Chapter 9, Problem 14P

Simplify the following expression:

1. (a) $\frac{\left(5-j6\right)-\left(2+j8\right)}{\left(-3+j4\right)\left(5-j\right)+\left(4-j6\right)}$
2. (b) $\frac{\left(240\angle {75}^{\circ }+160\angle -{30}^{\circ }\right)\left(60-j80\right)}{\left(67+j84\right)\left(20\angle {32}^{\circ }\right)}$
3. (c) ${\left(\frac{10+j20}{3+j4}\right)}^2\sqrt{\left(10+j5\right)\left(16-j20\right)}$

To determine

Find out the value of the given expression in polar form, rectangular form.

Answer to Problem 14P

The value of the given expression in polar form, rectangular form is $\underset{\_}{0.7788\angle 169.71°}$ and $\underset{\_}{-0.7663+j0.13912}$ respectively.

Explanation of Solution

Given data:

The given expression is,

$\frac{\left(5-j6\right)-\left(2+j8\right)}{\left(-3+j4\right)\left(5-j\right)+\left(4-j6\right)}$

Calculation:

Simplify the given expression as follows,

$\begin{array}{l}\frac{\left(5-j6\right)-\left(2+j8\right)}{\left(-3+j4\right)\left(5-j\right)+\left(4-j6\right)}=\frac{5-j6-2-j8}{-15+j3+j20-j^{2}4+4-j6}\\ \frac{\left(5-j6\right)-\left(2+j8\right)}{\left(-3+j4\right)\left(5-j\right)+\left(4-j6\right)}=\frac{3-j14}{-15+j3+j20+4+4-j6}\left\{\because j^2=-1\right\}\\ \frac{\left(5-j6\right)-\left(2+j8\right)}{\left(-3+j4\right)\left(5-j\right)+\left(4-j6\right)}=\frac{3-j14}{-7+j17}\\ \frac{\left(5-j6\right)-\left(2+j8\right)}{\left(-3+j4\right)\left(5-j\right)+\left(4-j6\right)}=\frac{14.318\angle 282.095°}{18.385\angle 112.380°}\end{array}$

Re-write the value as follows in polar form.

$\frac{\left(5-j6\right)-\left(2+j8\right)}{\left(-3+j4\right)\left(5-j\right)+\left(4-j6\right)}\cong 0.7788\angle 169.71°$

Re-write the value as follows in rectangular form.

$\begin{array}{l}\frac{\left(5-j6\right)-\left(2+j8\right)}{\left(-3+j4\right)\left(5-j\right)+\left(4-j6\right)}=-0.76627+j0.13912\\ \frac{\left(5-j6\right)-\left(2+j8\right)}{\left(-3+j4\right)\left(5-j\right)+\left(4-j6\right)}\cong -0.7663+j0.13912\end{array}$

Conclusion:

Thus, the value of the given expression in polar form, rectangular form is $\underset{\_}{0.7788\angle 169.71°}$ and $\underset{\_}{-0.7663+j0.13912}$ respectively.

To determine

Find out the value of the given expression.

Answer to Problem 14P

The value of the given expression in polar form, rectangular form is $\underset{\_}{11.71\angle 260.55°}$ and $\underset{\_}{-1.922-11.55j}$ respectively.

Explanation of Solution

Given data:

The given expression is,

$\frac{\left(240\angle 75°+160\angle -30°\right)\left(60-j80\right)}{\left(67+j84\right)\left(20\angle 32°\right)}$

Calculation:

Simplify the given expression as follows,

$\begin{array}{l}\frac{\left(240\angle 75°+160\angle -30°\right)\left(60-j80\right)}{\left(67+j84\right)\left(20\angle 32°\right)}=\frac{\left(62.116+j231.82+138.56-j80\right)\left(60-j80\right)}{\left(107.44\angle 51.42°\right)\left(20\angle 32°\right)}\\ \frac{\left(240\angle 75°+160\angle -30°\right)\left(60-j80\right)}{\left(67+j84\right)\left(20\angle 32°\right)}=\frac{\left(200.676+j151.82\right)\left(60-j80\right)}{\left(107.44\angle 51.42°\right)\left(20\angle 32°\right)}\\ \frac{\left(240\angle 75°+160\angle -30°\right)\left(60-j80\right)}{\left(67+j84\right)\left(20\angle 32°\right)}=\frac{12040.56+j9109.2-j16054.08-j^{2}12145.6}{2148.8\angle 83.42°}\\ \frac{\left(240\angle 75°+160\angle -30°\right)\left(60-j80\right)}{\left(67+j84\right)\left(20\angle 32°\right)}=\frac{12040.56+j9109.2-j16054.08+12145.6}{2148.8\angle 83.42°}\end{array}$

Simplify the equation as follows.

$\begin{array}{l}\frac{\left(240\angle 75°+160\angle -30°\right)\left(60-j80\right)}{\left(67+j84\right)\left(20\angle 32°\right)}=\frac{24186.16-j6944.88}{2148.8\angle 83.42°}\\ \frac{\left(240\angle 75°+160\angle -30°\right)\left(60-j80\right)}{\left(67+j84\right)\left(20\angle 32°\right)}=\frac{25163.49\angle 343.97°}{2148.8\angle 83.42°}\end{array}$

Re-write the value as follows in polar form.

$\begin{array}{l}\frac{\left(240\angle 75°+160\angle -30°\right)\left(60-j80\right)}{\left(67+j84\right)\left(20\angle 32°\right)}=11.7104\angle 260.55°\\ \frac{\left(240\angle 75°+160\angle -30°\right)\left(60-j80\right)}{\left(67+j84\right)\left(20\angle 32°\right)}\cong 11.71\angle 260.55°\end{array}$

Re-write the value as follows in rectangular form.

$\frac{\left(240\angle 75°+160\angle -30°\right)\left(60-j80\right)}{\left(67+j84\right)\left(20\angle 32°\right)}=-1.922-11.55j$

Conclusion:

Thus, the value of the given expression in polar form, rectangular form is $\underset{\_}{11.71\angle 260.55°}$ and $\underset{\_}{-1.922-11.55j}$ respectively.

To determine

Find out the value of the given expression in both polar form and rectangular form.

Answer to Problem 14P

The value of the given expression in polar form, rectangular form is $\underset{\_}{338.4\angle -8.22°}$ and $\underset{\_}{334.923-j48.382}$ respectively.

Explanation of Solution

Given data:

The given expression is,

${\left(\frac{10+j20}{3+j4}\right)}^2\sqrt{\left(10+j5\right)\left(16-j20\right)}$

Calculation:

Simplify the given expression as follows,

$\begin{array}{l}{\left(\frac{10+j20}{3+j4}\right)}^2\sqrt{\left(10+j5\right)\left(16-j20\right)}=\frac{{\left(10+j20\right)}^2}{{\left(3+j4\right)}^2}\sqrt{160-j200+j80-j^{2}100}\\ {\left(\frac{10+j20}{3+j4}\right)}^2\sqrt{\left(10+j5\right)\left(16-j20\right)}=\frac{{10}^2-{20}^2+j2\left(10\right)\left(20\right)}{3^2-4^2+j2\left(3\right)\left(4\right)}\sqrt{160-j200+j80-j^{2}100}\\ {\left(\frac{10+j20}{3+j4}\right)}^2\sqrt{\left(10+j5\right)\left(16-j20\right)}=\frac{-300+j400}{-7+j24}\sqrt{160-j200+j80+100}\\ {\left(\frac{10+j20}{3+j4}\right)}^2\sqrt{\left(10+j5\right)\left(16-j20\right)}=\frac{500\angle 126.869°}{25\angle 106.26°}\sqrt{286.36\angle 335.22°}\end{array}$

Simplify the above equation as follows.

$\begin{array}{l}{\left(\frac{10+j20}{3+j4}\right)}^2\sqrt{\left(10+j5\right)\left(16-j20\right)}=\frac{500\angle 126.869°}{25\angle 106.26°}\sqrt{286.36\angle -24.78°}\\ {\left(\frac{10+j20}{3+j4}\right)}^2\sqrt{\left(10+j5\right)\left(16-j20\right)}=20\angle 20.609°\sqrt{286.36}\angle \left(\frac{-24.78°}{2}\right)\\ {\left(\frac{10+j20}{3+j4}\right)}^2\sqrt{\left(10+j5\right)\left(16-j20\right)}=\left(20\angle 20.609°\right)\left(16.922\angle -12.39°\right)\\ {\left(\frac{10+j20}{3+j4}\right)}^2\sqrt{\left(10+j5\right)\left(16-j20\right)}=338.44\angle -8.219°\end{array}$

The approximate equivalent value in polar form,

${\left(\frac{10+j20}{3+j4}\right)}^2\sqrt{\left(10+j5\right)\left(16-j20\right)}\cong 338.4\angle -8.22°$

The approximate equivalent value in rectangular form,

${\left(\frac{10+j20}{3+j4}\right)}^2\sqrt{\left(10+j5\right)\left(16-j20\right)}\cong 334.923-j48.382$

Conclusion:

Thus, the value of the given expression in polar form, rectangular form is $\underset{\_}{338.4\angle -8.22°}$ and $\underset{\_}{334.923-j48.382}$ respectively.