Chapter 6.6, Problem 112E

(a)

To determine

To calculate:

The forth roots of complex number.

Answer to Problem 112E

The forth roots of complex number $625i$ are

  $5\left(\cos {22.5}^{\circ }+i\sin {22.5}^{\circ }\right)$ , $5\left(\cos {112.5}^{\circ }+i\sin {112.5}^{\circ }\right)$

  $5\left(\cos {202.5}^{\circ }+i\sin {202.5}^{\circ }\right)$ and $5\left(\cos {292.5}^{\circ }+i\sin {292.5}^{\circ }\right)$

Explanation of Solution

Given information:

The complex number is $625i$

Formula used:

The formula to calculate the roots of complex number $z=r\left(\cos \theta +i\sin \theta \right)$ 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:

Now first write the complex number in the standard from of the trigonometric function,

  $x^4=625\left(\cos {90}^{\circ }+i\sin {90}^{\circ }\right)$

The formula is given as,

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

From the formula to find the roots of the complex number. It is written as,

  ${625}^{\frac{1}{4}}\left(\cos \frac{1}{4}\left({90}^{\circ }+{360}^{\circ k}\right)\right)+i\sin \frac{1}{4}\left({90}^{\circ }+{360}^{\circ k}\right)$

Now, find the forth root of the complex number as

  $k=0,1,2,3$

Now, find for $k=0$

The roots are $5\left(\cos {22.5}^{\circ }+i\sin {22.5}^{\circ }\right)$

For $k=1$ ,

The roots are,

  $5\left(\cos {112.5}^{\circ }+i\sin {112.5}^{\circ }\right)$

For $k=2$

The roots are,

  $5\left(\cos {202.5}^{\circ }+i\sin {202.5}^{\circ }\right)$

The roots for $k=3$ are,

  $5\left(\cos {292.5}^{\circ }+i\sin {292.5}^{\circ }\right)$

(b)

To determine

To calculate:

The roots in standard form

Answer to Problem 112E

The roots in standard form are $4.6+1.9i,-1.9+4.6i,-4.6-1.9i$ and $1.9-4.6i$

Explanation of Solution

Given information:

The roots of complex number,

  $5\left(\cos {22.5}^{\circ }+i\sin {22.5}^{\circ }\right)$ ,

  $5\left(\cos {112.5}^{\circ }+i\sin {112.5}^{\circ }\right)$ ,

  $5\left(\cos {202.5}^{\circ }+i\sin {202.5}^{\circ }\right)$ ,

  $5\left(\cos {292.5}^{\circ }+i\sin {292.5}^{\circ }\right)$

Calculation:

Now write the standard form of all the given roots of trigonometric form.

Consider For,$5\left(\cos {22.5}^{\circ }+i\sin {22.5}^{\circ }\right)$ ,

The standard form is given as,

As the value of $\cos {22.5}^{\circ }$

  $=$ $0.92$ and $\sin {22.5}^{\circ }=0.38$

  $5\left(\cos {22.5}^{\circ }+i\sin {22.5}^{\circ }\right)$

  $\begin{array}{l}=5\left(0.92+0.38i\right)\\ =4.6+1.9i\end{array}$

Now the standard form of $5\left(\cos {112.5}^{\circ }+i\sin {112.5}^{\circ }\right)$

  $\begin{array}{l}5\left(\cos {112.5}^{\circ }+i\sin {112.5}^{\circ }\right)\\ =5\left(-0.38+0.92i\right)\\ =-1.9+4.6i\end{array}$

Now the standard form of $5\left(\cos {202.5}^{\circ }+i\sin {202.5}^{\circ }\right)$

  $\begin{array}{l}5\left(\cos {202.5}^{\circ }+i\sin {202.5}^{\circ }\right)\\ =5\left(-0.92-0.38i\right)\\ =-4.6-1.9i\end{array}$

The standard form of $5\left(\cos {292.5}^{\circ }+i\sin {292.5}^{\circ }\right)$

  $\begin{array}{l}5\left(\cos {292.5}^{\circ }+i\sin {292.5}^{\circ }\right)\\ =5\left(0.38-0.92i\right)\\ =1.9-4.6i\end{array}$

(c)

To determine

To graph:

The roots in graphical form.

Explanation of Solution

Given information:

The Roots of complex number,

  $5\left(\cos {22.5}^{\circ }+i\sin {22.5}^{\circ }\right)$ ,

  $5\left(\cos {112.5}^{\circ }+i\sin {112.5}^{\circ }\right)$ ,

  $5\left(\cos {202.5}^{\circ }+i\sin {202.5}^{\circ }\right)$ ,

  $5\left(\cos {292.5}^{\circ }+i\sin {292.5}^{\circ }\right)$

Graph:

  

Interpretation:

The graph of the forth roots of complex number $625i$ is sketched as above:

The forth roots of the complex number $625i$ are $5\left(\cos {22.5}^{\circ }+i\sin {22.5}^{\circ }\right)$ , $5\left(\cos {112.5}^{\circ }+i\sin {112.5}^{\circ }\right)$ $5\left(\cos {202.5}^{\circ }+i\sin {202.5}^{\circ }\right)$ and $5\left(\cos {292.5}^{\circ }+i\sin {292.5}^{\circ }\right)$