A First Course in Linear Algebra, 2017a

330 Complex Numbers
Exercise 6.1.5 If z,w are complex numbers prove zw = z w and then show by induction that z 1 ···z m =
z 1 ···z m . Also verify that ∑ m k=1 z k = ∑ m k=1 z k. In words this says the conjugate of a product equals the
product of the conjugates and the conjugate of a sum equals the sum of the conjugates.
Exercise 6.1.6 Suppose p(x)=a n x n + a n−1 x n−1 + ···+ a 1 x + a 0 where all the a k are real numbers. Suppose
also that p(z)=0 for some z ∈ C. Show it follows that p(z)=0 also.
Exercise 6.1.7 I claim that 1 = −1. Here is why.
−1 = i 2 = √ −1 √ −1 =
√
(−1) 2 = √ 1 = 1
This is clearly a remarkable result but is there something wrong with it? If so, what is wrong?
6.2 Polar Form
Outcomes
A. Convert a complex number from standard form to polar form, and from polar form to standard
form.
In the previous section, we identified a complex number z = a+bi with a point (a,b) in the coordinate
plane. There is another form in which we can express the same number, called the polar form. The polar
form is the focus of this section. It will turn out to be very useful if not crucial for certain calculations as
we shall soon see.
Suppose z = a + bi is a complex number, and let r = √ a 2 + b 2 = |z|. Recall that r is the modulus of z
. Note first that
( a
) ( ) 2 b 2
+ = a2 + b 2
r r r 2 = 1
and so ( a
r
, b )
r is a point on the unit circle. Therefore, there exists an angle θ (in radians) such that
cosθ = a r ,sinθ = b r
In other words θ is an angle such that a = r cosθ and b = r sinθ, thatisθ = cos −1 (a/r) and θ =
sin −1 (b/r). We call this angle θ the argument of z.
We often speak of the principal argument of z. This is the unique angle θ ∈ (−π,π] such that
cosθ = a r ,sinθ = b r
The polar form of the complex number z = a + bi = r (cosθ + isinθ) is for convenience written as:
where θ is the argument of z.
z = re iθ