Chapter 6.6, Problem 21E

To determine

To identify: The trigonometric form of the complex number $3+\sqrt{3}i$ .

Answer to Problem 21E

  $3+\sqrt{3}i=\pm 2\sqrt{3}\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right)$ .

Explanation of Solution

Given information:

The complex number $3+\sqrt{3}i$ .

Concept Involved:

The general complex number is the form of $x+iy$ .

Here x is a real number and y is a complex number.

Use the concept of De Moivre’s theorem to change the given number in the trigonometric form such that,

  ${\left[r\left(\cos \theta +i\sin \theta \right)\right]}^n=r^n\left(\cos n\theta +\sin n\theta \right)$

Sketch the complex number $3+\sqrt{3}i$ , where $x=3$ and $y=\sqrt{3}$ .

  

Consider $3+\sqrt{3}i={\left[r\left(\cos \theta +i\sin \theta \right)\right]}^n$

On comparing both sides,

  $\begin{array}{l}n=1\\ r\cos \theta =3\\ r\sin \theta =\sqrt{3}\end{array}$

On squaring and adding above both equations,

  $\begin{array}{c}{r}^2{\cos }^2\theta +r^2{\sin }^2\theta =3^2+{\left(\sqrt{3}\right)}^2\\ r^2\left({\cos }^2\theta +{\sin }^2\theta \right)=9+3\end{array}$

Since,

  ${\cos }^2\theta +{\sin }^2\theta =1$

Then,

  $\begin{array}{c}{r}^2=12\\ r=\pm 2\sqrt{3}\end{array}$

And

  $\begin{array}{l}\frac{r\sin \theta }{r\cos \theta }=\frac{\sqrt{3}}{3}\\ \tan \theta =\frac{1}{\sqrt{3}}\\ \theta ={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)\\ \theta =\frac{\pi }{6}\end{array}$

Therefore,

  $\begin{array}{c}3+\sqrt{3}i={\left[r\left(\cos \theta +i\sin \theta \right)\right]}^n\\ ={\left[\pm 2\sqrt{3}\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right)\right]}^1\\ =\pm 2\sqrt{3}\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right)\end{array}$

Conclusion:

The trigonometric form of the complex number $3+\sqrt{3}i$ is

  $\pm 2\sqrt{3}\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right)$ .