Chapter 5.4, Problem 65PPS

a.

To determine

Draw a graph for the given imaginary number with line segment from the origin.

Explanation of Solution

Given:

The given imaginary number is $3+4i$ .

Calculation:

Plot the real part of the number on the x-axis and the imaginary part on the y-axis.

  $\left(x,y\right)=\left(3,4\right)$

Origin $=\left(0,0\right)$

For the line segment, find the slope $m=\frac{y_2-y_1}{x_2-x_1}$ .

  $x_1=0,y_1=0,x_2=3,y_2=4$

  $m=\frac{0-4}{0-3}=\frac{4}{3}$

Use point-slope form for the equation of line.

  $y-y_1=m\left(x-x_1\right)$

  $\begin{array}{l}y-0=\frac{4}{3}\left(x-0\right)\\ \Rightarrow y=\frac{4}{3}x,\left(0\le x\le 3\right)\end{array}$

Hence the graph of the imaginary number and the line segment is given below.

  

b.

To determine

Draw a graph for the given imaginary number with line segment from the origin.

Explanation of Solution

Given:

The given imaginary number is $-2-5i$ .

Calculation:

Plot the real part of the number on the x-axis and the imaginary part on the y-axis.

  $\left(x,y\right)=\left(-2,-5\right)$

Origin $=\left(0,0\right)$

For the line segment, find the slope $m=\frac{y_2-y_1}{x_2-x_1}$ .

  $x_1=0,y_1=0,x_2=-2,y_2=-5$

  $m=\frac{0-\left(-5\right)}{0-\left(-2\right)}=\frac{5}{2}$

Use point-slope form for the equation of line.

  $y-y_1=m\left(x-x_1\right)$

  $\begin{array}{l}y-0=\frac{5}{2}\left(x-0\right)\\ \Rightarrow y=\frac{5}{2}x,\left(-2\le x\le 0\right)\end{array}$

Hence the graph of the imaginary number and the line segment is given below.

  

c.

To determine

Find the fourth point of the given parallelogram.

Explanation of Solution

Given:

The given vertices of the parallelogram are $\left(-2,-5\right)$ , $\left(0,0\right)$ and $\left(3,4\right)$ .

Calculation:

In a parallelogram, diagonals bisect each other.

Let the fourth vertex be $C\left(x,y\right)$ .

  $A\left(3,4\right)$

  $B\left(-2,-5\right)$

  $D\left(0,0\right)$

Now,

  $\begin{array}{l}\left(\frac{0+x}{2},\frac{0+y}{2}\right)=\left(\frac{-2+3}{2},\frac{-5+4}{2}\right)\\ \left(\frac{x}{2},\frac{y}{2}\right)=\left(\frac{1}{2},-\frac{1}{2}\right)\\ \frac{x}{2}=\frac{1}{2}\\ \Rightarrow x=1\\ \frac{y}{2}=-\frac{1}{2}\\ \Rightarrow y=-1\end{array}$

The fourth point $C\left(1,-1\right)$ .

Graph of the parallelogram is given below.

  

d.

To determine

Find the complex number which represents by the point $C$ .

Answer to Problem 65PPS

The complex number represented by the point $C$ is $1-i$ .

Explanation of Solution

Given:

The given vertices of the parallelogram are $\left(-2,-5\right)$ , $\left(0,0\right)$ and $\left(3,4\right)$ .

Calculation:

In a parallelogram, diagonals bisect each other.

Let the fourth vertex be $C\left(x,y\right)$ .

  $A\left(3,4\right)$

  $B\left(-2,-5\right)$

  $D\left(0,0\right)$

Now,

  $\begin{array}{l}\left(\frac{0+x}{2},\frac{0+y}{2}\right)=\left(\frac{-2+3}{2},\frac{-5+4}{2}\right)\\ \left(\frac{x}{2},\frac{y}{2}\right)=\left(\frac{1}{2},-\frac{1}{2}\right)\\ \frac{x}{2}=\frac{1}{2}\\ \Rightarrow x=1\\ \frac{y}{2}=-\frac{1}{2}\\ \Rightarrow y=-1\end{array}$

The fourth point $C\left(1,-1\right)$ .

The complex number represented by the point $C$ is $1-i$ .

Graph of the parallelogram is given below.

  

The points A,B and C are the vertices of the parallelogram.