Chapter 9.8, Problem 21E

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 21E

  ${\left(-2+i\right)}^{\frac{1}{4}}=1.2078-1.1464i$

Explanation of Solution

Given:

The given complex number is ${\left(-2+i\right)}^{\frac{1}{4}}$

Calculation:

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

  $r=\sqrt{{\left(-2\right)}^2+{(1)}^2}=\sqrt{5}$

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

  $\begin{array}{l}\theta =\arctan \left(\frac{1}{-2}\right)\\ \theta =-0.46\text{ rad}\end{array}$

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

  $\begin{array}{l}{\left(-2+i\right)}^{\frac{1}{4}}={\left[\sqrt{5}\left(\cos \left(-0.46\right)+i\sin \left(-0.46\right)\right)\right]}^{\frac{1}{4}}\\ {\left(-2+i\right)}^{\frac{1}{4}}=\sqrt[8]{5}\left(\cos \left(\frac{-0.46}{4}\right)+i\sin \left(\frac{-0.46}{4}\right)\right)\\ =1.22(0.99-.12i)\\ {\left(-2+i\right)}^{\frac{1}{4}}=1.2078-1.1464i\\ \end{array}$