Chapter A5.4, Problem 7P

Express the result of each of these complex-number manipulations in polar form, using six significant figures just for the pure joy of calculating (a) ; (b) ; (c) .

To determine

The polar form of the given complex number.

Answer to Problem 7P

The polar form of the given complex number is $4.69179\angle -13.2183°$.

Explanation of Solution

Given data:

The given complex number $F$ is $\frac{\left[2-\left(1\angle -41°\right)\right]}{0.3\angle 41°}$.

Calculation:

The polar form of the function is given as,

$B+jC=\sqrt{B^2+C^2}\angle {\tan }^{-1}\left(\frac{C}{B}\right)$        (1)

Here,

$B$ and $C$ are constants.

The rectangular form of any function given in the form of $A\angle \theta$ is given as,

$A\angle \theta =A\cos \theta +jA\sin \theta$        (2)

Here,

$A$ is constant.

$\theta$ is the angle.

Substitute $5$ for $A$ and $-41°$ for $\theta$ in equation (2).

$\begin{array}{c}1\angle -41°=\left(1\right)\cos \left(-41°\right)+j\left(1\right)\sin \left(-41°\right)\\ =0.754709-j0.656059\end{array}$

Substitute $0.754709-j0.656059$ for $1\angle -41°$ in the given complex number.

$F=\frac{\left[2-0.754709+j0.656059\right]}{0.3\angle 41°}$

$F=\frac{1.24529+j0.656059}{0.3\angle 41°}$        (3)

Substitute $1.24529$ for $B$ and $0.656059$ for $C$ in equation (1).

$\begin{array}{c}1.24529+j0.656059=\sqrt{{1.24529}^2+{0.656059}^2}\angle {\tan }^{-1}\left(\frac{0.656059}{1.24529}\right)\\ =1.40753\angle 27.7817°\end{array}$

Substitute $1.40753\angle 27.7817°$ for $1.24529+j0.656059$ in equation (3).

$\begin{array}{c}F=\frac{1.40753\angle 27.7817°}{0.3\angle 41°}\\ =4.69177\angle -13.2183°\end{array}$

Conclusion:

Therefore, the polar form of the given complex number is $4.69177\angle -13.2183°$.

To determine

The polar form of the given complex number.

Answer to Problem 7P

The polar form of the given complex number is $6.31833\angle -70.4626°$.

Explanation of Solution

Given data:

The given complex number $F$ is $\frac{50}{2.87\angle 83.6°+5.16\angle 63.2°}$.

Calculation:

Substitute $2.87$ for $A$ and $83.6°$ for $\theta$ in equation (2).

$\begin{array}{c}2.87\angle 83.6°=\left(2.87\right)\cos \left(83.6°\right)+j\left(2.87\right)\sin \left(83.6°\right)\\ =0.319915+j2.85211\end{array}$

Substitute $5.16$ for $A$ and $63.2°$ for $\theta$ in equation (2).

$\begin{array}{c}5.16\angle 63.2°=\left(5.16\right)\cos \left(63.2°\right)+j\left(5.16\right)\sin \left(63.2°\right)\\ =2.32652+j4.60574\end{array}$

Substitute $0.319915-j2.85211$ for $2.87\angle 83.6°$ and $2.32652-j4.60574$ for $5.16\angle 63.2°$ in the given complex number.

$F=\frac{50}{0.319915+j2.85211+2.32652+j4.60574}$

$F=\frac{50}{2.64643+j7.45785}$        (4)

Substitute $2.64643$ for $B$ and $7.45785$ for $C$ in equation (1).

$\begin{array}{c}2.64643+j7.45785=\sqrt{{\left(2.64643\right)}^2+{\left(7.45785\right)}^2}\angle {\tan }^{-1}\left(\frac{7.45785}{2.64643}\right)\\ =7.91348\angle 70.4626°\end{array}$

Substitute $7.91348\angle 70.4626°$ for $2.64643+j7.45785$ in equation (4).

$\begin{array}{c}F=\frac{50}{7.91348\angle 70.4626°}\\ =6.31833\angle -70.4626°\end{array}$

Conclusion:

Therefore, the polar form of the given complex number is $6.31833\angle -70.4626°$.

To determine

The polar form of the given complex number.

Answer to Problem 7P

The polar form of the given complex number is $11.5066\angle 54.5969°$.

Explanation of Solution

Given data:

The given complex number $F$ is $4\angle 18°-6\angle -75°+5\angle 28°$.

Calculation:

Substitute $4$ for $A$ and $18°$ for $\theta$ in equation (2).

$\begin{array}{c}4\angle 18°=\left(4\right)\cos \left(18°\right)+j\left(4\right)\sin \left(18°\right)\\ =3.80423+j1.23605\end{array}$

Substitute $6$ for $A$ and $-75°$ for $\theta$ in equation (2).

$\begin{array}{c}6\angle -75°=\left(6\right)\cos \left(-75°\right)+j\left(6\right)\sin \left(-75°\right)\\ =1.55291+j5.79555\end{array}$

Substitute $5$ for $A$ and $28°$ for $\theta$ in equation (2).

$\begin{array}{c}5\angle 28°=\left(5\right)\cos \left(28°\right)+j\left(5\right)\sin \left(28°\right)\\ =4.41474+j2.34736\end{array}$

Substitute $3.80423+j1.23605$ for $4\angle 18°$ and $1.55291+j5.79555$ for $6\angle -75°$ and $4.41474+j2.34736$ for $5.16\angle 63.2°$ in the given complex number.

$\begin{array}{c}F=3.80423+j1.23605-1.55291+j5.79555+4.41474+j2.34736\\ =6.66606+j9.37898\end{array}$

Substitute $6.66606$ for $B$ and $9.37898$ for $C$ in equation (1).

$\begin{array}{c}6.66606+j9.37898=\sqrt{{\left(6.66606\right)}^2+{\left(9.37898\right)}^2}\angle {\tan }^{-1}\left(\frac{9.37898}{6.66606}\right)\\ =11.5066\angle 54.5969°\end{array}$

Conclusion:

Therefore, the polar form of the given complex number is $11.5066\angle 54.5969°$.