Chapter 9.8, Problem 20E

To determine

To find the principal root and express the result in form of $(a+bi)$ with $a$ and $b$ rounded to nearest hundredth.

Answer to Problem 20E

  ${\left(-1\right)}^{\frac{1}{4}}=\sqrt{2}+i\sqrt{2}$

Explanation of Solution

Given:

The given number is ${\left(-1\right)}^{\frac{1}{4}}$

Calculation:

Writing as complex number

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

Using De Moiver’ theorem as follows to find the principal root as follows

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