Chapter 14.4, Problem 28WE

a.

To determine

To find: The complex number $wz$ .

Answer to Problem 28WE

  $wz=14\left(\cos {90}^0+i\sin {90}^0\right)$

Explanation of Solution

Given information:

Two complex numbers are given as −

  $w=2\left(\cos {60}^0+i\sin {60}^0\right)\text{ and }z=7\left(\cos {30}^0+i\sin {30}^0\right)$ .

Concepts used: Multiplication and division of two complex numbers.

If $w=a\left(\cos \alpha +i\sin \alpha \right)\text{ and }z=b\left(\cos \beta +i\sin \beta \right)$ then −

  $\begin{array}{l}wz=ab\left[\cos \left(\alpha +\beta \right)+i\sin \left(\alpha +\beta \right)\right]\\ \frac{w}{z}=\frac{a}{b}\left[\cos \left(\alpha -\beta \right)+i\sin \left(\alpha -\beta \right)\right]\end{array}$

Calculation:

Two complex numbers are given as −

  $w=2\left(\cos {60}^0+i\sin {60}^0\right)\text{ and }z=7\left(\cos {30}^0+i\sin {30}^0\right)$

  $\begin{array}{l}\text{So, }wz=2\cdot 7\left[\cos \left({60}^0+{60}^0\right)+i\sin \left({30}^0+{30}^0\right)\right]\\ \Rightarrow wz=14\left(\cos {90}^0+i\sin {90}^0\right)\end{array}$

Hence, desired complex numbers are −

  $wz=14\left(\cos {90}^0+i\sin {90}^0\right)$

b.

To determine

To find: The complex number $wz$ in $x+iy$ form.

Answer to Problem 28WE

  $wz=14i$ .

Explanation of Solution

Given information:

Same as part $a$ of this exercise.

Calculation:

From part $a$ of this exercise, $wz=14\left(\cos {90}^0+i\sin {90}^0\right)$

Converting into $x+iy$ form,

  $\begin{array}{l}wz=14\left(\cos {90}^0+i\sin {90}^0\right)\\ \Rightarrow wz=14\left(0+i\right)\\ \Rightarrow wz=14i\end{array}$

c.

To determine

To find: The complex number $w\text{ and }z$ in $x+iy$ form.

Answer to Problem 28WE

  $w=1+\sqrt{3}i\text{ and }z=\frac{7\sqrt{3}}{2}+\frac{7}{2}i$ .

Explanation of Solution

Given information:

Same as part $a$ of this exercise.

Calculation:

Two complex numbers are given as −

  $w=2\left(\cos {60}^0+i\sin {60}^0\right)\text{ and }z=7\left(\cos {30}^0+i\sin {30}^0\right)$

Converting into $x+iy$ form,

  $\begin{array}{l}w=2\left(\cos {60}^0+i\sin {60}^0\right)\\ \Rightarrow w=2\left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)\\ \Rightarrow w=1+\sqrt{3}i\\ \text{Similarly, }z=7\left(\cos {30}^0+i\sin {30}^0\right)\\ \Rightarrow z=7\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)\\ \Rightarrow z=\frac{7\sqrt{3}}{2}+\frac{7}{2}i\end{array}$

Hence, given complex numbers in desired form are $w=1+\sqrt{3}i\text{ and }z=\frac{7\sqrt{3}}{2}+\frac{7}{2}i$

d.

To determine

To find: The complex number $wz$ in $x+iy$ form.

Answer to Problem 28WE

  $wz=14i$ .

Explanation of Solution

Given information:

Same as part $a$ of this exercise.

Calculation:

From part $c$ of this exercise, $w=1+\sqrt{3}i\text{ and }z=\frac{7\sqrt{3}}{2}+\frac{7}{2}i$

  $\begin{array}{l}wz=\left(1+\sqrt{3}i\right)\left(\frac{7\sqrt{3}}{2}+\frac{7}{2}i\right)\\ \Rightarrow wz=\frac{7\sqrt{3}}{2}+\frac{7}{2}i+\frac{21}{2}i-\frac{7\sqrt{3}}{2}\\ \Rightarrow wz=14i\end{array}$