Complex numbersi or j

Complex numbers

An Imaginary Number, when squared, gives a negative result.

imaginary2 = negative

Complex numbers

i = √-1 i is used in maths

Butj is used in electronics and

engineering (because i is already used as a symbol for current)

Complex numbers

i = √-1

i2 = -1

i3 = -√-1

i4 = 1

i5 = √-1

Complex numbers

Example What is i6 ?

i6 = i4 × i2 = 1 × -1

= -1

Adding complex numbers

•(4 +j3) + (3 + j5)•4 +j3 + 3 + j5

•7 + j8

Adding complex numbers

•(3 +j6) + (2 – j3)•3 +j6 + 2 – j3

•5 + j3

Subtracting complex numbers

•(6 +j8) - (2 + j3)•6 +j8 - 2 – j3

•4 + j5Note the change of

sign

Multiplying complex numbers

Example 1,

6(3 +j4) =

18 + j24

Example 2

j8 + 3(3 – j2) =

j8 + 9 – j6

j2 + 9

Multiplying complex numbers

(3 + j2)(4 + j)Use F.O.I.L.

(3x4) + (3xj) + (j2 x4) + (j2 x j)12 + j3 + j8 + j22

j2 = -1

12 +j11 – 210 + j11

Multiplying complex numbers

(5 - j2)(2 + j2)Use F.O.I.L.

(5 x 2) + (5 x j2) - (j2 x2) - (-j2 x j2)10 + j10 – j4 - j24

j2 = -1

10 – j6 + 414 - j10

Multiplying complex numbers

(4 - j2)(3 - j)Use F.O.I.L.

(4 x 3) - (4 x j) - (j2 x3) + (j2 x j)12 – j4 – j6 + j22

j2 = -1

12 - j10 – 210 - j10

Multiplying a conjugate pair

(4 - j2)(4 + j2)Use F.O.I.L.

(4 x 4) + (4 x j2) - (j2 x 4) - (j2 x j2)16 + j8 – j8 - j24

j2 = -1

16 + 420

Dividing complex numbers

(2 +6j)/2j =2/2j + 6j/2j =

1/j +3J-1 + 3

Dividing complex numbers

(6 + j3)/ (3+j2)Multiply by the conjugate of the denominator

(6 + j3) x (3 – j2)(3 + j2) (3 - j2)

18 – j12 +j9 –j269 – j6 + j6 –j24

Dividing complex numbers

18 - j3 + 69 – j24=24 – j3

9+424 – j3

13

Argand Diagrams

Imaginary axis y

Real axis x

Z = x +yj

Argand Diagrams

r

r = √(x2 + y2)

Argand Diagrams

Φ

tanΦ = y/x

Argand Diagrams

Imaginary axis y

Real axis x

Z = x +yj

r

Φ

yj = r sinΦ

x = r cosΦ

Example

•Argand diagrams are used to calculate

impedance in RLC circuits

Example

The impedance of a circuit is given by the complex number 3 +j4

Construct the Argand diagram for 3 +j4

Example

Imaginary axis y

Real axis x

Z = 3 +j4

j4

3

r

Example

From the Argand diagram derive the expression for the impedance in polar

form

Example

Imaginary axis y

Real axis x

Z = 3 +j4

j4

3

r

r = √(32 + 42) = √(9 + 16) √(25) = 5

Example

Imaginary axis y

Real axis x

Z = 3 +j4

j4

3

r

tanΦ = 4/3Φ = 53.13

Example

Imaginary axis y

Real axis x

Z = 3 +j4

j4

3

r

AnswerZ = 5 53.13

Multiplying and dividing polar form

6∟20° x 4∟30°Multiply the length (modulus) and add the argument (angle)

= 24∟50°

9∟10° / 3∟40° = 9/3 ∟(10°-40°) divide the length (modulus) and subtract the argument (angle)

= 3∟-30°

Argand diagrams as phasor diagrams

The voltage of a circuit is given as V = 3 + j3

and the current drawn is given as I = 8 + j2

Find the phase difference between V and I

Find the power (VI.cosФ)

Argand diagrams as phasor diagrams

Voltage = √ (32 + 32) = √18 = 4.24 Volts

Current = √(82 + 22) = √ 68 = 8.25 amps

Argand diagrams as phasor diagrams

Voltage phase angle tanΦ = 3/3 =1, Φ = 45o

Current phase angle tanΦ = 2/8 =0.25,

Φ = 14.0o

•

Argand diagrams as phasor diagrams

Phase difference between V and I = 45o - 14.0o = 31o

power = VIcosΦ4.24 x 8.25 cos31o

4.24 x 8.25 x .86= 30 watts