Chapter 6.6, Problem 69E

(a)

To determine

To calculate:

The trigonometric form of following complex number:

  $\frac{3+4i}{1-\sqrt{3}i}$

Answer to Problem 69E

The trigonometric from of following complex number is $10\left(\cos 0.927+i\sin 0.927\right)\left(\cos \left(-\frac{\pi }{3}\right)+i\sin \left(-\frac{\pi }{3}\right)\right)$

Explanation of Solution

Given information:

  $\frac{3+4i}{1-\sqrt{3}i}$

Calculation:

Consider,

  $\begin{array}{l}{z}_1=\left(3+4i\right)\\ z_2=\left(1-\sqrt{3}i\right)\end{array}$

For $z_1=\left(3+4i\right)$ :

Compare $z_1=\left(3+4i\right)$ with $z=a+bi$ :

  $\begin{array}{l}a=3\\ b=4\end{array}$

The modulus is calculated as:

  $\begin{array}{l}r=\sqrt{a^2+b^2}\\ r=\sqrt{{\left(3\right)}^2+{\left(4\right)}^2}\\ r=\sqrt{9+16}\\ r=\sqrt{25}\\ r=5\end{array}$

The argument is calculated as:

  $\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{b}{a}\right)\\ \theta ={\tan }^{-1}\left(\frac{4}{3}\right)\\ \theta ={\tan }^{-1}\left(1.33\right)\\ \theta =0.927\end{array}$

The trigonometric form of the complex number is calculated as:

  $\begin{array}{l}{z}_1=r\left(\cos \theta +i\sin \theta \right)\\ z_1=5\left(\cos \left(0.927\right)+i\sin \left(0.927\right)\right)\end{array}$

For $z_2=1-\sqrt{3}i$ :

Compare $z_2=1-\sqrt{3}i$ with $z=a+bi$ :

  $\begin{array}{l}a=1\\ b=-\sqrt{3}\end{array}$

The modulus is calculated as:

  $\begin{array}{l}r=\sqrt{a^2+b^2}\\ r=\sqrt{{\left(1\right)}^2+{\left(-\sqrt{3}\right)}^2}\\ r=\sqrt{1+3}\\ r=2\end{array}$

The argument is calculated as:

  $\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{b}{a}\right)\\ \theta ={\tan }^{-1}\left(\frac{-\sqrt{3}}{1}\right)\\ \theta ={\tan }^{-1}\left(-\sqrt{3}\right)\\ \theta =-\frac{\pi }{3}\end{array}$

The trigonometric form of the complex number is calculated as:

  $\begin{array}{l}{z}_2=r\left(\cos \theta +i\sin \theta \right)\\ z_2=2\left(\cos \left(-\frac{\pi }{3}\right)+i\sin \left(-\frac{\pi }{3}\right)\right)\end{array}$

Therefore, trigonometric form for the given complex number is calculated as:

  $\begin{array}{l}\frac{3+4i}{1-\sqrt{3}i}=\left[5\left(\cos \left(0.927\right)+i\sin \left(0.927\right)\right)\right]\left[2\left(\cos \left(-\frac{\pi }{3}\right)+i\sin \left(-\frac{\pi }{3}\right)\right)\right]\\ \frac{3+4i}{1-\sqrt{3}i}=10\left(\cos 0.927+i\sin 0.927\right)\left(\cos \left(-\frac{\pi }{3}\right)+i\sin \left(-\frac{\pi }{3}\right)\right)\end{array}$

(b)

To determine

To calculate:

The product of complex number of the trigonometric form from subpart (a)

Answer to Problem 69E

The product of complex number of the trigonometric form is $\left[5\left(\cos \left(0.927\right)+i\sin \left(0.927\right)\right)\right]\left[2\left(\cos \left(-\frac{\pi }{3}\right)+i\sin \left(-\frac{\pi }{3}\right)\right)\right]=\frac{5}{2}\left[\cos \left(60.92\right)+i\sin \left(60.92\right)\right]$

Explanation of Solution

Given information:

  $\left[5\left(\cos \left(0.927\right)+i\sin \left(0.927\right)\right)\right]\left[2\left(\cos \left(-\frac{\pi }{3}\right)+i\sin \left(-\frac{\pi }{3}\right)\right)\right]$

Formula used:

The division rule for two complex numbers:

  $\begin{array}{l}\frac{z_1}{z_2}=\frac{r_1\left(\cos {\theta }_1+i\sin {\theta }_1\right)}{r_2\left(\cos {\theta }_2+i\sin {\theta }_2\right)}\\ \frac{z_1}{z_2}=\frac{r_1}{r_2}\cos \left({\theta }_1-{\theta }_2\right)+i\sin \left({\theta }_1-{\theta }_2\right)\end{array}$

Calculation:

The division of complex number is calculated as:

  $\left[5\left(\cos \left(0.927\right)+i\sin \left(0.927\right)\right)\right]\left[2\left(\cos \left(-\frac{\pi }{3}\right)+i\sin \left(-\frac{\pi }{3}\right)\right)\right]$

On comparing it with formula:

  $r_1=5,r_2=2,{\theta }_1=0.927,{\theta }_2=-\frac{\pi }{3}$

Now,

  $\begin{array}{l}\left[5\left(\cos \left(0.927\right)+i\sin \left(0.927\right)\right)\right]\left[2\left(\cos \left(-\frac{\pi }{3}\right)+i\sin \left(-\frac{\pi }{3}\right)\right)\right]=\frac{5}{2}\left[\cos \left(\left(0.927\right)-\left(-\frac{\pi }{3}\right)\right)+i\sin \left(\left(0.927\right)-\left(-\frac{\pi }{3}\right)\right)\right]\\ \left[5\left(\cos \left(0.927\right)+i\sin \left(0.927\right)\right)\right]\left[2\left(\cos \left(-\frac{\pi }{3}\right)+i\sin \left(-\frac{\pi }{3}\right)\right)\right]=\frac{5}{2}\left[\cos \left(60.92\right)+i\sin \left(60.92\right)\right]\end{array}$

(c)

To determine

To calculate:

The product of complex number of the standard form

Answer to Problem 69E

The product of complex number of the standard form is $\frac{3-4\sqrt{3}}{4}+\frac{4+3\sqrt{3}}{4}i$ .

Explanation of Solution

Given information:

  $\frac{3+4i}{1-\sqrt{3}i}$

Calculation:

Know that:

The division of given two vectors $\left(3+4i\right)$ and $\left(1-\sqrt{3}i\right)$ is calculated as:

  $\begin{array}{l}\frac{3+4i}{1-\sqrt{3}i}=\left(\frac{3+4i}{1-\sqrt{3}i}\right)\left(\frac{1+\sqrt{3}i}{1+\sqrt{3}i}\right)\\ \frac{3+4i}{1-\sqrt{3}i}=\frac{3+4i+3\sqrt{3}i+4\sqrt{3}{i}^2}{{\left(1\right)}^2-{\left(\sqrt{3}i\right)}^2}\left[\begin{array}{l}\because {\left(a\right)}^2-{\left(b\right)}^2=\left(a+b\right)\left(a-b\right)\\ i^2=-1\end{array}\right]\\ \frac{3+4i}{1-\sqrt{3}i}=\frac{3+\left(4+3\sqrt{3}\right)i+4\sqrt{3}{i}^2}{1-3\left(-1\right)}\\ \frac{3+4i}{1-\sqrt{3}i}=\frac{3+\left(4+3\sqrt{3}\right)i-4\sqrt{3}}{4}\\ \frac{3+4i}{1-\sqrt{3}i}=\frac{3-4\sqrt{3}}{4}+\frac{4+3\sqrt{3}}{4}i\end{array}$