Complex Number Exercise # 4

EXERCISE # 4  Old questions Q.1  The complex numbers z1, z2 and z3 satisfying Sol. Q.3 3 z1  z 3 1 =  +i 2 z3  z 2 2 = (i)n = 1 = (i)4K For all complex numbers z1, z2 satisfying |z1| = 12 and |z2 – 3 – 4i| = 5, the minimum value of |z1 – z2| is- Sol. z z 1 i 3 [C] 1 3 = 2 z 2  z3  z n2 n = 4k Option (D) is correct answer. z1  z 3 1  i 3 = are the vertices of a triangles 2 z 2  z3 which is (A) of area zero (B) right angled isosceles (C) equilateral (D) obtuse angled isosceles z1n  Applying rotation theorem, we have z1  z 3 z z = 1 3 ei 2 / 3 z3  z 2 z3  z 2 A(z1) Q.4 (A) 0 (B) 2 (C) 7 (D) 17 [B] Given : | z1 | = 12 and |z2 – 3 – 4i| = 5 use, |z1 – z2| | z1 | – | z2 |  |z1 – (z2 – 3 – 4i) – (3 + 4i)|  |z1| – |z2 – 3 – 4i| – |3 + 4i|  12 – 5 – 5 2  Option (B) is correct answer. Let  = – 3 1 + i . Then the value of the 2 2 1 1 1 determinant 1  1   1 2 2 B(z2) z1  z 3  1 i 3   z3  z 2 2 2  Option (C) is correct answer. Sol.[B]  = – 1 Q.2 Sol. If z1 and z2 be the nth roots of unity which subtend right angle at the origin. Then n must be of the form (A) 4 k + 1 (B) 4k + 2 (C) 4k + 3 (D) 4k [D] xn – 1 = (x – 1) (x – 1) (x – 2) (x – 3) (x – 4) …… (x – n – 1)  | z1 | = 1 and | z2 | = 1 1  Q.5 n   = (i)n  Sol. (B) 3( – 1) (D) 3 (1 – ) i 3 1 + 2 2 1 1 1   z |z |  1 = 1 e i / 2 = e i / 2 = i z2 | z2 | z   1  z2  (A) 3 (C) 32 C(z3) 2 is 4  2 1 2  2  4 1 1 1 = 1  2 1 2  = 1 (2 – 4) – 1( – 2) + 1(2 – ) = 2 – –  + 2 + 2 =  = 32 – 3 = 3 ( – 1)  Option (B) is correct answer. Let a complex number  be a root of the equation zp+q – zp – zq + 1 = 0, where p, q are distinct primes. Show that either 1 +  + 2 + ..... + p – 1 = 0 or 1 +  + 2 + ...... + q – 1 = 0, but not both together. zp + q – zp – zq + 1 = 0 30  Q.6  zp. zq – zp – zq + 1 = 0  zp (zq – 1) – 1(zq – 1) = 0  (zp – 1) (zq – 1) = 0  Either zp – 1 = 0 or zq – 1 = 0 when zp – 1 = (z – 1) (z – 1) (z – 2) (z – 3) … (z – p – 1) where 1, 1, 2, …. , p – 1 are root of zq – 1  sum of root 1 + 1 + 2 + …. + p – 1 = 0 when zq – 1 = 0  zq – 1 = (z – 1) (z – 1) (z – 2) (z – 3) (z – 4) …. (z – q – 1)  Sum of roots = 1 + 1 + 2 + 3 + … + q – 1 = 0  1 + 1 + 2 + 3 + …. + q – 1 = 0 Hence, Proved. Sol. 1  z1 z 2 z1  z 2 (C) Sol. 1 | z  1| (B) 2 2  (1 – z1 z 2 – z1 z2 + z1 z1 . z2 z 2 ) < (z1 z1 – z2 z1 – z1 z 2 + z2 z 2 )  (1 – z1 z 2 – z1 z2 + |z1|2 |z2|2) < (|z1|2 + |z2|2 –z2 z1 – z1 z 2 )  1 + |z1|2. |z2|2 < |z1|2 + |z2|2  1 + |z1|2 |z2|2 – |z1|2 – |z2|2 < 0  |z1|2 (|z2|2 – 1) – 1 (|z2|2 – 1) < 0  (|z2|2 – 1) (|z1|2 – 1) < 0 For |z2|2 – 1 < 0 and |z1|2 – 1 > 0  Either |z1|2 – 1 > 0 and |z2|2 – 1 < 0 or 2 |z2| – 1 < 0 and |z1|2 – 1 > 0  Either |z1|2 > 1 and |z2|2 < 1 or 2 |z2| < 1 < |z1|2  Either |z1| < 1 < |z2| or |z2| < 1 < |z1| Hence, Proved. 1 | z  1| 2 Given : | z | = 1, z  1 z 1 = z 1 Multiplying by ( z + 1) in numerator and denominator, we have z 1 z 1 = × z 1 z 1 [D] = (z  1)  ( z  1) zz  z  z  1 = (z  1)(z  1) | z  1 |2 =  = | z |2  z  z  1 | z 1| 2 = Q.8 1  z  z 1 n r 1 zz Sol. | z  1 |2 1  z1 z 2 z1  z 2 Given, ai ; i = 1, 2, 3, ………. |ar| < 2 n  a .z r r 1 r 1 Let z1 & z2 are two complex numbers such that |z1| < 1 < |z2|. Prove that Let ai, i = 1, 2, 3, ........ are complex numbers such that |ar| < 2, then prove that there is no complex number such that |z| < 1/3 and  a r z r = 1. | z  1 |2 Since, z – z = Purely imaginary  Real part of  = 0  Option (D) is correct answer. Q.7 <1  (1 – z1 z 2 ) (1 – z1 z2) < (z1 – z2) ( z1 – z 2 ) (D) 0 | z  1| 2 2  |1 – z1 z 2 |2 < |z1 – z2|2 w=? (A) <1 squaring both sides, we have z 1 then real part of z 1 If |z| = 1, z  –1 and w = 1  z1 z 2 z1  z 2 < 1.  a1z + a2z2 + a3z3 + a4.z4 + a4z4 + a5.z5 + …. + ar.zr + ….. + anzn = 1 Taking modulus of both sides, we get. |a1z + a2z2 + a3z3 + … + arzr + … + anzn | = 1 Using property |z1 + z2|  |z1| + |z2| 31  |a1 z| + |a2 z2| + |a3 z3| + |a4z4| + |a5z5| + … + |arzr| + …. + |anzn|  1 Since, |ar| < 2  2(|z| + |z|2 + |z|3 + |z|4 + |z|5 + |z|6 + … + |z|r + … + |z|n ) 1  |z| + |z|2 + |z|3 + |z|4 + |z|5 + … + |z|r + … + |z|n  1/2 It forms a G.P. with common ratio |z|   | z | (| z | 1) 1  | z | 1 2  | z |n 1  | z | 1  | z | 1 2  n  | z |n + 1 – | z |   | z |n + 1 – –  1 1 |z|– 2 2 Q.10 3 1 |z|– 2 2 |z|– |z| Find the centre and radius of the circle formed by all the points represented by z = x + iy satisfying the relation Sol. z = k ; (k  1) z  1 2 2 | z |n + 1 > × 3 3 2 Squaring both sides, we get 2 1 + | z |n + 1 3 3  (z – ) ( z –  ) = k2 (z – ) ( z –  ) |z – a|2 = k2 |z – |2  z z – z –  z +   i.e. there will be no complex number if |z| < 1/3 = k2 (z z –  z – z  +   n and  a r .z r = 1  |z|2 –  z – z +  = k2 |z|2 – k2 z – k2  z – k 2  r 1 Q.9 Sol. If  is cube root of unity (  1) then the least value of n, where n is positive integer such that (1+ 2)n = (1 + 4)n is(A) 2 (B) 3 (C) 5 (D) 6 [B] Since,  is cube root of unity i 3 1 =– + 2 2 2 = – z = k; (k  1) where z   and  are constant complex number given by  = 1 + i2, = 1 + i2. 3 1 | z | + | z |n + 1  – 2 2 3 1 | z | – | z |n + 1  2 2 1 + 4 + 22 = 1 + 2 + 2 1 +  + 2 +2 = 1 +  + 2 +  2 =  which is not true. Take n = 3  (1 + 2)3 = (1 + )3  (1 + 2) .(1 + 2)2 = (1 + ).(1 + )2  (1 + 2).(1 + 2)2 = (1 + ).(1 + )2  (1 + 2).2 = (1 + )    2 + 4 = + 2 2 + 3 =  + 2 2 +  =  + 2  L.H.S. = R.H.S.  Least value of n is 3  Option (B) is correct answer  |z|2 (k2 – 1) + z (  – k2  ) + z ( – k2) + (k2   –   ) = 0  |z|2 (k2 – 1) + z(  – k2  ) + z ( – k2) + (k2  –  ) 2     k2    + k  2   k 2 1   k 1     |z|2 + z   k 2       =0 +  2  i 3 1 – 2 2 1++  (1 + 2)n 2 =0 = (1 + 4)n = (1 + 3)n = (1 + )n Let n = 2  (1 + 2)2 = (1 + )2  k 1  We can compare above equation with general. Equation of circle |z|2 + az + az + b = 0 Centre : ( – a) Radius = | a |2 b 32  X2 = |a + b + c2|2 aa  b = = (a + b + c2) (a + b  + c  2)  k 2     ;k1 Centre :  2   k 1  Radius = (k 2  )(k 2    ) (k 2  1) 2  ( k 2     ) (k 2  1) (k   )(k    )  (k  1)(k      2 = = a2 + ab + ac2 + ab  + b2  + 2 2 b c2.  + ac  2 + bc  2 + c2(  )2 = a2 + b2 |  |2 + c2 ||4 + ab ( +  ) + bc (2  +   2 ) + ac(2 +  2 ) 2 (k 2  1) 2 Since,  = – 4 2 2 4 2 2 = k      k  k       (k    k    k      ) k 4     k 2  k 2      k 4   k 2    k 2     (k 2  1) = = = = = = k 2 (           (k 2  1) k           (k 2  1) k k (k  1) k |    |2 (k  1) (k  1) |– | ; k  1 Hence, require answer is – Centre = Radius = Q.11 Sol.  2   2 = – i 3 1 i 3 1 – – + =–1 2 2 2 2 2 . + . 2  1 i 3  1 i 3     =    2  2   2 2   1 i 3  1 i 3     +    2  2   2 2 2 k i 3 i 3 1 1 + – – =–1 2 2 2 2 (  )(   ) 2 2 +  =– (   )  (    ) (k 2  1) i 3 1 1 i 3 +   = – + 2 2 2 2 2 = – (k 2  1) = i 3 i 3 1 +   = – 2 2 2 k 2   k 2 1 k k 1 2 ;k1  is cube root of unity,   1 2 = – = 1 3 i 2 3 1 3 2i 3      4 4 4 4 4 4 = 1 3 – =–1 2 2 i 3 1 + 2 2 1 i 3 – 2 2 2  X2 = a2 + b2 + c2 – ab – bc – ac = | – | ; k  1 a, b, c are variable integers not all equal and   1,  is cube root of unity, then minimum value of x = |a + b + c2| is (A) 0 (B) 1 (C) 2 (D) 3 [B] Given : a, b, c are variable integers not all equal Take, = –  1 i 3    +  1  i 3  =    2  2 2  2    2  1 {(a – b)2 + (b – c)2 + (c – a)2} 2 Take a = b such that b  c and a  c X2 = 1 {0 + (a – c)2 + (c – a)2} 2 X2 = 1 {2 × (a – c)2} 2 X2 = (a – c)2 Since, (Difference of integer numbers)2  1 i.e. (a – c)2  1  X2  1  Minimum value of X2 = 1  Minimum value of |a + b+ c2| is 1.  Option (B) is correct answer. 33 Q.12 Four points P (–1, 0), Q (1, 0), R ( 2 –1, 2 ), S ( 2 –1, – 2 ) are given on a complex plane then equation of the locus of the shaded region excluding the boundaries  z0 = z1  z 3 = 1 + 0i. 2  z3 = 2 – z1 z3 = 2 – 2 – i 3 z3 = – 3 ( 2 –1, 2 ) R P (–1, 0) Q (1, 0) ( 2 –1,– 2 ) Applying rotaion theorem at point D S  4  (B) |z + 1| > 2 & |arg (z + 1)| < 2  (C) |z 1| > 2 & |arg (z  1)| < 4  (D) |z 1| > 2 & |arg (z  1)| < 2 (A) |z + 1| > 2 & |arg (z + 1)| < Sol.  z1  z 4 = cos /2 + + i sin /2 z 4  z3  z1  z 4 =i z 4  z3  z1 – z4 = i z4 – i z3  z1 + iz3 = z4 (1 + i) z1  iz 3 1 i  z4 = 2  i 3  i ( i 3 ) 1 i  z4 = (1  i) 2i 3  3 × (1  i) (1  i)  z4 = ( 2  3  i 3 )(1  i) 2  z4 = 2  3  i 3  2i  i 3  3 2   < Arg (z + 1) < 4 4  | Arg (z + 1)| < 4  z4 = 2  2 3  2i 2 A square having one vertex as 2 + 3 i is circumscribed on the circle |z – 1| = 2. Then find the other vertices of square. Given |z – 1| = 2 M is the mid point of diagonals AC and BD. z3  z 2 |z z | = 3 2 ei/2 z 2  z1 | z 2  z1 | [A] Shaded region is 'the outside region of circle |z + 1| = 2 So |z + 1| > 2 /4 Also Sol. z1  z 4 |z z | = 1 4 ei/2 z 4  z3 | z 4  z3 |  z4 = /4 Q.13  A(z1) (2 + i 3 ) B(z2)  z4 = 1 + 3 –i Also, coni method or rotation theorem at point B.  z3  z 2 = cos /2 + i sin /2 z 2  z1 z3 – z2 = i (z2 – z1) z3 – z2 = iz2 – iz1 z3 + iz1 = z2 (1 + i) M z2 = z0(1 + 0i) z2 = C(z3) D(z4) z 3  iz 1 1 i z 3  iz 1 (1  i)  (1  i) (1  i) 34 z2 =  i 3  i( 2  i 3 )  i(1  i 3 )  2  i 3 2 P(z)  3 i3  A    3 2  2 3  2i z2 = 2 z2 = 1 – North z 3  iz 1  iz 3  z1 2 Imaginary axis z2 = West 145º O 3 +i  2 2 East Real axis Hence, required veritus are z2 = (1 – z3 = – i z4 = (1 + Q.14 Sol. 3)+i South 3 zA = |zA| ei/4 3 ) – i Ans If w = +i where   0 and z  1 satisfies the  w  wz  condition that   is purely real, then  1 z  set of z is – (A) {z : z = z } (B) {z : |z| = 1, z 1} (C) {z : z  1} (D) {z : |z| = 1} w  wz [B] is purely real 1 z w  wz w  wz  = 1 z 1 z  (w  wz)(1  z) = (1  z)(w  wz)  (w  w) (| z |2 – 1) = 0 As ww | z | = 1 and z  1 zA = 3ei/4 Applying Rotation theorem at A, we have 0  3e i / 4 z  3e = i 4 3 i/2 e 4  z – 3 ei/4 = – 3 × 4 × e i / 2 × ei/4 3  z – 3 ei/4 = + 4 × i ei/4  z = (3 + 4i) ei/4  Option (D) is the correct answer. Q.16 If |z| = 1 and z  ±1, then all the values of z 1 z2 lie on - (A) a line not passing through the origin Q.15 A man walks a distance of 3 units from the origin towards the north-east (N 45º E) direction. From there, he walks a distance of 4 units towards the north-west (N 45º W) direction to reach a point P. Then the position of P in the Argand plane is(A) 3ei/4 + 4i (B) (3 – 4i) ei/4 i/4 (C) (4 + 3i) e (D) (3 + 4i) ei/4 Sol. [D] Sol. (B) |z| = 2 (C) the x–axis (D) the y–axis [D] Given | z | = 1 and z ± 1 z 1 z 2 = z (1  z )(1  z ) × (1  z )(1  z ) (1  z)(1  z) = = = z(1  z  z  z z ) | 1  z |2 | 1  z |2 z(1  z z ) | 1  z |2 | 1  z |2 (z  z z . z ) | 1  z |2 | 1  z |2 35 = = A:y1 (z  z | z |2 ) |1 z | |1 z | 2 2 (z  z ) | 1  z |2 | 1  z |2 B : (x – 2)2 + (y – 1)2 = 9 C:x+y= = Purely Imaginary   Option (D) is the correct answer. Q.17 A particle P starts from the point z0 = 1 + 2i, z 2 (–1,1) P where i = – 1 . It moves first horizontally away from origin by 5 units and then vertically away from origin by 3 units to reach a point z1. From z1 the particles moves 2 (2, 1) (5, 1) Q 2 2 units in the direction of the vector î + ˆj and then it moves There is only one common point which satisfies to all the three given curves  through an angle in anticlockwise direction 2 Sol. on a circle with centre at origin, to reach a point z2. The point z2 is given by (A) 6 + 7i (B) – 7 + 6i (C) 7 + 6i (D) – 6 + 7i [D] Q.19 Sol. Let z1 = (–1, 1) and z2 = (5, 1). P (7, 6) Z (6, 5) 2 | z + 1 – i |2 + | z – 5 – i |2 = | z1 – z2 |2 = (6)2 = 45º 36. 3 Z0 (1, 2) Q.20 (6, 2) In diagram you can see P point is given by complex No. 7 + 6i and now it is rotated by Let z be any point in A  B  C. Then, |z + 1 – i|2 + |z – 5 – i|2 lies between (A) 25 and 29 (B) 30 and 34 (C) 35 and 39 (D) 40 and 44 [C]  2 Sol. Let Z be any point in A  B  C and let w be any point satisfying |w – 2 – i| < 3. [IIT 2008] Then |z| – |w| + 3 lies between (A) –6 and 3 (B) –3 and 6 (C) –6 and 6 (D) –3 and 9 [D] We know that, ||z|–|w|||z–w| angle in anticlockwise, since so it will be i (7 +  –3 < | z | – | w | + 3 < 9. 6i)  z2= – 6 + 7i. Paragraph for Question Nos. 18 to 22 Q.21 15  Im (z 2m 1 ) at  = 2° is - Let A, B, C be three sets of complex numbers as defined below A = {z : Imz  1} B = {z : |z – 2 – i| = 3} m 1 1 sin 2 1 (C) 2 sin 2 (A) C = { z : Re ((1 – i) z) = 2 } Q.18 Sol. The number of elements in the set A  B  C is(A) 0 (B) 1 (C) 2 (D)  [B] Let z = cos  + i sin . Then the value of Sol. 1 3 sin 2 1 (D) 4 sin 2 (B) [D] Z = cos + i sin Z2m–1 = cos(2m–1) + i sin(2m–1) Given that, 36 Im Z 2m–1 = sin + sin3 + …. + sin29 (P)  2  sin 15.  2 2   = sin ( + (15 –1) ) 2  2  sin    2  = Q.22 Sol. 1 h  k2 =2 2 sin(15) 2 sin  h2  k2 = x2 + y2 = 1 =   = 2º 4 sin 2º 1 2 1 4 Circle (Q) | |z + 2| – |z –2| = 3 Let z = x + iy be a complex number where x and y are integers. Then the area of the rectangle whose vertices are the roots of the equation z z 3 + z z3 = 350 is (A) 48 (B) 32 (C) 40 (D) 80 [A] Let Z = x + iy  (x2 + y2) (x2 – y2 – 2ixy) + (x2 + y2) (x2 – y2 + 2ixy) = 350 2 2 2  2(x + y ) (x – y2) = 350  (x2 + y2) (x2 – y2) = 175  x2 + y2 = 25 x2 – y2 = 7 On solving x=±4 y=±3  Area of rectangle = 8 × 6 = 48 (–2, 0) (2, 0) hyperbola having focus (–2, 0) and (2, 0) (R) x = 3 1 t2 1 t2 ,y= 2t 1 t2 x2 + y2 = 1 ellipse 3 (S) Parabola, hyperbola (T) Let z = x + iy (x + 1)2 – y2 = x2 + y2 + 1 2x – y2 = y2 y2 = x parabola Q.23 Column – I (A) Circle (B) Parabola (C) Ellipse Column – II (P) The locus of the point (h, k) for which the line hx + ky = 1 touches the circle x2 + y2 = 4 (Q) Points z in the complex plane satisfying |z + 2| – |z –2| = ± 3 (R) Points of the conic have parametric representation  1 t 2   , y = 2t x = 3  2  1 t2 1 t  (D) Hyperbola Sol. A  P; (S) The eccentricity of the conic lies in the interval 1x< (T) Points z in the complex plane satisfying Re (z + 1)2 = | z |2 + 1 B  S,T; C  R; D  Q, S; Q.24 Let z1 and z2 be two distinct complex numbers let z = (1–t) z1 + tz2 for some real number t with 0 < t < 1. If Arg (w) denotes the principal argument of a nonzero complex number w, then (A) | z – z1 | + | z – z2 | = | z1 – z2 | (B) Arg (z – z1) = Arg (z – z2) (C) Sol. z  z1 z  z1 z 2  z1 z 2  z1 =0 (D) Arg (z – z1) = Arg (z2 – z1) [A,C,D] t= z  z1 z 2  z1 So, z  z1 = t eio V t  (0, 1) z 2  z1 So Geometrically z1 z z2 37 So option A, C, D are true. Q.25 2 2 .  i sin 3 3 Then the number of distinct complex number z Let  be the complex number cos satisfying Sol. Q.26 (B) z 1  2  z  2 1 2 1 z (C) (D) Q.27 [Note : Here z takes values in the complex plane and Im z and Re z denote, respectively, the imaginary part and the real part of z] [IIT-2010] Column – I so its equation will be = 0 is equal to [1] On solving the determinant It become z3 = 0 So no. of solutions = 1 |Z + 4| + |Z – 4| = 10 it will represent on ellipse having foci (–4, 0), (4, 0) whose eccentricity is 4/5 Let w = 2ei. 3 5 z= cos + i sin  2 2 Let w = ei Z = ei + e–i = 2 cos . If z is any complex number satisfying | z – 3 – 2i|  2, then the minimum value of | 2z – 6 + 5i| is Sol. (3, 2) B Column – II (A) The set of points z (P) an ellipse with satisfying Min value eccentricity 4/5 |z – i|z|| = |z + i|z|| is (Q) the set of points satisfying |z + 4| + |z –4| = 10 is If |w| = 2, then the satisfying set of points z = w – Q.28 1 w is contained in or equal to of points z = w + 1 w 5i 5 =2× =5 2 2 Let  = ei2/3, and a, b, c, x, y, z be non-zero complex numbers such that a + b + c = x, a + b + c2 = y, a + b2 + c = z. Then the value of | x |2  | y | 2  | z | 2 | a |2  | b |2  | c |2 is Sol. wrong question if  = ei 2 / 3 then ans. is 3. If  = ei  / 3 then no integral solution is possible. Q.29 Let z be a complex number such that the imaginary part of z is nonzero and a = z2 + z + 1 is real. Then a cannot take the value (S) the set of points z satisfying |Re z|  2 is contained in or equal to (T) the set of points z satisfying |z|  3 Sol. = Min 2 z – 3  (R) the set of points z |Im z|  1 (D) If |w| =1, then the set 5  P  3,–  2  z satisfying Im z = 0 contained in or equal to (C) A So, Min of |2z – 6 + 5i| = PA contained in or equal to (B) The set of points z x 2 y2 =1  25 9 A  Q, R ; BP; C  P, S, T ; D  Q, R, S, T (A) Let |Z| = r  r  R Z  ir = 1Which is the equation of line Z  ir of perpendicular bisector of y = r & y = – r that is y = 0 (A) –1 (B) 1 3 (C) 1 2 (D) 3 4 Sol. [D] As a is real So a = a  z2 + z + 1 = z 2 + z + 1  (z – z ) (z + z +1) = 0 38 As z is imaginary So z – z  0 z+ z +1=0  z + z = –1  z = x + iy x= 1 2 so a = (x + iy)2 + (x + iy) + 1 = (x2 + x + 1 – y2) + (2x + 1)yi  x = – a= 1  2 3 – y2 4 so a < 3  y2 > 0 4 So a  3 4 EXERCISE # 4 1. (C) 2. (D) 3. (B) 10. Centre of the circle = – z0 = 4. (B)   k 2 1 k 2 6. (D) 9. (B) , Radius of the circle = r = k (  ) 1 k 2 13. (0, – 3 ), (1 – 3 , 1), (1 + 3 , – 1) 14. (B) 15. (D) 19. (C) 23. A P; B  S, T; 20. (D) 16. (D) 11. (B) 12. (A) 17. (D) 18. (B) C  R; D  Q, S 21. (D) 22. (A) 24. (A, C, D) 25. 1 26. A Q,R; B  P; C  P,S,T; D  Q, R, S, T 27. 5 29. D 28. 3 39