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Complex numbers

Complex numbers. i or j . Complex numbers. An Imaginary Number, when squared, gives a negative result. imaginary 2 = negative. Complex numbers. i = √-1 i is used in maths But j is used in electronics and engineering (because i is already used as a symbol for current).

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Complex numbers

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  1. Complex numbers i or j

  2. Complex numbers An Imaginary Number, when squared, gives a negative result. imaginary2 = negative

  3. Complex numbers i = √-1 i is used in maths But j is used in electronics and engineering (because i is already used as a symbol for current)

  4. Complex numbers i = √-1 i2 = -1 i3 = -√-1 i4 = 1 i5= √-1

  5. Complex numbers Example What is i6 ? i6= i4 × i2 = 1 × -1 = -1

  6. Adding complex numbers • (4 +j3) + (3 + j5) • 4 +j3 + 3 + j5 • 7 + j8

  7. Adding complex numbers • (3 +j6) + (2 – j3) • 3 +j6 + 2 – j3 • 5 + j3

  8. Subtracting complex numbers • (6 +j8) - (2 + j3) • 6 +j8 - 2 – j3 • 4 + j5 Note the change of sign

  9. Multiplying complex numbers Example 1, 6(3 +j4) = 18 + j24 Example 2 j8 + 3(3 – j2) = j8 + 9 – j6 j2 + 9

  10. 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 – 2 10 + j11

  11. 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 + 4 14 - j10

  12. 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 – 2 10 - j10

  13. 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 + 4 20

  14. Dividing complex numbers (2 +6j)/2j = 2/2j + 6j/2j = 1/j +3 J-1 + 3

  15. 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 –j26 9 – j6 + j6 –j24

  16. Dividing complex numbers 18 - j3 + 6 9 – j24= 24 – j3 9+4 24 – j3 13

  17. Argand Diagrams Imaginary axis y Z = x +yj Real axis x

  18. Argand Diagrams r = √(x2 + y2) r

  19. Argand Diagrams tanΦ = y/x Φ

  20. Argand Diagrams Imaginary axis y x = r cosΦ Z = x +yj • r yj = r sinΦ Φ Real axis x

  21. Example • Argand diagrams are used to calculate impedance in RLC circuits

  22. Example The impedance of a circuit is given by the complex number 3 +j4 Construct the Argand diagram for 3 +j4

  23. Example Imaginary axis y Z = 3 +j4 3 r j4 Real axis x

  24. Example From the Argand diagram derive the expression for the impedance in polarform

  25. Example r = √(32+ 42) = √(9 + 16) √(25) = 5 Imaginary axis y Z = 3 +j4 3 r j4 Real axis x

  26. Example tanΦ = 4/3 Φ = 53.13 Imaginary axis y Z = 3 +j4 3 r j4 Real axis x

  27. Example Answer Z = 5 53.13 Imaginary axis y Z = 3 +j4 3 r j4 Real axis x

  28. 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°

  29. 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)

  30. Argand diagrams as phasor diagrams Voltage = √ (32 + 32) = √18 = 4.24 Volts Current = √(82 + 22) = √ 68 = 8.25 amps

  31. 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

  32. Argand diagrams as phasor diagrams Phase difference between V and I = 45o - 14.0o= 31o power = VIcosΦ 4.24 x 8.25 cos31o 4.24 x 8.25 x .86 = 30 watts

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