Chapter 8.3, Problem 45E

To determine

To calculate: To write the complex number in polar form with argument $\theta$ between $0$ and $2\pi$

Answer to Problem 45E

Complex number in polar form is $z=3\sqrt{2}\left(\cos \frac{3\pi }{4}+isin\frac{3\pi }{4}\right)$

Explanation of Solution

Given information: Complex numbers is $z=3i\left(1+i\right)$

Formula Used:

Complex number is a number that can be expressed in the form of $a+bi$ , where $a$ and $b$ are real numbers

  $z=a+bi$

Polar form of the complex number is given as

  $z=r\left(\cos \theta +isin\theta \right)$

Where,

  $r=\left|z\right|=\sqrt{a^2+b^2}$

  $\tan \theta =\frac{b}{a}$

Calculation:

Complex number is given as

  $z=3i\left(1+i\right)$

On simplifying,

  $\begin{array}{l}z=3i+3i^2\\ z=3i+3\left(-1\right)\\ z=-3+3i\end{array}$

Polar form of the complex number is given as

  $z=r\left(\cos \theta +isin\theta \right)----(1)$

Calculating the value of $r$ as follows:

  $r=\left|z\right|=\sqrt{a^2+b^2}$

Substituting the values,

  $\begin{array}{l}r=\sqrt{{\left(-3\right)}^2+{\left(3\right)}^2}\\ r=\sqrt{9+9}\\ r=\sqrt{18}\\ r=3\sqrt{2}\end{array}$

Calculating the value of $\theta$ as follows:

  $\tan \theta =\frac{b}{a}$

Substituting the values,

  $\begin{array}{l}\tan \theta =\frac{3}{-3}\\ \tan \theta =-1\\ \theta ={\tan }^{-1}-1\end{array}$

Here, solution lies in second quadrant

  $\begin{array}{l}\theta =\pi -\frac{\pi }{4}\\ \theta =\frac{3\pi }{4}\end{array}$

Now, substituting the values of $r=3\sqrt{2}$ and $\theta =\frac{3\pi }{4}$ in equation (1) to get the complex number in polar form

  $z=3\sqrt{2}\left(\cos \frac{3\pi }{4}+isin\frac{3\pi }{4}\right)$

Conclusion:

Hence, complex number in polar form is $z=3\sqrt{2}\left(\cos \frac{3\pi }{4}+isin\frac{3\pi }{4}\right)$