Chapter 9.8, Problem 22E

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 22E

  ${\left(4-i\right)}^{\frac{1}{3}}=1.584-0.1328i$

Explanation of Solution

Given:

The given complex number is ${\left(4-i\right)}^{\frac{1}{3}}$

Calculation:

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

  $r=\sqrt{4^2+{(-1)}^2}=\sqrt{17}$

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

  $\begin{array}{l}\theta =\arctan \left(\frac{-1}{4}\right)\\ \theta =-0.25\text{ rad}\end{array}$

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

  $\begin{array}{l}{\left(4-i\right)}^{\frac{1}{3}}={\left[\sqrt{17}\left(\cos \left(-0.25\right)+i\sin \left(-0.25\right)\right)\right]}^{\frac{1}{3}}\\ {\left(4-i\right)}^{\frac{1}{3}}=\sqrt[6]{17}\left(\cos \left(\frac{-0.25}{3}\right)+i\sin \left(\frac{-0.25}{3}\right)\right)\\ \approx 1.60(0.99-0.083i)\\ {\left(4-i\right)}^{\frac{1}{3}}=1.584-0.1328i\\ \end{array}$