Complex Numbers tmh solution

2 Complex Numbers LEVEL 2

fi (1 – z ) (w – wz ) = (1 – z) ( w – wz )

56. As ABCD is a parallelogram, mid point of AC = mid point of BD 1 1 fi ( z1 + z3 ) = ( z2 + z4 ) 2 2 fi z4 = z1 + z3 – z2 57. |a + bw + cw2|2 = (a + bw + c w ) (a + b w + cw) 2

2

2

= a + b + c – bc – ca – ab 1 2 2 2 = ÈÎ(b - c ) + (c - a ) + (a - b ) ˘˚ 2 As a, b, c are integers and at least two of them are unequal, we get,

fi (w – w ) (1 – z z ) = 0 As w π w , we get z z = 1 Thus, set of values of z is {z: |z| = 1, z π 1}. 60. zw = |z|2 fi zw = z z fi w = z Thus, |z – z | + |w + w | = 4 = |z – z | + | z + z| = 4 = |2iy| + |2x| = 4 = |x| + |y| = 2 y

(b – c)2 + (c – a)2 + (a – b) ≥ 2.

2

Thus, |a + bw + cw2 |2 ≥ 1 fi |a + bw + cw2| ≥ 1 Least value 1 is attained when a = 2, b = 1, c = 1.

-2

0

2

x

58. We have AP = AB = AQ = 2 -2

Thus, for the shaded region |z + 1| > 2

p Ê 2 ˆ = Also, –BAQ = tan -1 Á ˜ 4 Ë 2 - 1 + 1¯ and –BAP = – p / 4 Hence, for the shaded region |z + 1| > 2 w - wz 59. As 1- z w - wz = 1- z

and |arg (z + 1)| < p / 4

Fig. 2.49

This represents a square. See Fig. 2.49. 61. Let z = x + iy, then x = 1 – t, y =

t2 + t + 2

is purely real,

fi t = 1 – x and y2 = t2 + t + 2 = (t + 1/2)2 + 7/4

w - wz 1- z

fi y2 = (x – 3/2)2 + 7/4.

2.2  JEE Main Mathematics—Level 2, Hints and Solutions

This represents a hyperbola. 62. w + w = = =

az1 - bz2 az1 - bz2 + az1 + bz2 az1 + bz2

z2 z1

(az1 - bz2 ) (az1 + bz2 ) + (az1 - bz2 ) (az1 + bz2 ) az1 + bz2

2

a2 z1 z1 - b2 z2 z2

= 0 [ a |z1| = b |z2|] 2 az1 + bz2 fi w lies on the imaginary axis.

Fig. 2.50

63. |z| + |1 – z| + |z – 2| ≥ max {|z|, |(1 – z) + (z – 2)|, |z + (1 – z)| + |z – 2|, |1 – z| + |z – (z – 2)|} = 2 The value 2 is attained when z = 1 64. See Theory. 65. |z2 – z3|2 + |z1|2 = |z2 – z3|2 + | –z2 – z3|2 fi |z2 – z3|2 + 1 = 2|(|z2|2 + |z3|2) = 4 fi |z2 – z3| =

3

Similarly, |z3 – z1| = |z1 – z2| =

3

Thus, z1, z2, z3 are vertices of an equilateral triangle. 66. 1 + w + w2 + º + wn – 1 1 - wn 1 - cos p + i sin p = = 1- w 1 - cos (p / n) - i sin (p / n) = =

2 2 sin 2 (p 2n) - 2 i sin (p 2n) cos (p 2n)

2 - 2 i sin(p 2 n) [cos (p 2 n) + i sin (p 2 n)]

cos (p 2n) - i sin (p 2n) Êpˆ = 1 + i cot Á ˜ = Ë 2n ¯ - i sin (p 2n)

Alternative Solution Distance of z1 = 2(cosq + isinq) from x + y = 4 2, is 2 cos q + 2 sin q - 4 2 2 =

2 ( 2 2 – (cosq + sinq))

Maximum possible value of cosq + sinq is 2 . 69. Let z1 = r (cosq + isinq). Equation of tanget to x2 + y2 = r2 at (rcosq, rsinq) is x cosq + y sinq = r Ê z + z ˆ Ê z1 + z1 ˆ Ê z - z ˆ Ê z1 - z1 ˆ fi Á + = r2 Ë 2 ˜¯ ÁË 2 ˜¯ ÁË 2i ˜¯ ÁË 2i ˜¯ fi z z1 + z z1 = 2r 2 z z fi + =2. z1 z1 70. Using |z1| + |z2| ≥ |z1 – z2|, we get |z – 2 + 3i| + |z – 1 + i| ≥ |–1 + 2i| =

5

71. |z – a2| + |z – 2a| = 3 will represent an ellipse if

67. |z1 + z2|2 = |z1 – z2|2

|a2 – 2a| < 3

fi |z1|2 + |z2|2 + z1 z2 + z1 z2

¤ –3 < a2 – 2a < 3

= |z1|2 + |z2|2 – z1 z2 – z1 z2

¤ –2 < (a – 1)2 < 4

z z fi 2(z1 z2 + z1 z2) = 0 fi 1 + 2 = 0 z1 z2 68. z1 lies on the circle |z| = 2 and z2 lies on the line x+y= 4 2

¤ (a – 1)2 < 4 ¤ –1 < a < 3 ¤ aŒ(0, 3)

Distance of x + y = 4 2 from (0, 0) is 4 Thus, minimum distance between z1 and z2 is 2.

72. As AB = BC = CA, we get 2|z| = |1| = |1 – 2z| 1 1 1 fi |z| = and |z – | = 2 2 2 fi z is the point of intersection of circles

Complex Numbers  2.3

|z| = 1/2 and |z – 1/2| = 1/2 1 1 + 3i fiz= 4 73. |z + 1| + |z – 3| £ 10 represents the ellipse with focii at (–1, 0), and (3, 0) and length of major axis 10. Its centre is (1, 0), and its equation is

(

)

( x - 1) y 2 + =1 25 19 Any point on the ellipse is P (1 + 5 cosq, 2

fi |z|2 + 1 – z – z + 2[|z|2 + 1 – z w – z w] = 3[|z|2 + 1 – z w 2 – z w2] fi (3w – 2w2 – 1) z + (3w2 – 2w – 1) z = 0 which represents a straight line. 75. As R.H.S is real, L.H.S must be real. Ê 1 + i ˆ Ê i - i2 ˆ Also, Á ˜ =Á ˜ even. Ë 1 - i ¯ Ë 1 - i ¯

n

= in is real when n is

19 sinq ). Its distance from A (7, 0) is given by AP2 = (5cosq – 6)2 + 19 sin2q    = 6(5 – cosq)2 – 95. fi 1 £ AP2 < 121 fi 1 £ AP £ 11. 2

2

1 + x2 1 Ê 1ˆ = Á x + ˜ > 1. For x π 1, x¯ 2x 2Ë we get only possible value of x is 1. As x > 0 and

22

74. |z – 1| + 12| z – w| = 3|z – w |

\ RHS = 1, thus, least value of n is 4.