Section 5.8. (i) ( 3 + i)(14 2i) = ( 3)(14 2i) + i(14 2i) = {( 3)14 ( 3)(2i)} + i(14) i(2i) = ( i) + (14i + 2) = i.

Size: px

Start display at page:

Download "Section 5.8. (i) ( 3 + i)(14 2i) = ( 3)(14 2i) + i(14 2i) = {( 3)14 ( 3)(2i)} + i(14) i(2i) = ( i) + (14i + 2) = i."

Matthew Hensley
6 years ago
Views:

1 1. Section 5.8 (i) ( 3 + i)(14 i) ( 3)(14 i) + i(14 i) {( 3)14 ( 3)(i)} + i(14) i(i) ( 4 + 6i) + (14i + ) i. (ii) + 3i 1 4i ( + 3i)(1 + 4i) (1 4i)(1 + 4i) (( + 3i) + ( + 3i)(4i) i i. (iii) (1 + i) 1 i 1 + 4i + (i) 1 i 1 + 4i i 1 i 1 i ( 3 + 4i)(1 + i) 7 + i i.. (i) iz + ( 10i)z 3z + i z(i + 10i 3) i z( 1 9i) i z i 1 + 9i i(1 9i) i 8 9 i. 41 (ii) The coefficient determinant is 1 + i i 1 + i 3 + i (1 + i)(3 + i) ( i)(1 + i) + i 0. Hence Cramer s rule applies: there is a unique solution given by 3i i + i 3 + i 3 11i z 1 + 5i + i + i 1 + i 3i 1 + i + i 6 + 7i 19 8i w. + i + i 5 54

2 (1 + i) + + (1 + i) 99 (1 + i)100 1 (1 + i) 1 Now (1 + i) i. Hence (1 + i)100 1 i i { (1 + i) }. (1 + i) 100 (i) i ( 1) Hence i { (1 + i) } i( 50 1) ( )i. 4. (i) Let z 8 6i and write zx+iy, where x and y are real. Then z x y + xyi 8 6i, so x y 8 and xy 6. Hence y 3/x, x 3 8, x so x 4 + 8x 9 0. This is a quadratic in x. Hence x 1 or 9 and consequently x 1. Hence x 1, y 3 or x 1 and y 3. Hence z 1 3i or z 1 + 3i. (ii) z (3 + i)z i 0 has the solutions z (3 + i ± d)/, where d is any complex number satisfying d (3 + i) 4(4 + 3i) 8 6i. Hence by part (i) we can take d 1 3i. Consequently z 3 + i ± (1 3i) (i) The number lies in the first quadrant of the complex plane. 4 + i Also Arg (4 + i) α, where tan α 1/4 and 0 < α < π/. Hence α tan 1 (1/4). i or 1 + i. y 4 + i α x 55

3 (ii) The number lies in the third quadrant of the complex plane. 3 i 3 i 1 ( 3) + ( 1) Also Arg ( 3 i ) π + α, where tan α 1 / 3 1/3 and 0 < α < π/. Hence α tan 1 (1/3). (iii) The number lies in the second quadrant of the complex plane. 1 + i ( 1) + 5. Also Arg ( 1+i) π α, where tan α and 0 < α < π/. Hence α tan 1. (iv) The number lies in the second quadrant of the complex plane. 1 + i i 3 1 ( 1) + ( 3) Also Arg ( i) π α, where tan α 3 / 1 3 and 0 < α < π/. Hence α π/3. 6. (i) Let z (1 + i)(1 + 3i)( 3 i). Then 3 i α y 1 + i y α i α y x x x z 1 + i 1 + 3i 3 i ( 3) ( 3) + ( 1) Arg z Arg (1 + i) + Arg (1 + 3) + Arg ( 3 i) (mod π) 56

4 Hence Arg z 5 1 π 4 + π 3 π and the polar decomposition of z is z 4 ( cos 5π ) 5π + i sin. 1 1 (ii) Let z (1+i)5 (1 i 3) 5 ( 3+i) 4. Then z (1 + i) 5 (1 i ( ) 3) ( 3 + i) 4 4 7/. Arg z Arg (1 + i) 5 + Arg (1 3i) 5 Arg ( 3 + i) 4 (mod π) 5Arg (1 + i) + 5Arg (1 3i) 4Arg ( 3 + i) 5 π π π π 11π 1 1. Hence Arg z 11π 1 and the polar decomposition of z is ( z 7/ cos 11π ) 11π + i sin (i) Let z (cos π 4 + i sin π 4 ) and w 3(cos π 6 + i sin π 6 ). (Both of these numbers are already in polar form.) (a) zw 6(cos ( π 4 + π 6 ) + i sin ( π 4 + π 6 )) 6(cos 5π 5π 1 + i sin 1 ). (b) z w 3 (cos ( π 4 π 6 ) + i sin ( π 4 π 6 )) 3 (cos π 1 + i sin π 1 ). (c) w z 3 (cos ( π 6 π 4 ) + i sin ( π 6 π 4 )) 3 π π (cos ( 1 ) + i sin ( 1 )). (d) z5 5 (cos ( 5π w 3 4 π 6 ) + i sin ( 5π 4 π 6 )) 3 9 (cos 11π 1 (a) (1 + i) i, so + i sin 11π 1 ). (1 + i) 1 (i) 6 6 i 6 64(i ) 3 64( 1)

5 (b) ( 1 i ) i, so ( 1 i ) 6 ( (1 i ) ) 3 ( i) 3 1 i 3 1 i 1 i i. 8. (i) To solve the equation z 1 + 3i, we write 1 + 3i in modulus argument form: 1 + 3i (cos π 3 + i sin π 3 ). Then the solutions are z k ( ( π 3 cos + kπ ) + i sin Now k 0 gives the solution Clearly z 1 z 0. z 0 (cos π 6 + i sin π 6 ) ( π 3 + kπ )), k 0, 1. ( ) 3 + i 3 + i. (ii) To solve the equation z 4 i, we write i in modulus argument form: Then the solutions are ( π z k cos + kπ ) + i sin 4 ( π ) Now cos +kπ 4 cos ( π 8 + kπ ), so ( π z k cos i cos π + i sin π. 8 + kπ ( cos π + i sin π ( π + kπ 4 ) ( π + sin i k (cos π 8 + i sin π 8 ). ), k 0, 1,, 3. ) 8 + kπ ) k π (cos 8 + i sin π 8 ) 58

6 Geometrically, the solutions lie equi spaced on the unit circle at arguments π 8, π 8 + π 5π 8, π 8 + π 9π 8, π 8 + 3π 13π 8. Also z z 0 and z 3 z 1. (iii) To solve the equation z 3 8i, we rewrite the equation as Then z 1, i z 3 1. i 1 + 3i, or 1 3i. Hence z i, 3 + i or 3 + i. Geometrically, the solutions lie equi spaced on the circle z, at arguments π 6, π 6 + π 3 5π 6, π 6 + π 3 3π. (iv) To solve z 4 i, we write i in modulus argument form: ( i 3/ cos π ) π + i sin. 4 4 Hence the solutions are z k 3/8 cos ( π 4 + kπ 4 ) ( π + i sin 4 + kπ 4 We see the solutions can also be written as ( z k 3/8 i k cos π ) π + i sin ( /8 i k cos π 16 i sin π ). 16 ), k 0, 1,, 3. Geometrically, the solutions lie equi spaced on the circle z 3/8, at arguments π 16, π 16 + π 7π 16, π 16 + π 15π 16, π π 3π 16. Also z z 0 and z 3 z 1. 59

7 9. + i 1 + i 1 + i 1 + i i + i 1 + i R R (1 + i)r 1 R 3 R 3 ir 1 R 1 R 1 R R 3 R 3 R 1 i i R 1 R 1 R R ir 1 i The last matrix is in reduced row echelon form. 10. (i) Let p l + im and z x + iy. Then 1 i i 1 + i 1 i 1 i. pz + pz (l im)(x + iy) + (l + im)(x iy) 1 i (lx + liy imx + my) + (lx liy + imx + my) (lx + my). Hence pz + pz n lx + my n. (ii) Let w be the complex number which results from reflecting the complex number z in the line lx + my n. Then because p is perpendicular to the given line, we have w z tp, t R. (a) Also the midpoint w+z of the segment joining w and z lies on the given line, so w + z w + z p + p n, w + z w + z p + p n. (b) Taking conjugates of equation (a) gives w z tp. (c) Then substituting in (b), using (a) and (c), gives w tp z + tp p + p n 60

8 and hence pw + pz n. (iii) Let p b a and n b a. Then z a z b z a z b (z a)(z a) (z b)(z b) (z a)(z a) (z b)(z b) zz az za + aa zz bz zb + bb (b a)z + (b a)z b a pz + pz n. Suppose z lies on the circle z a z b and let w be the reflection of z in the line pz + pz n. Then by part (ii) pw + pz n. Taking conjugates gives pw + pz n and hence z n pw p (a) Substituting for z in the circle equation, using (a) gives n pw p a λ b n pw pa n pw pb. n pw p (b) However n pa b a (b a)a bb aa ba + aa b(b a) bp. Similarly n pb ap. Consequently (b) simplifies to λ bp pw ap pw b w a w w b w a which gives w a 1 λ. w b 61

9 11. Let a and b be distinct complex numbers and 0 < α < π. (i) When z 1 lies on the circular arc shown, it subtends a constant angle α. This angle is given by Arg (z 1 a) Arg (z 1 b). However z1 a Arg z 1 b Arg (z 1 a) Arg (z 1 b) + kπ α + kπ. It follows that k 0, as 0 < α < π and π < Arg θ π. Hence z1 a Arg α. z 1 b Similarly if z lies on the circular arc shown, then z a Arg γ (π α) α π. z b Replacing α by π α, we deduce that if z 4 lies on the circular arc shown, then z4 a Arg π α, z 4 b while if z 3 lies on the circular arc shown, then z3 a Arg α. z 3 b The straight line through a and b has the equation z (1 t)a + tb, 6

10 where t is real. Then 0 < t < 1 describes the segment ab. Also z a z b t t 1. Hence z a z b is real and negative if z is on the segment a, but is real and positive if z is on the remaining part of the line, with corresponding values z a Arg π, 0, z b respectively. (ii) Case (a) Suppose z 1, z and z 3 are not collinear. Then these points determine a circle. Now z 1 and z partition this circle into two arcs. If z 3 and z 4 lie on the same arc, then z3 z 1 z4 z 1 Arg Arg ; z 3 z z 4 z whereas if z 3 and z 4 lie on opposite arcs, then z3 z 1 Arg α z 3 z and Hence in both cases ( z3 z 1 Arg / z ) 4 z 1 z 3 z z 4 z In other words, the cross ratio z4 z 1 Arg α π. z 4 z z3 z 1 z4 z 1 Arg Arg z 3 z z 4 z 0 or π. z 3 z 1 z 3 z / z 4 z 1 z 4 z is real. (b) If z 1, z and z 3 are collinear, then again the cross ratio is real. The argument is reversible. (mod π) (iii) Assume that A, B, C, D are distinct points such that the cross ratio r z 3 z 1 z 3 z / z 4 z 1 z 4 z is real. Now r cannot be 0 or 1. Then there are three cases: 63

11 (i) 0 < r < 1; (ii) r < 0; (iii) r > 1. Case (i). Here r + 1 r 1. So z 4 z 1 z3 z z 4 z z 3 z z4 z 1 z3 z 1. z 4 z z 3 z 1 Multiplying both sides by the denominator z 4 z z 3 z 1 gives after simplification or z 4 z 1 z 3 z + z z 1 z 4 z 3 z 4 z z 3 z 1, (a) AD BC + AB CD BD AC. Case (ii). Here 1 + r 1 r. This leads to the equation (b) BD AC + AD BC+ AB CD. Case (iii). Here r r. This leads to the equation (c) BD AC + AB CD AD BC. Conversely if (a), (b) or (c) hold, then we can reverse the argument to deduce that r is a complex number satisfying one of the equations r + 1 r 1, 1 + r 1 r, r r, from which we deduce that r is real. 64

or i 2 = -1 i 4 = 1 Example : ( i ),, 7 i and 0 are complex numbers. and Imaginary part of z = b or img z = b

or i 2 = -1 i 4 = 1 Example : ( i ),, 7 i and 0 are complex numbers. and Imaginary part of z = b or img z = b 1 A- LEVEL MATHEMATICS P 3 Complex Numbers (NOTES) 1. Given a quadratic equation : x 2 + 1 = 0 or ( x 2 = -1 ) has no solution in the set of real numbers, as there does not exist any real number whose