{2}+i\sin \frac{120+2\pi \times 0}{2}\right)\\ z_1=\sqrt{5}\left(\cos {60}^{\circ }+i\sin {60}^{\circ }\right)\end{array}$

Now for $k=1,$

  $\begin{array}{l}{z}_2=\sqrt{5}\left(\cos \frac{120+2\pi \times 1}{2}+i\sin \frac{120+2\pi \times 1}{2}\right)\\ z_2=\sqrt{5}\left(\cos \left(60+\pi \right)\right)+i\sin \left(60+\pi \right)\\ z_2=\sqrt{5}\left(\cos \left(60+180\right)\right)+i\sin \left(60+180\right)\\ z_2=\sqrt{5}\left(\cos 240+i\sin 240\right)\end{array}$

(b)

To determine

To calculate:

The roots in standard from

Answer to Problem 107E

The roots in standard form are $-\frac{\sqrt{5}}{2}-i\frac{\sqrt{15}}{2}$

Explanation of Solution

Given information:

The roots are,

  $\sqrt{5}\left(\cos {60}^{\circ }+i\sin {60}^{\circ }\right)$ And $\sqrt{5}\left(\cos 240+i\sin 240\right)$

Calculation:

Consider the roots,

  $\sqrt{5}\left(\cos {60}^{\circ }+i\sin {60}^{\circ }\right)$ And $\sqrt{5}\left(\cos 240+i\sin 240\right)$

Now, first write the standard form of $\sqrt{5}\left(\cos {60}^{\circ }+i\sin {60}^{\circ }\right)$

  $\begin{array}{l}=\sqrt{5}\left(\cos {60}^{\circ }+i\sin {60}^{\circ }\right)\\ =\sqrt{5}\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)\\ =\frac{\sqrt{5}}{2}+i\frac{\sqrt{15}}{2}\end{array}$

Now, the standard form of $\sqrt{5}\left(\cos 240+i\sin 240\right)$

  $\begin{array}{l}=\sqrt{5}\left(\cos 240+i\sin 240\right)\\ =\sqrt{5}\left(\cos 180+60+i\sin 180+60\right)\end{array}$

As the cos and sine are negative in the third quadrant,

  $=\sqrt{5}\left(-\cos 60-i\sin 60\right)$

Put the values of $\cos 60$ and $\sin 60$ in the above complex number,

  $\begin{array}{l}=\sqrt{5}\left(-\cos 60-i\sin 60\right)\\ =\sqrt{5}\times \left(-\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)\\ =\left(-\frac{\sqrt{5}}{2}-i\frac{\sqrt{15}}{2}\right)\end{array}$

(c)

To determine

To graph:

The roots of the given complex number

Explanation of Solution

Given information:

Square Roots of complex number are

  $\sqrt{5}\left(\cos {60}^{\circ }+i\sin {60}^{\circ }\right)$ And $\sqrt{5}\left(\cos 240+i\sin 240\right)$

Graph:

  

Interpretation:

From sub-part (a):

The graph of the square roots of complex number is sketched as:

  $\sqrt{5}\left(\cos {60}^{\circ }+i\sin {60}^{\circ }\right)$ and $\sqrt{5}\left(\cos 240+i\sin 240\right)$