Theorem 1.2.12: Multiplying complex numbers geometrically |
If z = r1 cis(s) and w = r2 cis(t) in polar coordinates thenz*w = r1 r2 cis(s + t)andzn = rn cis(ns) |
The formulas means:
Multiplying vectors z and w
The proof is an exercise in trig identities. Recall that:
(1) cos(x + y) = cos(x)cos(y)-sin(x)sin(y)
(2) sin(x + y) = sin(x)cos(y)+cos(x)sin(y)
Now
z*w = r1 cis(s) r2 cis(t) =
= r1(cos(s)+ i sin(s)) r2(cos(t) + i sin(t)) =
= r1 r2 (cos(s)+ i sin(s))(cos(t) + i sin(t)) =
= r1 r2 (cos(s) cos(t) - sin(s)sin(t)) + i(cos(s)sin(t) + sin(s)cos(t)) =
= r1 r2 (cos(s+t) + i sin(s+t)) =
= r1 r2 cis(s+t)
which proves the first assertion. As for the second, first note that
z2 = z*z = r cis(t) r cis(t) = r2 cis(2t)
by our previous result. Now you should be able to finish the proof by carefully writing down an induction proof yourself.