Example 1.2.17 (d): Finding roots geometrically
Find both square-roots of 3i-2 by (a) using polar coordinates and (b) using rectangular coordinates and a formula from the previous section. Confirm that both methods result in the same answers.
To use our general formula, we need to find length and angle of our vector:
Arg(3i-2)=-arctan(3/2)= 2.158798931 (2nd quadrant)
Therefore the two roots are:
z1 = 131/4 cis(2.158798931/2) =
= 131/4 cis(1.079399466) =
= 0.8959774747 + 1.674149229 i
z2 = 131/4 cis((2.158798931+2)/2) =
= 131/4 cis(4.220992119) =
= -0.8959774756 - 1.674149228 i
The fact that z1 = -z2 is no accident - can you explain it?
It is up to you to confim this answer using an alternate formula. And while we're at it, why not try the good old quadratic formula? It should result in the same answer.