4.2 COMPLEX NUMBERS 123
Clearly this argument generalizes to multiplication by any complex number.
Multiplication by the complex number rCis e is a counterclockwise rota
tion by e followed by stretching by the fa ctor r.
So we have a third way to think about complex numbers. Every complex number is
simultaneously a point, a vector, and a geometric transformation, namely the rotation
and stretching above!
4.2.5 Division. It is easy now to determine the geometric meaning of z/w, where
z = rCisa and w = sCisf3. Let v = z/w = tCisy. Then vw = z. Using the rules for
multiplication, we have
t s = r, y + f3 = a,
and consequently
Thus
t=r/s, y=a-f3.
The geometric meaning of division by rCis e is clockwise rotation by e
(counterclockwise rotation by -e) fo llowed by stretching by the factor 1/ r.
4.2.6 De Moivre's Theorem. An easy consequence of the rules for multiplication and
division is this lovely trigonometric identity, true for any integer n, positive or negative,
and any real e:
(cos e + i sin et = cosne + i sinne.
4.2.7 Exponential Form. The algebraic behavior of argument under multiplication and
division should remind you of exponentiation. Indeed, Euler's formula states that
Cis e = eiO ,
where e = 2. 71828 ... is the familiar natural logarithm base that you have encountered
in calculus. This is a useful notation, somewhat less cumbersome than Cis e, and
quite profound, besides. Most calculus and complex analysis textbooks prove Euler's
formula using the power series for e, sin x, and cosx, but this doesn't really give much
insight as to why it is true. This is a deep and interesting issue that is beyond the scope
of this book. Consult [29] for a thorough treatment, and try Problem 4.2.29 below.
4.2.8 Easy Practice Exercises. Use the above to verify the following.
(a) Izwl = Izllwl and Iz/wl = Izl/Iwl.
(b) Re z =! (z + z) and 1m z = d:z (z -z).
(c) zz= Iz 12.
(d) The midpoint of the line segment joining the complex numbers z, w is (z+w)/2.
Make sure you can visualize this!
(e) z+w = z+w and zw = zw and z/w = z/w.
(f) (1 +i)lO = 32i and (1- i v'3)^5 = 16(1 +iV3).
(g) Show by drawing a picture that z = {} (1 + i) satisfies z^2 = i.