Complex Analysis - Maths KU

Note: The roots z1 and z2 are conjugates.
See Problem 20 in Exercises
1.6.
√
Interpretation of in this text
☞
☞
1.6 Applications 37
equations can be complex as well as real.For example, any second degree, or
quadratic, polynomial equation can solved by completing the square.In the
general case, ax 2 + bx + c = 0, where the coefficients a �= 0,b, and c are real,
completion of the square in x yields the quadratic formula:
x = −b ± √ b2 − 4ac
. (1)
2a
When the discriminant b 2 − 4ac is negative, the roots of the equation are
complex.For example, by (1) the two roots of x 2 − 2x +10=0are
x = −(−2) ± � (−2) 2 − 4(1)(10)
2(1)
= 2 ± √ −36
. (2)
2
In beginning courses the imaginary unit i is written i = √ −1 and the assumption
is made that the laws of exponents hold so that a number such as √ −36
can be written √ −36 = √ 36 √ −1=6i.Let us denote the two complex roots
in (2) as z1 =1+3i and z2 =1− 3i.
Quadratic Formula The quadratic formula is perfectly valid when the
coefficients a �= 0,b, and c of a quadratic polynomial equation az 2 +bz +c =0
are complex numbers.Although the formula can be obtained in exactly the
same manner as (1), we choose to write the result as
z = −b +(b2 − 4ac) 1/2
. (3)
2a
Notice that the numerator of the right-hand side of (3) looks a little different
than the traditional −b ± √ b2 − 4ac given in (1).Bear in mind that when
b2 − 4ac �= 0, the symbol � b2 − 4ac �1/2 represents the set of two square roots
of the complex number b2 −4ac.Thus, (3) gives
√
two complex solutions.From
this point on we reserve the use of the symbol
to real numbers where √ a
denotes the nonnegative root of the real number a ≥ 0.The next example
illustrates the use of (3).
EXAMPLE 1 Using the Quadratic Formula
Solve the quadratic equation z2 +(1−i)z − 3i =0.
Solution From (3), with a =1,b=1− i, and c = −3i we have
z = −(1 − i) + [(1 − i)2 − 4(−3i)] 1/2
2
= 1
�
−1+i + (10i)
2
1/2�
. (4)
To compute (10i) 1/2 we use (4) of Section 1.4 with r = √ 10, θ = π/2, and
n =2,k=0,k= 1.The two square roots of 10i are:
w0 = √ �
10 cos π π
�
+ i sin =
4 4
√ �
1
10 √2 + 1
�
√ i =
2 √ 5+ √ 5i