Chapter 20
Complex numbers
20.1 Cartesian complex numbers
There are several applications of complex numbers
in science and engineering, in particular in electrical
alternating current theory and in mechanical vector
analysis.
There are two main forms of complex number –
Cartesian form and polar form – and both are
explained in this chapter.
If we can add, subtract, multiply and divide complex
numbers in both forms and represent the numbers on
an Argand diagram then a.c. theory and vector analysis
become considerably easier.
(i) If the quadratic equation x 2 + 2x + 5 = 0 is
solved using the quadratic formula then,
x =
−2 ±
[(2) 2 − (4)(1)(5)]
2(1)
=
−2 ±
√
[−16]
2
=
−2 ±
√
[(16)(−1)]
2
=
−2 ±
√
16
√
−1
2
=
−2 ± 4
√
−1
2
= −1 ± 2
√ −1
It is not possible to evaluate
√
−1 in real
terms. However, if an operator j is defined as
j =
√
−1 then the solution may be expressed as
x =−1 ± j 2.
(ii) −1 + j 2 and −1 − j 2 are known as complex
numbers. Both solutions are of the form a + jb,
‘a’ being termed the real part and jb the
imaginary part. A complex number of the form
a + jb is called Cartesian complex number.
(iii) In pure mathematics the symbol i is used to
indicate
√
−1 (i being the first letter of the word
imaginary). However i is the symbol of electric
current in engineering, and to avoid possible confusion the next letter in the alphabet, j , is used to
represent
√
−1.
Problem 1. Solve the quadratic equation
x 2 + 4 = 0.
Since x
2
+ 4 =0 then x
2
=−4 and x =
√
−4.
i.e., x =
[(−1)(4)] =
(−1)
√
4 = j (±2)
= ± j2, (since j =
√
−1)
(Note that ± j 2 may also be written ±2 j).
Problem 2. Solve the quadratic equation
2x
2
+ 3x + 5 = 0.
Using the quadratic formula,
x =
−3 ±
[(3) 2 − 4(2)(5)]
2(2)
=
−3 ±
√
−31
4
=
−3 ±
√
(−1)
√
31
4
=
−3 ± j
√
31
4
Hence x =−
3
4
± j
√
31
4
or −0.750 ± j1.392,
correct to 3 decimal places.
(Note, a graph of y = 2x 2 + 3x + 5 does not cross
the x-axis and hence 2x 2 + 3x + 5 = 0 has no real
roots.)
Complex numbers
20.1 Cartesian complex numbers
There are several applications of complex numbers
in science and engineering, in particular in electrical
alternating current theory and in mechanical vector
analysis.
There are two main forms of complex number –
Cartesian form and polar form – and both are
explained in this chapter.
If we can add, subtract, multiply and divide complex
numbers in both forms and represent the numbers on
an Argand diagram then a.c. theory and vector analysis
become considerably easier.
(i) If the quadratic equation x 2 + 2x + 5 = 0 is
solved using the quadratic formula then,
x =
−2 ±
[(2) 2 − (4)(1)(5)]
2(1)
=
−2 ±
√
[−16]
2
=
−2 ±
√
[(16)(−1)]
2
=
−2 ±
√
16
√
−1
2
=
−2 ± 4
√
−1
2
= −1 ± 2
√ −1
It is not possible to evaluate
√
−1 in real
terms. However, if an operator j is defined as
j =
√
−1 then the solution may be expressed as
x =−1 ± j 2.
(ii) −1 + j 2 and −1 − j 2 are known as complex
numbers. Both solutions are of the form a + jb,
‘a’ being termed the real part and jb the
imaginary part. A complex number of the form
a + jb is called Cartesian complex number.
(iii) In pure mathematics the symbol i is used to
indicate
√
−1 (i being the first letter of the word
imaginary). However i is the symbol of electric
current in engineering, and to avoid possible confusion the next letter in the alphabet, j , is used to
represent
√
−1.
Problem 1. Solve the quadratic equation
x 2 + 4 = 0.
Since x
2
+ 4 =0 then x
2
=−4 and x =
√
−4.
i.e., x =
[(−1)(4)] =
(−1)
√
4 = j (±2)
= ± j2, (since j =
√
−1)
(Note that ± j 2 may also be written ±2 j).
Problem 2. Solve the quadratic equation
2x
2
+ 3x + 5 = 0.
Using the quadratic formula,
x =
−3 ±
[(3) 2 − 4(2)(5)]
2(2)
=
−3 ±
√
−31
4
=
−3 ±
√
(−1)
√
31
4
=
−3 ± j
√
31
4
Hence x =−
3
4
± j
√
31
4
or −0.750 ± j1.392,
correct to 3 decimal places.
(Note, a graph of y = 2x 2 + 3x + 5 does not cross
the x-axis and hence 2x 2 + 3x + 5 = 0 has no real
roots.)