The Student Room Group

Regular polygons in the complex plane.

When solving equations such as z^n+w=0, where w is a known complex number, i've noticed that whenever shown in an argand diagram the points representing z seem to form the vertices of a regular polygon. Does this always occur for any known integer n, where n>2?
I might have a go at proving the general case if I get bored with my revision. So far i've only shown that any two adjacent points representing z are of equal distance from each other, and that all points are equidistant from the origin. What else would I need to show for it to be a correct proof? In other words, what other properties are required for something to be a regular polygon? (other than the sides to be all the same length)
Reply 1
Gaz031

When solving equations such as z^n+w=0, where w is a known complex number, i've noticed that whenever shown in an argand diagram the points representing z seem to form the vertices of a regular polygon. Does this always occur for any known integer n, where n>2?


Yes

Gaz031

I might have a go at proving the general case if I get bored with my revision. So far i've only shown that any two adjacent points representing z are of equal distance from each other, and that all points are equidistant from the origin. What else would I need to show for it to be a correct proof?


That's actually enough.

A regular polygon has equal length sides and equal internal angles, but you can deduce this from the fact that they are of equal length and equidistant from the origin. This guarantees that the triangles OAB where O is the origin and A and B are consecutive roots are all congruent triangles
Reply 2
RichE
Yes
That's actually enough.

A regular polygon has equal length sides and equal internal angles, but you can deduce this from the fact that they are of equal length and equidistant from the origin. This guarantees that the triangles OAB where O is the origin and A and B are consecutive roots are all congruent triangles

Thanks for that. It's easiest to do it by letting z^n=r(cos(t+2kpi)+isin(t+2kpi)) so points in the argand diagram have coordinates: x=[r^(1/n)][cos(t/n+2kpi/n], y=[r^(1/n)][cos(t/n+2kpi/n)]. k is the only variable distinguishing between the points so using trigonometric identities it's simple to show the distance to the origin and distance between adjacent points are independent of k.
Thanks for the explanation of the angles, as all lengths in the triangles are constant it stands to reason the internal angles are all constant :smile: