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### Complex Numbers 1 - Math Academy - JC H2 maths A levels

• 2. Basic operations Deﬁnition Let i = √ −1. A complex number is a number of the form z = x + iy, where x, y ∈ R x is the real part of z, denoted by Re(z). y is the imaginary part of z, denoted by Im(z). Powers of i i0 1 i i i2 −1 i3 −i i4 1 i5 i i6 −1 i7 −i The pattern 1, i, −1, −i, . . . will repeat itself. c⃝ 2015 Math Academy www.MathAcademy.sg 2
• 3. Basic operations Deﬁnition Let i = √ −1. A complex number is a number of the form z = x + iy, where x, y ∈ R x is the real part of z, denoted by Re(z). y is the imaginary part of z, denoted by Im(z). Powers of i i0 1 i i i2 −1 i3 −i i4 1 i5 i i6 −1 i7 −i The pattern 1, i, −1, −i, . . . will repeat itself. c⃝ 2015 Math Academy www.MathAcademy.sg 3
• 4. Basic operations Deﬁnition Let i = √ −1. A complex number is a number of the form z = x + iy, where x, y ∈ R x is the real part of z, denoted by Re(z). y is the imaginary part of z, denoted by Im(z). Powers of i i0 1 i i i2 −1 i3 −i i4 1 i5 i i6 −1 i7 −i The pattern 1, i, −1, −i, . . . will repeat itself. c⃝ 2015 Math Academy www.MathAcademy.sg 4
• 5. Operations of Complex Numbers (i) Equality a + bi = c + di ⇐⇒ a = c AND b = d (ii) Addition/Subtraction (a + bi) + (c + di) = (a + c) + (b + d)i (a + bi) − (c + di) = (a − c) + (b − d)i (iii) Multiplication (a + bi)(c + di) = ac + adi + bci + bdi2 = (ac − bd) + (ad + bc)i (iv) Division (Multiply denominator by its complex conjugate) a + bi c + di = (a + bi)(c − di) (c + di)(c − di) = (a + bi)(c − di) c2 + d2 c⃝ 2015 Math Academy www.MathAcademy.sg 5
• 6. Operations of Complex Numbers (i) Equality a + bi = c + di ⇐⇒ a = c AND b = d (ii) Addition/Subtraction (a + bi) + (c + di) = (a + c) + (b + d)i (a + bi) − (c + di) = (a − c) + (b − d)i (iii) Multiplication (a + bi)(c + di) = ac + adi + bci + bdi2 = (ac − bd) + (ad + bc)i (iv) Division (Multiply denominator by its complex conjugate) a + bi c + di = (a + bi)(c − di) (c + di)(c − di) = (a + bi)(c − di) c2 + d2 c⃝ 2015 Math Academy www.MathAcademy.sg 6
• 7. Operations of Complex Numbers (i) Equality a + bi = c + di ⇐⇒ a = c AND b = d (ii) Addition/Subtraction (a + bi) + (c + di) = (a + c) + (b + d)i (a + bi) − (c + di) = (a − c) + (b − d)i (iii) Multiplication (a + bi)(c + di) = ac + adi + bci + bdi2 = (ac − bd) + (ad + bc)i (iv) Division (Multiply denominator by its complex conjugate) a + bi c + di = (a + bi)(c − di) (c + di)(c − di) = (a + bi)(c − di) c2 + d2 c⃝ 2015 Math Academy www.MathAcademy.sg 7
• 8. Operations of Complex Numbers (i) Equality a + bi = c + di ⇐⇒ a = c AND b = d (ii) Addition/Subtraction (a + bi) + (c + di) = (a + c) + (b + d)i (a + bi) − (c + di) = (a − c) + (b − d)i (iii) Multiplication (a + bi)(c + di) = ac + adi + bci + bdi2 = (ac − bd) + (ad + bc)i (iv) Division (Multiply denominator by its complex conjugate) a + bi c + di = (a + bi)(c − di) (c + di)(c − di) = (a + bi)(c − di) c2 + d2 c⃝ 2015 Math Academy www.MathAcademy.sg 8
• 9. Operations of Complex Numbers (i) Equality a + bi = c + di ⇐⇒ a = c AND b = d (ii) Addition/Subtraction (a + bi) + (c + di) = (a + c) + (b + d)i (a + bi) − (c + di) = (a − c) + (b − d)i (iii) Multiplication (a + bi)(c + di) = ac + adi + bci + bdi2 = (ac − bd) + (ad + bc)i (iv) Division (Multiply denominator by its complex conjugate) a + bi c + di = (a + bi)(c − di) (c + di)(c − di) = (a + bi)(c − di) c2 + d2 c⃝ 2015 Math Academy www.MathAcademy.sg 9
• 10. Complex conjugate z∗ If z = x + iy, then its complex conjugate z∗ is deﬁned as z∗ = x − iy Observe the following: zz∗ = x2 + y2 Thus, the product of a complex number and its complex conjugate is a real number. Proof: zz∗ = (x + yi)(x − yi) = x2 − (yi)2 = x2 − (−y2 ) = x2 + y2 c⃝ 2015 Math Academy www.MathAcademy.sg 10
• 11. Complex conjugate z∗ If z = x + iy, then its complex conjugate z∗ is deﬁned as z∗ = x − iy Observe the following: zz∗ = x2 + y2 Thus, the product of a complex number and its complex conjugate is a real number. Proof: zz∗ = (x + yi)(x − yi) = x2 − (yi)2 = x2 − (−y2 ) = x2 + y2 c⃝ 2015 Math Academy www.MathAcademy.sg 11
• 12. Complex conjugate z∗ If z = x + iy, then its complex conjugate z∗ is deﬁned as z∗ = x − iy Observe the following: zz∗ = x2 + y2 Thus, the product of a complex number and its complex conjugate is a real number. Proof: zz∗ = (x + yi)(x − yi) = x2 − (yi)2 = x2 − (−y2 ) = x2 + y2 c⃝ 2015 Math Academy www.MathAcademy.sg 12
• 13. Example (1) Express the following in the form x + iy, where x, y ∈ R. a) z = (3 − 3i) + (−3 + i) c) z = (2 + 2i)(1 − 3i) e) z = 2+10i 5−3i c⃝ 2015 Math Academy www.MathAcademy.sg 13
• 14. Example (1) Express the following in the form x + iy, where x, y ∈ R. b) z = (2 − √ 2i) − (1 − 3i) d) z = ( √ 3 + 2i)( √ 3 − 2i) f) z = 3+2i 4−i c⃝ 2015 Math Academy www.MathAcademy.sg 14
• 15. Example (2) Given that (2 − i)2 + (3λ + i)(µ − i) + 5i = 10, ﬁnd the exact values of λ and µ. (2 − i)2 + (3λ + i)(µ − i) + 5i = 10 (4 − 4i − 1) + (3λµ − 3λi + µi + 1) + 5i = 10 4 + 3λµ + i(1 − 3λ + µ) = 10 Comparing real-coeﬃcients, 4 + 3λµ = 10 λµ = 2 . . . (1) Comparing imaginary-coeﬃcients, 1 − 3λ + µ = 0 µ = 3λ − 1 . . . (2) Substituting (2) into (1), λ(3λ − 1) = 2 3λ2 − λ − 2 = 0 λ = 1 or − 2 3 =⇒ µ = 2 or − 3 c⃝ 2015 Math Academy www.MathAcademy.sg 15
• 16. Example (2) Given that (2 − i)2 + (3λ + i)(µ − i) + 5i = 10, ﬁnd the exact values of λ and µ. (2 − i)2 + (3λ + i)(µ − i) + 5i = 10 (4 − 4i − 1) + (3λµ − 3λi + µi + 1) + 5i = 10 4 + 3λµ + i(1 − 3λ + µ) = 10 Comparing real-coeﬃcients, 4 + 3λµ = 10 λµ = 2 . . . (1) Comparing imaginary-coeﬃcients, 1 − 3λ + µ = 0 µ = 3λ − 1 . . . (2) Substituting (2) into (1), λ(3λ − 1) = 2 3λ2 − λ − 2 = 0 λ = 1 or − 2 3 =⇒ µ = 2 or − 3 c⃝ 2015 Math Academy www.MathAcademy.sg 16
• 17. Example (2) Given that (2 − i)2 + (3λ + i)(µ − i) + 5i = 10, ﬁnd the exact values of λ and µ. (2 − i)2 + (3λ + i)(µ − i) + 5i = 10 (4 − 4i − 1) + (3λµ − 3λi + µi + 1) + 5i = 10 4 + 3λµ + i(1 − 3λ + µ) = 10 Comparing real-coeﬃcients, 4 + 3λµ = 10 λµ = 2 . . . (1) Comparing imaginary-coeﬃcients, 1 − 3λ + µ = 0 µ = 3λ − 1 . . . (2) Substituting (2) into (1), λ(3λ − 1) = 2 3λ2 − λ − 2 = 0 λ = 1 or − 2 3 =⇒ µ = 2 or − 3 c⃝ 2015 Math Academy www.MathAcademy.sg 17
• 18. Example (2) Given that (2 − i)2 + (3λ + i)(µ − i) + 5i = 10, ﬁnd the exact values of λ and µ. (2 − i)2 + (3λ + i)(µ − i) + 5i = 10 (4 − 4i − 1) + (3λµ − 3λi + µi + 1) + 5i = 10 4 + 3λµ + i(1 − 3λ + µ) = 10 Comparing real-coeﬃcients, 4 + 3λµ = 10 λµ = 2 . . . (1) Comparing imaginary-coeﬃcients, 1 − 3λ + µ = 0 µ = 3λ − 1 . . . (2) Substituting (2) into (1), λ(3λ − 1) = 2 3λ2 − λ − 2 = 0 λ = 1 or − 2 3 =⇒ µ = 2 or − 3 c⃝ 2015 Math Academy www.MathAcademy.sg 18
• 19. Example (2) Given that (2 − i)2 + (3λ + i)(µ − i) + 5i = 10, ﬁnd the exact values of λ and µ. (2 − i)2 + (3λ + i)(µ − i) + 5i = 10 (4 − 4i − 1) + (3λµ − 3λi + µi + 1) + 5i = 10 4 + 3λµ + i(1 − 3λ + µ) = 10 Comparing real-coeﬃcients, 4 + 3λµ = 10 λµ = 2 . . . (1) Comparing imaginary-coeﬃcients, 1 − 3λ + µ = 0 µ = 3λ − 1 . . . (2) Substituting (2) into (1), λ(3λ − 1) = 2 3λ2 − λ − 2 = 0 λ = 1 or − 2 3 =⇒ µ = 2 or − 3 c⃝ 2015 Math Academy www.MathAcademy.sg 19
• 20. Example (2) Given that (2 − i)2 + (3λ + i)(µ − i) + 5i = 10, ﬁnd the exact values of λ and µ. (2 − i)2 + (3λ + i)(µ − i) + 5i = 10 (4 − 4i − 1) + (3λµ − 3λi + µi + 1) + 5i = 10 4 + 3λµ + i(1 − 3λ + µ) = 10 Comparing real-coeﬃcients, 4 + 3λµ = 10 λµ = 2 . . . (1) Comparing imaginary-coeﬃcients, 1 − 3λ + µ = 0 µ = 3λ − 1 . . . (2) Substituting (2) into (1), λ(3λ − 1) = 2 3λ2 − λ − 2 = 0 λ = 1 or − 2 3 =⇒ µ = 2 or − 3 c⃝ 2015 Math Academy www.MathAcademy.sg 20
• 21. Example (2) Given that (2 − i)2 + (3λ + i)(µ − i) + 5i = 10, ﬁnd the exact values of λ and µ. (2 − i)2 + (3λ + i)(µ − i) + 5i = 10 (4 − 4i − 1) + (3λµ − 3λi + µi + 1) + 5i = 10 4 + 3λµ + i(1 − 3λ + µ) = 10 Comparing real-coeﬃcients, 4 + 3λµ = 10 λµ = 2 . . . (1) Comparing imaginary-coeﬃcients, 1 − 3λ + µ = 0 µ = 3λ − 1 . . . (2) Substituting (2) into (1), λ(3λ − 1) = 2 3λ2 − λ − 2 = 0 λ = 1 or − 2 3 =⇒ µ = 2 or − 3 c⃝ 2015 Math Academy www.MathAcademy.sg 21
• 22. Example (2) Given that (2 − i)2 + (3λ + i)(µ − i) + 5i = 10, ﬁnd the exact values of λ and µ. (2 − i)2 + (3λ + i)(µ − i) + 5i = 10 (4 − 4i − 1) + (3λµ − 3λi + µi + 1) + 5i = 10 4 + 3λµ + i(1 − 3λ + µ) = 10 Comparing real-coeﬃcients, 4 + 3λµ = 10 λµ = 2 . . . (1) Comparing imaginary-coeﬃcients, 1 − 3λ + µ = 0 µ = 3λ − 1 . . . (2) Substituting (2) into (1), λ(3λ − 1) = 2 3λ2 − λ − 2 = 0 λ = 1 or − 2 3 =⇒ µ = 2 or − 3 c⃝ 2015 Math Academy www.MathAcademy.sg 22
• 23. Example (3) Find the square root of 15 + 8i. We want to solve for z such that z = √ 15 + 8i. Let z = x + yi. z = √ 15 + 8i z2 = 15 + 8i (x + yi)2 = 15 + 8i x2 − y2 + 2xyi = 15 + 8i Comparing real coeﬃcient, x2 − y2 = 15 . . . (1) Comparing imaginary coeﬃcient, 2xy = 8 y = 4 x . . . (2) Substitute (2) into (1), x2 − ( 4 x )2 = 15 x4 − 15x2 − 16 = 0 c⃝ 2015 Math Academy www.MathAcademy.sg 23
• 24. Example (3) Find the square root of 15 + 8i. We want to solve for z such that z = √ 15 + 8i. Let z = x + yi. z = √ 15 + 8i z2 = 15 + 8i (x + yi)2 = 15 + 8i x2 − y2 + 2xyi = 15 + 8i Comparing real coeﬃcient, x2 − y2 = 15 . . . (1) Comparing imaginary coeﬃcient, 2xy = 8 y = 4 x . . . (2) Substitute (2) into (1), x2 − ( 4 x )2 = 15 x4 − 15x2 − 16 = 0 c⃝ 2015 Math Academy www.MathAcademy.sg 24
• 25. Example (3) Find the square root of 15 + 8i. We want to solve for z such that z = √ 15 + 8i. Let z = x + yi. z = √ 15 + 8i z2 = 15 + 8i (x + yi)2 = 15 + 8i x2 − y2 + 2xyi = 15 + 8i Comparing real coeﬃcient, x2 − y2 = 15 . . . (1) Comparing imaginary coeﬃcient, 2xy = 8 y = 4 x . . . (2) Substitute (2) into (1), x2 − ( 4 x )2 = 15 x4 − 15x2 − 16 = 0 c⃝ 2015 Math Academy www.MathAcademy.sg 25
• 26. Example (3) Find the square root of 15 + 8i. We want to solve for z such that z = √ 15 + 8i. Let z = x + yi. z = √ 15 + 8i z2 = 15 + 8i (x + yi)2 = 15 + 8i x2 − y2 + 2xyi = 15 + 8i Comparing real coeﬃcient, x2 − y2 = 15 . . . (1) Comparing imaginary coeﬃcient, 2xy = 8 y = 4 x . . . (2) Substitute (2) into (1), x2 − ( 4 x )2 = 15 x4 − 15x2 − 16 = 0 c⃝ 2015 Math Academy www.MathAcademy.sg 26
• 27. Example (3) Find the square root of 15 + 8i. We want to solve for z such that z = √ 15 + 8i. Let z = x + yi. z = √ 15 + 8i z2 = 15 + 8i (x + yi)2 = 15 + 8i x2 − y2 + 2xyi = 15 + 8i Comparing real coeﬃcient, x2 − y2 = 15 . . . (1) Comparing imaginary coeﬃcient, 2xy = 8 y = 4 x . . . (2) Substitute (2) into (1), x2 − ( 4 x )2 = 15 x4 − 15x2 − 16 = 0 c⃝ 2015 Math Academy www.MathAcademy.sg 27
• 28. Example (3) Find the square root of 15 + 8i. We want to solve for z such that z = √ 15 + 8i. Let z = x + yi. z = √ 15 + 8i z2 = 15 + 8i (x + yi)2 = 15 + 8i x2 − y2 + 2xyi = 15 + 8i Comparing real coeﬃcient, x2 − y2 = 15 . . . (1) Comparing imaginary coeﬃcient, 2xy = 8 y = 4 x . . . (2) Substitute (2) into (1), x2 − ( 4 x )2 = 15 x4 − 15x2 − 16 = 0 c⃝ 2015 Math Academy www.MathAcademy.sg 28
• 29. Example (3) Find the square root of 15 + 8i. We want to solve for z such that z = √ 15 + 8i. Let z = x + yi. z = √ 15 + 8i z2 = 15 + 8i (x + yi)2 = 15 + 8i x2 − y2 + 2xyi = 15 + 8i Comparing real coeﬃcient, x2 − y2 = 15 . . . (1) Comparing imaginary coeﬃcient, 2xy = 8 y = 4 x . . . (2) Substitute (2) into (1), x2 − ( 4 x )2 = 15 x4 − 15x2 − 16 = 0 c⃝ 2015 Math Academy www.MathAcademy.sg 29
• 30. Example (3) Find the square root of 15 + 8i. We want to solve for z such that z = √ 15 + 8i. Let z = x + yi. z = √ 15 + 8i z2 = 15 + 8i (x + yi)2 = 15 + 8i x2 − y2 + 2xyi = 15 + 8i Comparing real coeﬃcient, x2 − y2 = 15 . . . (1) Comparing imaginary coeﬃcient, 2xy = 8 y = 4 x . . . (2) Substitute (2) into (1), x2 − ( 4 x )2 = 15 x4 − 15x2 − 16 = 0 c⃝ 2015 Math Academy www.MathAcademy.sg 30
• 31. Example (3) Find the square root of 15 + 8i. We want to solve for z such that z = √ 15 + 8i. Let z = x + yi. z = √ 15 + 8i z2 = 15 + 8i (x + yi)2 = 15 + 8i x2 − y2 + 2xyi = 15 + 8i Comparing real coeﬃcient, x2 − y2 = 15 . . . (1) Comparing imaginary coeﬃcient, 2xy = 8 y = 4 x . . . (2) Substitute (2) into (1), x2 − ( 4 x )2 = 15 x4 − 15x2 − 16 = 0 c⃝ 2015 Math Academy www.MathAcademy.sg 31
• 32. Substitute (2) into (1), x2 − ( 4 x )2 = 15 x4 − 15x2 − 16 = 0 From GC, x = 4 or −4. =⇒ y = 1 or −1. ∴ z = 4 + i or −4 − i. c⃝ 2015 Math Academy www.MathAcademy.sg 32
• 33. Substitute (2) into (1), x2 − ( 4 x )2 = 15 x4 − 15x2 − 16 = 0 From GC, x = 4 or −4. =⇒ y = 1 or −1. ∴ z = 4 + i or −4 − i. c⃝ 2015 Math Academy www.MathAcademy.sg 33
• 34. Substitute (2) into (1), x2 − ( 4 x )2 = 15 x4 − 15x2 − 16 = 0 From GC, x = 4 or −4. =⇒ y = 1 or −1. ∴ z = 4 + i or −4 − i. c⃝ 2015 Math Academy www.MathAcademy.sg 34
• 35. Substitute (2) into (1), x2 − ( 4 x )2 = 15 x4 − 15x2 − 16 = 0 From GC, x = 4 or −4. =⇒ y = 1 or −1. ∴ z = 4 + i or −4 − i. c⃝ 2015 Math Academy www.MathAcademy.sg 35
• 36. Substitute (2) into (1), x2 − ( 4 x )2 = 15 x4 − 15x2 − 16 = 0 From GC, x = 4 or −4. =⇒ y = 1 or −1. ∴ z = 4 + i or −4 − i. c⃝ 2015 Math Academy www.MathAcademy.sg 36
• 37. Example (4) The complex numbers p and q are such that p = 2 + ia, q = b − i, where a and b are real numbers. Given that pq = 13 + 13i, ﬁnd the possible values of a and b. [4] [a = 3 or 10, b = 5 or 3 2 ] pq = (2 + ia)(b − i) = 2b − 2i + abi + a = 2b + a + i(ab − 2) 13 + 13i = 2b + a + i(ab − 2) Comparing real and imaginary parts, 2b + a = 13 a = 13 − 2b − − − − − (1) ab − 2 = 13 ab = 15 − − − − − (2) Sub (1) into (2) (13 − 2b)b = 15 −2b2 + 13b − 15 = 0 b = 5 or 3 2 ∴ a = 3 or 10c⃝ 2015 Math Academy www.MathAcademy.sg 37
• 38. Example (4) The complex numbers p and q are such that p = 2 + ia, q = b − i, where a and b are real numbers. Given that pq = 13 + 13i, ﬁnd the possible values of a and b. [4] [a = 3 or 10, b = 5 or 3 2 ] pq = (2 + ia)(b − i) = 2b − 2i + abi + a = 2b + a + i(ab − 2) 13 + 13i = 2b + a + i(ab − 2) Comparing real and imaginary parts, 2b + a = 13 a = 13 − 2b − − − − − (1) ab − 2 = 13 ab = 15 − − − − − (2) Sub (1) into (2) (13 − 2b)b = 15 −2b2 + 13b − 15 = 0 b = 5 or 3 2 ∴ a = 3 or 10c⃝ 2015 Math Academy www.MathAcademy.sg 38
• 39. Example (5) Two complex numbers w and z are such that 2w + z = 12i and w∗ + 2z = −13 + 4i 2 − i . Find w and z, giving each answer in the form x + iy. 2w + z = 12i z = 12i − 2w —(1) w∗ + 2z = −13 + 4i 2 − i —(2) Sub (1) into (2): w∗ + 2(12i − 2w) = −13 + 4i 2 − i Let w = x + iy. Then w∗ = x − iy. (x − iy) + 2[12i − 2(x + iy)] = −13 + 4i 2 − ic⃝ 2015 Math Academy www.MathAcademy.sg 39
• 40. Example (5) Two complex numbers w and z are such that 2w + z = 12i and w∗ + 2z = −13 + 4i 2 − i . Find w and z, giving each answer in the form x + iy. 2w + z = 12i z = 12i − 2w —(1) w∗ + 2z = −13 + 4i 2 − i —(2) Sub (1) into (2): w∗ + 2(12i − 2w) = −13 + 4i 2 − i Let w = x + iy. Then w∗ = x − iy. (x − iy) + 2[12i − 2(x + iy)] = −13 + 4i 2 − ic⃝ 2015 Math Academy www.MathAcademy.sg 40
• 41. Example (5) Two complex numbers w and z are such that 2w + z = 12i and w∗ + 2z = −13 + 4i 2 − i . Find w and z, giving each answer in the form x + iy. 2w + z = 12i z = 12i − 2w —(1) w∗ + 2z = −13 + 4i 2 − i —(2) Sub (1) into (2): w∗ + 2(12i − 2w) = −13 + 4i 2 − i Let w = x + iy. Then w∗ = x − iy. (x − iy) + 2[12i − 2(x + iy)] = −13 + 4i 2 − ic⃝ 2015 Math Academy www.MathAcademy.sg 41
• 42. Example (5) Two complex numbers w and z are such that 2w + z = 12i and w∗ + 2z = −13 + 4i 2 − i . Find w and z, giving each answer in the form x + iy. 2w + z = 12i z = 12i − 2w —(1) w∗ + 2z = −13 + 4i 2 − i —(2) Sub (1) into (2): w∗ + 2(12i − 2w) = −13 + 4i 2 − i Let w = x + iy. Then w∗ = x − iy. (x − iy) + 2[12i − 2(x + iy)] = −13 + 4i 2 − ic⃝ 2015 Math Academy www.MathAcademy.sg 42
• 43. Example (5) Two complex numbers w and z are such that 2w + z = 12i and w∗ + 2z = −13 + 4i 2 − i . Find w and z, giving each answer in the form x + iy. 2w + z = 12i z = 12i − 2w —(1) w∗ + 2z = −13 + 4i 2 − i —(2) Sub (1) into (2): w∗ + 2(12i − 2w) = −13 + 4i 2 − i Let w = x + iy. Then w∗ = x − iy. (x − iy) + 2[12i − 2(x + iy)] = −13 + 4i 2 − ic⃝ 2015 Math Academy www.MathAcademy.sg 43
• 44. Example (5) Two complex numbers w and z are such that 2w + z = 12i and w∗ + 2z = −13 + 4i 2 − i . Find w and z, giving each answer in the form x + iy. 2w + z = 12i z = 12i − 2w —(1) w∗ + 2z = −13 + 4i 2 − i —(2) Sub (1) into (2): w∗ + 2(12i − 2w) = −13 + 4i 2 − i Let w = x + iy. Then w∗ = x − iy. (x − iy) + 2[12i − 2(x + iy)] = −13 + 4i 2 − ic⃝ 2015 Math Academy www.MathAcademy.sg 44
• 45. (x − iy) + 2[12i − 2(x + iy)] = −13 + 4i 2 − i ... (We group the terms with i together) (−6x − 5y + 24) + i(3x − 10y + 48) = −13 + 4i Comparing coeﬃcients, −6x − 5y + 24 = −13 . . . (3) 3x − 10y + 48 = 4 . . . (4) Solving the simultaneous equations (3) and (4), we have x = 2 and y = 5. Therefore, w = 2 + 5i. From (1), z = 12i − 2w = 12i − 2(2 + 5i) = −4 + 2i c⃝ 2015 Math Academy www.MathAcademy.sg 45
• 46. (x − iy) + 2[12i − 2(x + iy)] = −13 + 4i 2 − i ... (We group the terms with i together) (−6x − 5y + 24) + i(3x − 10y + 48) = −13 + 4i Comparing coeﬃcients, −6x − 5y + 24 = −13 . . . (3) 3x − 10y + 48 = 4 . . . (4) Solving the simultaneous equations (3) and (4), we have x = 2 and y = 5. Therefore, w = 2 + 5i. From (1), z = 12i − 2w = 12i − 2(2 + 5i) = −4 + 2i c⃝ 2015 Math Academy www.MathAcademy.sg 46
• 47. (x − iy) + 2[12i − 2(x + iy)] = −13 + 4i 2 − i ... (We group the terms with i together) (−6x − 5y + 24) + i(3x − 10y + 48) = −13 + 4i Comparing coeﬃcients, −6x − 5y + 24 = −13 . . . (3) 3x − 10y + 48 = 4 . . . (4) Solving the simultaneous equations (3) and (4), we have x = 2 and y = 5. Therefore, w = 2 + 5i. From (1), z = 12i − 2w = 12i − 2(2 + 5i) = −4 + 2i c⃝ 2015 Math Academy www.MathAcademy.sg 47
• 48. (x − iy) + 2[12i − 2(x + iy)] = −13 + 4i 2 − i ... (We group the terms with i together) (−6x − 5y + 24) + i(3x − 10y + 48) = −13 + 4i Comparing coeﬃcients, −6x − 5y + 24 = −13 . . . (3) 3x − 10y + 48 = 4 . . . (4) Solving the simultaneous equations (3) and (4), we have x = 2 and y = 5. Therefore, w = 2 + 5i. From (1), z = 12i − 2w = 12i − 2(2 + 5i) = −4 + 2i c⃝ 2015 Math Academy www.MathAcademy.sg 48
• 49. (x − iy) + 2[12i − 2(x + iy)] = −13 + 4i 2 − i ... (We group the terms with i together) (−6x − 5y + 24) + i(3x − 10y + 48) = −13 + 4i Comparing coeﬃcients, −6x − 5y + 24 = −13 . . . (3) 3x − 10y + 48 = 4 . . . (4) Solving the simultaneous equations (3) and (4), we have x = 2 and y = 5. Therefore, w = 2 + 5i. From (1), z = 12i − 2w = 12i − 2(2 + 5i) = −4 + 2i c⃝ 2015 Math Academy www.MathAcademy.sg 49
• 50. Example (6) By writing z = x + iy, x, y ∈ R, solve the simultaneous equations z2 + zw − 2 = 0 and z∗ = w 1 + i . [z = 1 − i, w = 2i, z = −1 + i, w = −2i] c⃝ 2015 Math Academy www.MathAcademy.sg 50
• 51. From second equation, w = z∗ (1 + i) Substitute into ﬁrst equation, z2 + z[z∗ (1 + i)] − 2 = 0 Let z = x + iy. (x + iy)2 + (x2 + y2 )(1 + i) − 2 = 0 x2 + 2xyi − y2 + x2 + y2 + (x2 + y2 )i − 2 = 0 2x2 − 2 + i(2xy + x2 + y2 ) = 0 Comparing coeﬃcient of real and imaginary parts, 2x2 − 2 = 0 x2 = 1 x = ±1 2xy + x2 + y2 = 0 (x + y)2 = 0 if x = 1, y = −1 if x = −1, y = 1 When z = 1 − i, w = z∗ (1 + i) = (1 + i)(1 + i) = 1 + 2i − 1 = 2i When z = −1 + i, w = z∗ (1 + i) = (−1 − i)(1 + i) = −2i c⃝ 2015 Math Academy www.MathAcademy.sg 51
• 52. Properties of Complex Conjugates (i) z + z∗ = 2Re(z) (ii) z − z∗ = 2iIm(z) (iii) (z∗ )∗ = z Bring ∗ into the brackets (a) (z1 ± z2)∗ = z∗ 1 ± z∗ 2 (b) (z1z2)∗ = z∗ 1 .z∗ 2 (c) ( z1 z2 )∗ = z∗ 1 z∗ 2 (d) (kz)∗ = kz∗ , where k ∈ R Example Prove (i), (ii) and (a) in the above properties. c⃝ 2015 Math Academy www.MathAcademy.sg 52
• 53. z + z∗ = 2Re(z) Let z = x + yi. Then z∗ = x − yi. c⃝ 2015 Math Academy www.MathAcademy.sg 53
• 54. z − z∗ = 2iIm(z) Let z = x + yi. Then z∗ = x − yi. c⃝ 2015 Math Academy www.MathAcademy.sg 54
• 55. (z1 ± z2)∗ = z∗ 1 ± z∗ 2 c⃝ 2015 Math Academy www.MathAcademy.sg 55
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