Chapter 1.6, Problem 59E

Solution

a.

To Find: the additive and multiplicative inverses of the given complex number.

The additive inverse of $2+i$ is $-2-i$ and multiplicative inverse of $2+i$ is $\frac{1}{2+i}$ .

Given:

  $z=2+i$ .

Concept Used:

The additive inverse of a complex number $z$ is a complex number $z_a$ such that $z+z_a=0$ . The multiplicative inverse of $z$ is a complex number $z_m$ such that $z\cdot z_m=1$ .

Calculation:

  $z=2+i$ .

For additive inverse of $z=2+i$ :

Let the $z_1$ be the required additive inverse.

  $\therefore z_1+z=0$ ,

  $\begin{array}{c}{z}_1+2+i=0\\ z_1=0-\left(2+i\right)\\ z_1=-2-i\end{array}$

Now, let the multiplicative inverse be $z_2$:

  $\therefore z_2\cdot z=1$ ,

  $\begin{array}{c}{z}_2\cdot \left(2+i\right)=1\\ z_2=\frac{1}{2+i}\end{array}$

Conclusion:

The additive inverse of $2+i$ is $-2-i$ and multiplicative inverse of $2+i$ is $\frac{1}{2+i}$ .

b.

To Find: the additive and multiplicative inverses of the given complex number.

The additive inverse of $5-i$ is $-5+i$ and multiplicative inverse of $5-i$ is $\frac{1}{5-i}$ .

Given: $z=5-i$

Concept Used:

The additive inverse of a complex number $z$ is a complex number $z_a$ such that $z+z_a=0$ . The multiplicative inverse of $z$ is a complex number $z_m$ such that $z\cdot z_m=1$ .

Calculation:

For additive inverse of $z=5-i$ :

Let the $z_1$ be the required additive inverse.

  $\therefore z_1+z=0$ ,

  $\begin{array}{c}{z}_1+5-i=0\\ z_1=0-\left(5-i\right)\\ z_1=-5+i\end{array}$

Now, let the multiplicative inverse be $z_2$ :

  $\therefore z_2\cdot z=1$ ,

  $\begin{array}{c}{z}_2\cdot \left(5-i\right)=1\\ z_2=\frac{1}{5-i}\end{array}$

Conclusion:

The additive inverse of $5-i$ is $-5+i$ and multiplicative inverse of $5-i$ is $\frac{1}{5-i}$ .

c.

To Find: the additive and multiplicative inverses of the given complex number.

The additive inverse of $-1+3i$ is $1-3i$ and multiplicative inverse of $-1+3i$ is $\frac{1}{-1+3i}$ .

Given: $z=-1+3i$

Concept Used:

The additive inverse of a complex number $z$ is a complex number $z_a$ such that $z+z_a=0$ . The multiplicative inverse of $z$ is a complex number $z_m$ such that $z\cdot z_m=1$ .

Calculation:

Fir additive inverse of $z=-1+3i$ :

Let the $z_1$ be the required additive inverse.

  $\therefore z_1+z=0$ ,

  $\begin{array}{c}{z}_1+(-1+3i)=0\\ z_1=0-\left(-1+3i\right)\\ z_1=1-3i\end{array}$

Now, let the multiplicative inverse be $z_2$ :

  $\therefore z_2\cdot z=1$ ,

  $\begin{array}{c}{z}_2\cdot \left(-1+3i\right)=1\\ z_2=\frac{1}{-1+3i}\end{array}$

Conclusion:

The additive inverse of $-1+3i$ is $1-3i$ and multiplicative inverse of $-1+3i$ is $\frac{1}{-1+3i}$ .