EdExcel FP1 Complex Numbers - GlynMathsAlevel

EdExcel Pure Mathematics 1 Complex Numbers Topic assessment 1. Solve the equation z² + 2z + 10 = 0. Find the modulus and argument of each root.

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2. The complex number α is given by α = –2 + 5i. (i) Write down the complex conjugate α*. (ii) Find the modulus and argument of α. α +α * (iii) Find in the form a + bi.

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3. Find the complex number z which satisfies (2 + i) z + (3 − 2i) z* = 32 .

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[3]

α

4. (i) Given that w = 1 + 2i, express w², w³ and w4 in the form a + bi. [5] 4 3 2 (ii) Given that w is a root of the equation z + pz + qz − 6 z + 65 = 0 , find the [5] values of p and q. (iii) Write down a second root of the equation. [1] (iv) Find the other two roots of the equation. [6] 5. Complex numbers α and β are given by π π⎞ 5π 5π ⎞ ⎛ ⎛ α = 2 ⎜ cos + i sin ⎟ , β = 4 2 ⎜ cos + i sin ⎟ 8 8 ⎠ 8 8⎠ ⎝ ⎝ (i) Write down the modulus and argument of each of the complex numbers α and β. Illustrate these two complex numbers on an Argand diagram. [3] (ii) Indicate a length on your diagram which is equal to β − α , and show that

β −α = 6 .

[3]

Show that z1 = 2 + i is one of the roots of the equation z² – 4z + 5 = 0. Find the other root, z2. 1 1 4 (ii) Show that + = . z1 z2 5 (iii) Show also that Im (z1² + z2²) = 0 and find Re (z1² – z2²).

6. (i)

[3] [3] [4]

Total 50 marks

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EdExcel FP1 Complex numbers Assessment solutions Solutions to Topic assessment 1. z 2 + 2 z + 10 = 0 −2 ± 2 2 − 4 × 1 × 10 2 −2 ± −36 = 2 −2 ± 6i = 2 = −1 ± 3i

Using the quadratic formula, z =

−1 + 3i = 1 2 + 3 2 = 10 (-1 + 3i) is in the second quadrant, 3 so arg( −1 + 3i) = arctan ⎛⎜ ⎞⎟ + π = 1.89 (3 s.f.) ⎝ −1 ⎠ −1 − 3i = 10 arg( −1 − 3i) = −1.89

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2. α = −2 + 5 i (i) α * = −2 − 5 i

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(ii) α = 2 2 + 5 2 = 29

5 ⎞ α is in the second quadrant, so arg α = π + arctan ⎛⎜ ⎟ ⎝ −2 ⎠ = π − 1.19 = 1.95 (3 s.f.)

α + α * −2 + 5 i − 2 − 5 i (iii) = α −2 + 5 i

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−4 −2 + 5 i −4( −2 − 5 i) = ( −2 + 5 i)( −2 − 5 i) 8 + 20i 8 20 i = = + 29 29 29 =

[3] 3. (2 + i)z + (3 − 2i)z * = 32 Let z = x + iy (2 + i)( x + iy ) + (3 − 2i)( x − iy ) = 32 2 x + ix + 2iy − y + 3 x − 2ix − 3iy − 2 y = 32 5 x − 3 y − ix − iy = 32

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EdExcel FP1 Complex numbers Assessment solutions Equating imaginary parts: Equating real parts:

−x − y = 0 ⇒ y = −x 5 x − 3 y = 32 ⇒ 5 x + 3 x = 32 ⇒ 8 x = 32 ⇒ x = 4, y = −4

so z = 4 – 4i.

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4. (i) w = 1 + 2i

w 2 = (1 + 2i)(1 + 2i) = 1 + 4i − 4 = −3 + 4i w 3 = ( −3 + 4i)(1 + 2i) = −3 − 2i − 8 = −11 − 2i w 4 = ( −11 − 2i)(1 + 2i) = −11 − 24i + 4 = −7 − 24i

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(ii) z + pz + qz − 6 z + 65 = 0 4

3

2

−7 − 24i + p( −11 − 2i) + q( −3 + 4i) − 6(1 + 2i) + 65 = 0

Equating real parts:

−7 − 11 p − 3q − 6 + 65 = 0

Equating imaginary parts:

11 p + 3q = 52 −24 − 2 p + 4q − 12 = 0

2 ×c + 3 ×d :

c

p − 2q = −18 d 25 p = 50 ⇒ p = 2, q = 10 [5]

(iii) A second root of the equation is w * = 1 − 2i .

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(iv)A quadratic factor is ( z − 1 − 2i )( z − 1 + 2i ) = ( z − 1) + 4 2

= z 2 − 2z + 1 + 4 = z 2 − 2z + 5 z 4 + 2 z 3 + 10z 2 − 6 z + 65 = ( z 2 − 2 z + 5 )( z 2 + 4z + 13) The other two roots are the roots of the quadratic equation z 2 + 4z + 13 = 0 −4 ± 16 − 4 × 1 × 13 z= 2 −4 ± −36 = 2 −4 ± 6i = 2 = −2 ± 3i The other two roots are -2 + 3i and -2 – 3i.

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EdExcel FP1 Complex numbers Assessment solutions 5. (i)

α = 2 ⎛⎜ cos ⎝ α =2

argα =

π 8

+ i sin

π⎞ ⎟ 8⎠

Im β

π

β −α

8

4 2

5π 5π ⎞ β = 4 2 ⎛⎜ cos + i sin ⎟ 8 8 ⎠ ⎝ β =4 2 5π arg β = 8 (ii) The triangle is a right-angled triangle, 2 so β − α = 2 2 + (4 2 )2

π

α

8

π

8

2 Re

[3]

= 4 + 32 = 36

β −α = 6 [3] 6. (i)

z 12 − 4z + 5 = (2 + i)2 − 4(2 + i) + 5 = 4 + 4i − 1 − 8 − 4i + 5 =0 The other root, z 2 = z 1 * = 2 − i

(ii)

1

z1

+

1

[3]

1 1 + z2 2 + i 2 − i 2 −i +2 +i = (2 + i)(2 − i) 4 = 4+1 4 = 5 =

[3]

(iii) z + z = (2 + i) + (2 − i) 2 1

2 2

2

2

= 4 + 4i − 1 + 4 − 4i − 1 =6 2 Im ( z 1 + z 22 ) = 0

z 12 − z 22 = (2 + i)2 − (2 − i)2 = 4 + 4i − 1 − 4 + 4i + 1 = 8i

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EdExcel FP1 Complex numbers Assessment solutions Re ( z 12 − z 22 ) = 0 [4]

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