Roots of Complex Number/Examples/Cube Roots of -11-2i/Mistake

Source Work

 * Chapter $1$: Complex Numbers
 * Supplementary Problems: Roots of Complex Numbers: $99$

This mistake can be seen in the 1981 printing of the second edition (1974) as published by Schaum: ISBN 0-070-84382-1

Mistake

 * Find the cube roots of $-11 - 2 i$.


 * Ans. ... $1 + 2 i, \dfrac 1 2 - \sqrt 3 + \paren {1 + \dfrac {\sqrt 3} 2} i, -\dfrac 1 2 - \sqrt 3 + \paren {\dfrac {\sqrt 3} 2 - 1} i$

Correction
The correct solution is:
 * $\set {1 + 2 i, -\dfrac 1 2 + \sqrt 3 + \paren {-1 - \dfrac {\sqrt 3} 2} i, -\dfrac 1 2 - \sqrt 3 + \paren {-1 + \dfrac {\sqrt 3} 2} i}$

as demonstrated in Cube Roots of $-11 - 2 i$.