## Example 1.2.17 (d): Finding roots geometrically |

Find both square-roots of |

To use our general formula, we need to find length and angle of our vector:

|3i-2|=13^{1/2}

Arg(3i-2)=-arctan(3/2)= 2.158798931 (2nd quadrant)

Therefore the two roots are:

z_{1}= 13^{1/4}cis(2.158798931/2) =

= 13^{1/4}cis(1.079399466) =

= 0.8959774747 + 1.674149229 i

z_{2}= 13^{1/4}cis((2.158798931+2)/2) =

= 13^{1/4}cis(4.220992119) =

= -0.8959774756 - 1.674149228 i

The fact that *z _{1} = -z_{2}* 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.