4.2 COMPLEX NUMBERS 1274.2. 11 For each positive integer n, there are n different nth roots of unity, namely
1 , S', S'^2 , S'^3 , ... , S'n-l,
whereS' = Cis
2n
n
Geometrically, the nth roots of unity are the vertices of a regular n-gon that is inscribedin the unit circle (the set {z E C : Izl = I}) with one vertex at 1.
4.2.12 Let S' = Cis^2 :: as above. Then for each positive integer n, and for each complex
number x,
(a) xn - 1 = (x - 1) (x - S') (x -S'^2 ) ". (x -S' n-l ),
(b) xn-^1 +xn-^2 + ... +x+ 1 = (x -S')(x-S'^2 ) ... (x - S'n-l),(c) I+S'+S'^2 +".+S'n-l=o.
Can you see why (c) is true without using the formula for the sum of geometric series?Some Applications
We will conclude this section with a few examples of interesting use of complex num
bers in several branches of mathematics, including trigonometry, geometry, and num
ber theory.
Example 4.2.13 Find a formula for tan(2a).Solution: This can of course be done in many ways, but the complex numbers
method is quite slick, and works easily with many other trig identities. The key idea is
that if z = x+ iy, then tan(argz) = y/x. Lett := tan a and
Now square z, and we getButz = 1 +it.
Z^2 = ( 1 + it)^2 = 1 - t^2 + 2it.
tan( argz^2 ) = --2t
1 - t^2 '
and of course argz^2 = 2a, so we conclude that
2tanatan2a = 2.
I-tan aExample 4.2. 14 (Putnam 1996) Let Cl and C 2 be circles whose centers are 10 units
apart, and whose radii are 1 and 3. Find, with proof, the locus of all points M for which
there exist points X on C 1 and Y on C 2 such that M is the midpoint of the line segment
XY.