Complex Analysis - Maths KU

38 Chapter 1 Complex Numbers and the Complex Plane
Note: The roots z1 and z2 are not
conjugates. See Problem 24 in Exercises
1.6.
☞
and
w1 = √ �
10 cos 5π
�
5π
+ i sin =
4 4
√ �
10 − 1 √ −
2 1
�
√ i = −
2 √ 5 − √ 5i.
Therefore, (4) gives two values:
z1 = 1
� �√ √ ��
−1+i + 5+ 5i
2
and z2 = 1
� �
−1+i + −
2
√ 5 − √ ��
5i .
These solutions of the original equation, written in the form z = a + ib, are
z1 = 1
�√ �
5 − 1 +
2
1
�√ �
5+1 i and z2 = −
2
1
�√ �
5+1 −
2
1
�√ �
5 − 1 i.
2
Factoring a Quadratic Polynomial By finding all the roots of a
polynomial equation we can factor the polynomial completely.This statement
follows as a corollary to an important theorem that will be proved in Section
5.5. For the present, note that if z1 and z2 are the roots defined by (3), then
a quadratic polynomial az 2 + bz + c factors as
az 2 + bz + c = a(z − z1)(z − z2). (5)
For example, we have already used (2) to show that the quadratic equation
x 2 − 2x + 10 = 0 has roots z1 = 1+3i and z2 =1− 3i.With a = 1, (5)
enables us to factor the polynomial x 2 − 2x + 10 using complex numbers:
x 2 − 2x +10=[x − (1+3i)] [x − (1 − 3i)] = (x − 1 − 3i)(x − 1+3i).
Similarly, the factorization of the quadratic polynomial in Example 1 is
z 2 +(1−i)z − 3i =(z− z1)(z − z2)
�
= z − 1
�√ �
5 − 1
2
− 1
�√ � ��
5+1 i z +
2
1
�√ �
5+1 +
2
1
�√ � �
5 − 1 i
2
Because a first course in calculus deals principally with real quantities, you
probably did not see any complex numbers until you took a course in differential
equations or in electrical engineering.
Differential Equations The first step in solving a linear second-order
ordinary differential equation ay ′′ + by ′ + cy = f(x) with real coefficients a,
b, and c is to solve the associated homogeneous equation ay ′′ + by ′ + cy =0.
The latter equation possesses solutions of the form y = e mx .To see this, we
substitute y = e mx ,y ′ = me mx ,y ′′ = m 2 e mx into ay ′′ + by ′ + cy =0:
ay ′′ + by ′ + cy = am 2 e mx + bme mx + ce mx = e mx � am 2 + bm + c � =0.
From e mx � am 2 + bm + c � = 0, we see that y = e mx is a solution of the homogeneous
equation whenever m is root of the polynomial equation am 2 + bm +