Higher Engineering Mathematics

Complex numbers

E

23

Complex numbers

23.1 Cartesian complex numbers

(i) If the quadratic equationx^2 + 2 x+ 5 =0is

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 operatorjis defined as

j=

√

−1 then the solution may be expressed as

x=− 1 ±j2.

(ii)− 1 +j2 and− 1 −j2 are known ascomplex

numbers. Both solutions are of the forma+jb,

‘a’ being termed thereal partandjbtheimag-

inary part. A complex number of the form

a+jbis calledcartesian complex number.

(iii) In pure mathematics the symboliis used to

indicate

√

−1(ibeing the first letter of the word

imaginary). Howeveriis 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.

Sincex^2 + 4 =0 thenx^2 =−4 andx=

√

−4.

i.e., x=

√

[(−1)(4)]=

√

(−1)

√

4 =j(±2)

=± j 2 , (sincej=

√

−1)

(Note that±j2 may also be written± 2 j).

Problem 2. Solve the quadratic equation

2 x^2 + 3 x+ 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

Hencex=−

3

4

± j

√

31

4

or− 0. 750 ± j 1. 392 ,

correct to 3 decimal places.

(Note, a graph ofy= 2 x^2 + 3 x+5 does not cross

thex-axis and hence 2x^2 + 3 x+ 5 =0 has no real

roots.)

Problem 3. Evaluate

(a)j^3 (b) j^4 (c)j^23 (d)

− 4

j^9

(a)j^3 = j^2 ×j=(−1)×j=−j, sincej^2 =− 1

(b)j^4 = j^2 ×j^2 =(−1)×(−1)= 1

(c)j^23 =j×j^22 =j×(j^2 )^11 =j×(−1)^11

=j×(−1)=−j

(d)j^9 = j×j^8 =j×(j^2 )^4 =j×(−1)^4

= j× 1 =j

Hence

− 4

j^9

=

− 4

j

=

− 4

j

×

−j

−j

=

4 j

−j^2

=

4 j

−(−1)

= 4 jorj 4