Chapter 11.3, Problem 20E

To determine

To find: the expression from complex number to polar form.

Answer to Problem 20E

  $\sqrt{26}\left(\text{cos}\left(3.34\right)\text{+isin}\left(3.34\right)\right).$

Explanation of Solution

Given:

  $\text{z=-5-i}$ .

Concept used:

Complex form: $\text{z=a+ib}$

In polar form: $\text{z=r}\left(\text{cosθ+isinθ}\right)$

Where, $\text{a=rcosθ}$ and $\text{b=rsinθ}$ .

Modulus: $\text{r=}\sqrt{{\text{a}}^{\text{2}}{\text{+b}}^{\text{2}}}$ .

Argument: ${\text{θ=tan}}^{\text{-1}}\frac{\text{b}}{\text{a}}$ .

Calculation:

Consider the complex number $\text{z=-5-i}$ .

The polar form of the complex number $\text{z=a+bi}$ is given by:

  $\text{z=r}\left(\text{cosθ+isinθ}\right)$ .

Here, $a=r\cos \theta$ , $\text{b=rsinθ}$ , $\text{r=}\sqrt{{\text{a}}^{\text{2}}{\text{+b}}^{\text{2}}}$ and ${\text{θ=tan}}^{\text{-1}}\frac{\text{b}}{\text{a}}$

Substituting $\text{a=-5 and b=-1}$

Since, the angle lies in the third quadrant value of $x$ and $y$ both are negative:

  $\begin{array}{l}{\text{θ=tan}}^{\text{-1}}\left(\frac{-1}{-5}\right)+\pi .\\ \Rightarrow 3.34rad.\end{array}$

So, the argument is $3.34$ radian.

Let’s find the modulus:

  $\begin{array}{l}\text{r=}\sqrt{{\text{a}}^{\text{2}}{\text{+b}}^{\text{2}}}\\ \Rightarrow \text{r=}\sqrt{{\left(\text{-5}\right)}^{\text{2}}\text{+}{\left(-1\right)}^{\text{2}}}=\sqrt{26}.\\ r=\sqrt{26.}\end{array}$

The modulus of the function is $\sqrt{26}$ .

Hence the polar form of complex number $\text{z=-5-i}$ is $\sqrt{26}\left(\text{cos}\left(3.34\right)\text{+isin}\left(3.34\right)\right).$