Chapter 8, Problem 31RE

a.

To determine

Graph the complex number in the complex plane.

  $5+3i$

Answer to Problem 31RE

The graph of complex number in the complex plane $5+3i$ is,

  

Explanation of Solution

Given:

  $5+3i$

Concept Used:

Desmos Graph.

According to the given details,

  $z=5+3i$

  

Conclusion:

Hence, graph is

  

b.

To determine

The modulus and argument.

  $5+3i$

Answer to Problem 31RE

The modulus and argument of $5+3i$ is $r=\sqrt{34},\theta =\arctan \left(\frac{3}{5}\right)$

Explanation of Solution

Given:

  $5+3i$

Concept Used:

Modulus and argument: $r=\sqrt{x^2+y^2}$

Polar form: $z=r\left(\cos \theta +i\sin \theta \right)$

Calculation:

The given complex number is $z=5+3i$ ,

So, write it in the polar form,

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

Now evaluate $r$ and $\theta$ , as know that the formula for these terms,

  $r=\sqrt{x^2+y^2}$

So, from the equation,

  $\begin{array}{l}x=5\\ y=3\end{array}$

Substituting these values in the above formula,

  $\begin{array}{l}r=\sqrt{{\left(5\right)}^2+{\left(3\right)}^2}\\ r=\sqrt{25+9}\\ r=\sqrt{34}\end{array}$

Now evaluate, $\theta$

  $\begin{array}{l}\theta =\arctan \left(\frac{3}{5}\right)\\ \theta =\arctan \left(\frac{3}{5}\right)\end{array}$

Conclusion:

Hence the modulus and argument of $5+3i$ is $r=\sqrt{34},\theta =\arctan \left(\frac{3}{5}\right)$ .

c.

To determine

The polar form of $5+3i$ .

Answer to Problem 31RE

The polar form of $5+3i$ is $z=\sqrt{34}\left(\cos \left(\arctan \left(\frac{3}{5}\right)\right)+i\sin \left(\arctan \left(\frac{3}{5}\right)\right)\right)$

Explanation of Solution

Given:

  $5+3i$

Concept Used:

Polar form: $z=r\left(\cos \theta +i\sin \theta \right)$

Calculation:

  $z=5+3i$

In polar form,

  $\begin{array}{l}z=r\left(\cos \theta +i\sin \theta \right)\\ r=\sqrt{34}\\ \theta =\arctan \left(\frac{3}{5}\right)\end{array}$

So now write these calculated values in standard equation of polar form,

  $z=\sqrt{34}\left(\cos \left(\arctan \left(\frac{3}{5}\right)\right)+i\sin \left(\arctan \left(\frac{3}{5}\right)\right)\right)$

Conclusion:

Hence, the polar form of $5+3i$ is $z=\sqrt{34}\left(\cos \left(\arctan \left(\frac{3}{5}\right)\right)+i\sin \left(\arctan \left(\frac{3}{5}\right)\right)\right)$