Maths Level 4 Module 2

Mathematics NQF Level 4 *see terms and conditions Downloaded from hlayiso.com Mathematics NQF Level 4

Complex numbers Topic 1 Downloaded from hlayiso.com Mathematics NQF Level 4

Solve problems using complex numbers Module 2 Downloaded from hlayiso.com Mathematics NQF Level 4

Learning activity 2.1 Downloaded from hlayiso.com Mathematics NQF Level 4

Solve for x and y if: 6. 3 x − 4i =y + 2i x − 4i = ( y + 2i )3 2 x − 4i = ( y + 2i )( y + 4 yi − 4) 3 2 2 2 x − 4i =y + 4 y i − 4 y + 2 y i + 8 yi − 8i 3 2 x − 4i = y + 6 y i − 4 y − 8 y − 8i i 2 = −1 Downloaded from hlayiso.com Learning activity 2.1

Solve for x and y if: 6. 3 x − 4i =y + 2i Continued ∴ x − 4i = y 3 + 6 y 2i − 12 y − 8i [Equate real parts] [Equate imaginary and parts] = y 3 − 12 y x ...(1) 2 −4 i = 6 y i − 8 i 2 −4= 6 y − 8 6 y2 = 4 y = ±0,816 Downloaded from hlayiso.com Learning activity 2.1

Solve for x and y if: 6. 3 x − 4i =y + 2i Continued 3 Substitute into (1): x = y − 12 y x=(−0,816)3 − 12(−0,816) = 9, 25 or 3 =x (0,816) − 12(0,816) = −9, 25 Downloaded from hlayiso.com Learning activity 2.1

8. x − yi = (i − 4) + 2 + i 2−i VIDEO Click above to play this video. Downloaded from hlayiso.com Learning activity 2.1

Learning activity 2.2 Downloaded from hlayiso.com Mathematics NQF Level 4

Solve for x if: 2 1. x + 25 = 0 2 x + 25 = 0 x 2 = −25 x =± −25 = ± 25 −1 i= −1   = ±5i Downloaded from hlayiso.com Learning activity 2.2

2 9. 4 x + 2 x + 15 = 0 VIDEO Click above to play this video. Downloaded from hlayiso.com Learning activity 2.2

Summative assessment Module 2 Downloaded from hlayiso.com Mathematics NQF Level 4

1. Solve for x if : 1.1) 5 x - 2i = 15 + yi 5 x − 2i = 15 + yi [Equate real parts] [Equate imaginary parts] ∴5x = 15 and −2i =yi ∴x =3 y = −2 Downloaded from hlayiso.com Summative assessment – Module 2

1. Solve for x if : 1.2) 2 x - yi = 2i + 1 2 2 x − yi = 4i + 4i + 1 2 x − yi =−4 + 4i + 1 [i 2 = −1] 2 x + 4 − 1 = yi + 4i ∴ 2x =−3 and 4i − yi = 3 ∴ x =− y = −4 2 Downloaded from hlayiso.com Summative assessment – Module 2

1. Solve for x if : 1.3) 4+ i y - i = 2-i x ∴ 4 x + ix = (2 − i )( y − i ) 2 ∴ 4 x + ix = 2 y − 2i − iy + i ∴ 4 x + ix = 2 y − (2 + y )i − 1 [i 2 = −1] ∴ 4x = 2 y −1 and [Equate real parts] [Equate imaginary parts] Downloaded from hlayiso.com Summative assessment – Module 2

1. Solve for x if : 1.3) 4+ i y - i (Continued) = 2-i x ∴ 4x − 2 y = −1 ... (1) and x =−2 − y ∴ x + y =−2 ... (2) 2 × (2) : 2 x + 2 y = −4 ...(3) (1) + (3): 4 x + 2 x =−1 − 4 −5 ∴6x = −5 ∴x = 6 Downloaded from hlayiso.com Summative assessment – Module 2

1. Solve for x if : 4+ i y - i 1.3) = (Continued) 2-i x Substitute into (2): y =−2 − x  −5  =−2 −    6  5 =−2 + 6 −12 + 5 −7 = ∴y = 6 Downloaded from hlayiso.com 6 Summative assessment – Module 2

2. Solve for x and y if z1 and z2 are identical, and: z1 = 4 + x − 2 yi and z2 =1 − y + 3 xi z1 = z2 ∴ 4 + x − 2 yi = 1 − y + 3 xi ∴4 + x =1− y and −2 y i = 3x i ∴ x + y =−3 ...(1) 3x + 2 y 0 =....(2) Downloaded from hlayiso.com Summative assessment – Module 2

2. Solve for x and y if z1 and z2 are identical, and: z1 = 4 + x − 2 yi and z2 =1 − y + 3 xi (1) × 2 : 2 x + 2 y = −6 ...(3) (2) − (3) : 3 x + 2 y 0 =....(2) 2x + 2 y = −6 ...(3) x=6 Substitute into (1) : y =−3 − 6 = −9 Downloaded from hlayiso.com Summative assessment – Module 2

3. Solve for x if: a) x2 + 81 = 0 2 x + 81 = 0 2 −81 ∴x = x =± 81 ⋅ −1 [ ab = a ⋅ b] ∴ x =±9i i 2 = -1 [Imaginary roots: conjugate pair] Downloaded from hlayiso.com Summative assessment – Module 2

3. Solve for x if: b) 4x5 − 64x = 0 5 4 x − 64 x 0 = 4 4 x( x − 16) 0 = [Difference of squares] 2 2 0 [Difference of squares] 4 x( x − 4)( x + 4) = 0 or ∴ 4x = 2 x −4=0 or 2 x +4=0 ∴x =0 ∴ x =±2 ∴ x =±2i [Real roots] Downloaded [Imaginary roots] from hlayiso.com Summative assessment – Module 2

3. Solve for x if: 3.1) 7x3 + 343x = 0 2 ∴ 7 x( x + 49) 0 = ∴7x = 0 or 2 x = −49 ∴x =0 x = ±7i Downloaded from hlayiso.com Summative assessment – Module 2

3. Solve for x if: 3.4) 2x2 + 18 = − 3x 2 x 2 + 3 x + 18 = 0 −b ± b 2 − 4ac [Quadratic formula] x= 2a −3 ± 9 − 4(2)(18) = 2(2) Downloaded from hlayiso.com Summative assessment – Module 2

3. Solve for x if: d) 2x2 + 18 = − 3x Continued −3 ± 11, 619i = 4 = −0, 75 ± 2,905i Downloaded from hlayiso.com Summative assessment – Module 2

Mathematics NQF Level 2 Downloaded from hlayiso.com Mathematics NQF Level 4