Roots of Complex Number/Examples/z^6 + 1 = root 3 i/Mistake

Source Work

 * Chapter $1$: Complex Numbers
 * Supplementary Problems: Roots of Complex Numbers: $97 \ \text{(b)}$

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

Mistake

 * Solve the equations ... $\text{(b)}$ $z^6 + 1 = \sqrt 3 i$


 * ''Ans. ... $\set {\sqrt [6] 2 \cis 40 \degrees, \sqrt [6] 2 \cis 1000 \degrees, \sqrt [6] 2 \cis 160 \degrees, \sqrt [6] 2 \cis 220 \degrees, \sqrt [6] 2 \cis 280 \degrees, \sqrt [6] 2 \cis 340 \degrees}$

''

Correction
The correct solution is:
 * $\set {\sqrt [6] 2 \cis 20 \degrees, \sqrt [6] 2 \cis 80 \degrees, \sqrt [6] 2 \cis 140 \degrees, \sqrt [6] 2 \cis 200 \degrees, \sqrt [6] 2 \cis 260 \degrees, \sqrt [6] 2 \cis 320 \degrees}$

as demonstrated in Roots of $z^6 + 1 = \sqrt 3 i$.

The mistake probably arose from mistaking $\dfrac {2 \pi} 3 = 240 \degrees$ when calculating the arguments in degrees.