Chapter 8.3, Problem 78E

To determine

To calculate: To find the indicated power of complex number using De Moivre’s Theorem

Answer to Problem 78E

  ${\left(3+\sqrt{3}i\right)}^4=-72+72\sqrt{3}i$

Explanation of Solution

Given information: Complex number is ${\left(3+\sqrt{3}i\right)}^4$

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

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

According to De Moivre’s Theorem,

  $z^n=r^n\left(\cos n\theta +isinn\theta \right)$

Calculation:

Complex number is given as

  ${\left(3+\sqrt{3}i\right)}^4$

Let us consider complex number:

  $z=3+\sqrt{3}i$

Polar form of complex number is

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

Calculating the value of $r$ :

  $\begin{array}{l}r=\sqrt{{\left(3\right)}^2+{\left(\sqrt{3}\right)}^2}\\ r=\sqrt{9+3}\\ r=\sqrt{12}\\ r=2\sqrt{3}\end{array}$

Calculating the value of $\theta :$

  $\begin{array}{l}\tan \theta =\frac{\sqrt{3}}{3}\\ \tan \theta =\frac{1}{\sqrt{3}}\\ \theta ={\tan }^{-1}\frac{1}{\sqrt{3}}\\ \theta =\frac{\pi }{6}\end{array}$

Hence, complex number in polar form is

  $z=2\sqrt{3}\left(\cos \left(\frac{\pi }{6}\right)+isin\left(\frac{\pi }{6}\right)\right)$

Now, ${\left(3+\sqrt{3}i\right)}^4$ is calculated using De Moivre’s Theorem as follows:

  $\begin{array}{l}{\left(3+\sqrt{3}i\right)}^4={\left(2\sqrt{3}\left(\cos \left(\frac{\pi }{6}\right)+isin\left(\frac{\pi }{6}\right)\right)\right)}^4\\ {\left(3+\sqrt{3}i\right)}^4={\left(2\sqrt{3}\right)}^4\left(\cos \left(4\times \frac{\pi }{6}\right)+isin\left(4\times \frac{\pi }{6}\right)\right)\\ {\left(3+\sqrt{3}i\right)}^4=\left(16\times 9\right)\left(\cos \left(\frac{2\pi }{3}\right)+isin\left(\frac{2\pi }{3}\right)\right)\\ {\left(3+\sqrt{3}i\right)}^4=144\left(\left(-\frac{1}{2}\right)+i\left(\frac{\sqrt{3}}{2}\right)\right)\\ {\left(3+\sqrt{3}i\right)}^4=144\left(\frac{-1+\sqrt{3}i}{2}\right)\\ {\left(3+\sqrt{3}i\right)}^4=72\left(-1+\sqrt{3}i\right)\\ {\left(3+\sqrt{3}i\right)}^4=-72+72\sqrt{3}i\end{array}$

Conclusion:

Hence, ${\left(3+\sqrt{3}i\right)}^4=-72+72\sqrt{3}i$