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# X2 t01 03 argand diagram (2013)

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### X2 t01 03 argand diagram (2013)

1. 1. The Argand Diagram Complex numbers can be represented geometrically on an Argand y (imaginary axis) Diagram. 4 3 2 1 -4 -3 -2 -1 -1 -2 -3 1 2 3 4 x (real axis)
2. 2. The Argand Diagram Complex numbers can be represented geometrically on an Argand y (imaginary axis) Diagram. 4 3 A=2 2 1 A -4 -3 -2 -1 1 2 3 4 x (real axis) -1 -2 -3
3. 3. The Argand Diagram Complex numbers can be represented geometrically on an Argand y (imaginary axis) Diagram. 4 3 A=2 2 B = -3i 1 A -4 -3 -2 -1 1 2 3 4 x (real axis) -1 -2 -3 B
4. 4. The Argand Diagram Complex numbers can be represented geometrically on an Argand y (imaginary axis) Diagram. 4 3 A=2 2 B = -3i C = -2 + i C 1 A -4 -3 -2 -1 1 2 3 4 x (real axis) -1 -2 -3 B
5. 5. The Argand Diagram Complex numbers can be represented geometrically on an Argand y (imaginary axis) Diagram. 4 3 A=2 2 B = -3i C = -2 + i C 1 A D=4-i -4 -3 -2 -1 1 2 3 4 x (real axis) -1 D -2 -3 B
6. 6. The Argand Diagram Complex numbers can be represented geometrically on an Argand y (imaginary axis) Diagram. 4 3 A=2 NOTE: Conjugates 2 B = -3i are reflected in the E real (x) axis C = -2 + i C 1 A D=4-i -4 -3 -2 -1 1 2 3 4 x (real axis) E=4+i -1 D -2 -3 B
7. 7. The Argand Diagram Complex numbers can be represented geometrically on an Argand y (imaginary axis) Diagram. 4 3 A=2 NOTE: Conjugates 2 B = -3i are reflected in the E real (x) axis C = -2 + i F C 1 A D=4-i -4 -3 -2 -1 1 2 3 4 x (real axis) E=4+i -1 D F = -(4 - i) -2 = - 4 + i NOTE: Opposites are -3 B rotated 180°
8. 8. The Argand Diagram Complex numbers can be represented geometrically on an Argand y (imaginary axis) Diagram. 4G NOTE: Multiply 3 A=2 NOTE: Conjugates by i results in 2 B = -3i 90° rotation are reflected in the E real (x) axis C = -2 + i F C 1 A D=4-i -4 -3 -2 -1 1 2 3 4 x (real axis) E=4+i -1 D F = -(4 - i) -2 = - 4 + i NOTE: G = i(4 - i) Opposites are -3 B rotated 180° = 1+4i
9. 9. The Argand Diagram Complex numbers can be represented geometrically on an Argand y (imaginary axis) Diagram. 4G NOTE: Multiply 3 A=2 NOTE: Conjugates by i results in 2 B = -3i 90° rotation are reflected in the E real (x) axis C = -2 + i F C 1 A D=4-i -4 -3 -2 -1 1 2 3 4 x (real axis) E=4+i -1 D F = -(4 - i) -2 = - 4 + i NOTE: G = i(4 - i) Opposites are -3 B rotated 180° = 1+4i Every complex number can be represented by a unique point on the Argand Diagram.
10. 10. Locus in Terms of Complex Numbers Horizontal and Vertical Lines
11. 11. Locus in Terms of Complex Numbers Horizontal and Vertical Lines y c x Im z   c
12. 12. Locus in Terms of Complex Numbers Horizontal and Vertical Lines y c y x Im z   c k x Re z   k
13. 13. y R Circles zz  R 2 R R R x
14. 14. y R Circles zz  R 2 R R x R e.g.Find and describe the locus of points in the Argand diagram; (i)  z  4  i  z  4  i   49
15. 15. y R Circles zz  R 2 R R x R e.g.Find and describe the locus of points in the Argand diagram; (i)  z  4  i  z  4  i   49 z z  (4  i ) z  (4  i ) z  (4  i )(4  i )  49
16. 16. y R Circles zz  R 2 R R x R e.g.Find and describe the locus of points in the Argand diagram; (i)  z  4  i  z  4  i   49 z z  (4  i ) z  (4  i ) z  (4  i )(4  i )  49 z z  4( z  z )  iz  iz  (4  i )(4  i )  49 z z  4( z  z )  iz  iz  (4  i )(4  i )  49   z z  4( z  z )  iz  iz  (4  i )(4  i )  49 x 2  y 2  8 x  2 y  16  1  49
17. 17. y R Circles zz  R 2 R R x R e.g.Find and describe the locus of points in the Argand diagram; (i)  z  4  i  z  4  i   49 z z  (4  i ) z  (4  i ) z  (4  i )(4  i )  49 z z  4( z  z )  iz  iz  (4  i )(4  i )  49 z z  4( z  z )  iz  iz  (4  i )(4  i )  49   z z  4( z  z )  iz  iz  (4  i )(4  i )  49 x 2  y 2  8 x  2 y  16  1  49 x 2  8 x  16  y 2  2 y  1  49
18. 18. y R Circles zz  R 2 R R x R e.g.Find and describe the locus of points in the Argand diagram; (i)  z  4  i  z  4  i   49 z z  (4  i ) z  (4  i ) z  (4  i )(4  i )  49 x 2  8 x  16  y 2  2 y  1  49 z z  4( z  z )  iz  iz  (4  i )(4  i )  49  x  4    y  1 z z  4( z  z )  iz  iz  (4  i )(4  i )  49   z z  4( z  z )  iz  iz  (4  i )(4  i )  49 x 2  y 2  8 x  2 y  16  1  49 2 2  49
19. 19. y R Circles zz  R 2 R R x R e.g.Find and describe the locus of points in the Argand diagram; (i)  z  4  i  z  4  i   49 z z  (4  i ) z  (4  i ) z  (4  i )(4  i )  49 x 2  8 x  16  y 2  2 y  1  49 z z  4( z  z )  iz  iz  (4  i )(4  i )  49  x  4    y  1 z z  4( z  z )  iz  iz  (4  i )(4  i )  49   z z  4( z  z )  iz  iz  (4  i )(4  i )  49 x 2  y 2  8 x  2 y  16  1  49 2 2 Locus is a circle centre :  4 ,  1 radius : 7 units  49
20. 20. y R Circles  z    z     R 2 zz  R 2 R R x Locus is a circle centre ω radius R R e.g.Find and describe the locus of points in the Argand diagram; (i)  z  4  i  z  4  i   49 z z  (4  i ) z  (4  i ) z  (4  i )(4  i )  49 x 2  8 x  16  y 2  2 y  1  49 z z  4( z  z )  iz  iz  (4  i )(4  i )  49  x  4    y  1 z z  4( z  z )  iz  iz  (4  i )(4  i )  49   z z  4( z  z )  iz  iz  (4  i )(4  i )  49 x 2  y 2  8 x  2 y  16  1  49 2 2 Locus is a circle centre :  4 ,  1 radius : 7 units  49
21. 21. 1 1 (ii )   1 z z
22. 22. 1 1 (ii )   1 z z z  z  zz
23. 23. 1 1 (ii )   1 z z z  z  zz 2x  x 2  y 2
24. 24. 1 1 (ii )   1 z z z  z  zz 2x  x 2  y 2 x2  2 x  y 2  0 ( x  1) 2  y 2  1
25. 25. 1 1 (ii )   1 z z z  z  zz 2x  x 2  y 2 x2  2 x  y 2  0 ( x  1) 2  y 2  1 Locus is a circle centre: 1,0  radius: 1 unit
26. 26. 1 1 (ii )   1 z z z  z  zz Locus is a circle centre: 1,0  radius: 1 unit 2x  x 2  y 2 x2  2 x  y 2  0 y ( x  1) 2  y 2  1 1 2 x
27. 27. 1 1 (ii )   1 z z z  z  zz Locus is a circle centre: 1,0  radius: 1 unit 2x  x 2  y 2 x2  2 x  y 2  0 y ( x  1) 2  y 2  1 excluding the point (0,0) 1 2 x
28. 28. 1 1 (ii )   1 z z z  z  zz Locus is a circle centre: 1,0  radius: 1 unit 2x  x 2  y 2 x2  2 x  y 2  0 y ( x  1) 2  y 2  1 excluding the point (0,0) 1 2 x Cambridge: Exercise 1C; 1 ace, 2 bdf, 3, 4 ace, 5 bdfh, 6, 10 to 15