Chapter 9, Problem 26RE

To determine

To find : the complex cube roots of 27.

Answer to Problem 26RE

  $z_0,z_1,z_{2,}$ are the cube roots of 27.

Explanation of Solution

Given information: the standard form is a+ib

Calculation:

now using D’moivres theorem we find the three roots ,

  $\Rightarrow 27=27(\cos 0+i\sin 0)$ are given by,

  $\Rightarrow z_k=\sqrt[3]{27}(\cos \frac{0+2k\pi }{3}+i\sin \frac{0+2k\pi }{3})$

  $=3(\cos \frac{2k\pi }{3}+i\sin \frac{2k\pi }{3})$ when k=0,1,2,

Putting k =0,1,2

  $\begin{array}{l}\Rightarrow z_0=3(\cos \frac{2\pi \times 0}{3}+i\sin \frac{2\pi \times 0}{3})\\ \Rightarrow z_1=3(\cos \frac{2\pi \times 1}{3}+i\sin \frac{2\pi \times 1}{3})\\ \Rightarrow z_2=3(\cos \frac{2\pi \times 2}{3}+i\sin \frac{2\pi \times 2}{3})\end{array}$

Thus $z_0,z_1,z_{2,}$ are the cube roots of 27.