Chapter 9.7, Problem 2GCE

To determine

To approximate the fourth roots of $i$ .

Answer to Problem 2GCE

  ${\left(i\right)}^{\frac{1}{4}}=\left(\cos \left(\frac{\pi }{8}\right)+i\sin \left(\frac{\pi }{8}\right)\right)$

Explanation of Solution

Given:

The fourth roots of $i$ .

Calculation:

Using graphing calculator, the fourth roots of $i$ can be approximated and these are as follows-

Determine $r=\sqrt{x^2+y^2}$

  $r=\sqrt{{\left(0\right)}^2+{(1)}^2}=1$

Now, determine $\theta =\arctan \left(\frac{y}{x}\right)$

  $\begin{array}{l}\theta =\arctan \left(\frac{1}{0}\right)\\ \theta =\frac{\pi }{2}\end{array}$

Now using De Moiver’ theorem as follows to find the principal root as follows

  $\begin{array}{l}i=1\left(\cos \left(\frac{\pi }{2}\right)+i\sin \left(\frac{\pi }{2}\right)\right)\\ {\left(i\right)}^{\frac{1}{4}}={\left(1\right)}^{\frac{1}{4}}\left(\cos \left(\frac{1}{4}\cdot \frac{\pi }{2}\right)+i\sin \left(\frac{1}{4}\cdot \frac{\pi }{2}\right)\right)\\ {\left(i\right)}^{\frac{1}{4}}=\left(\cos \left(\frac{\pi }{8}\right)+i\sin \left(\frac{\pi }{8}\right)\right)\end{array}$