Complex Number and Quadratic Solved Problems

Page # 6 COMPLEX NUMBER AND QUADRATIC EQUATION SOLVED PROBLEMS Ex.1 Find the number of real roots of the equation e sin x – e– sin x – 4 = 0 Sol. Let e sin x = y then given equation reduces to 1 – 4 = 0 y – 4y – 1 = 0 y –  y2 4  20 = 2  5  y = 2 = 4.23, – 0.23 But y = e sin x is never negative. So y = e sin x = 4.23  sin x = log 4.23 > 1 which is not possible. Hence the equation has no real root. Ex.2 Sol. Ex.3 Sol. ab+ ca – bc = 0  a = Show that the Both roots of the equation ( x – b) ( x– c) + ( x– c) ( x– a) + (x– a) ( x– b) = 0 are real The given equation can be written in the following form : 3x2 – 2 (a+ b+ c) x + (ab+ bc + ca) = 0 Here discriminant = 4(a + b+ c)2 – 12 (ab + bc + ca) = 4[(a2 + b2 + c2) – (ab + bc + ca)] > 0 [  a2 + b2 + c2 > ab + bc+ ca]  Both roots are real. If a < b < c < d, then show that roots of (x–a)(x–c) + 2 (x–b)(x–d)= 0 are real and unequal Here 3x2–(a+c+2b+2d)x + (ac+2bd) = 0  Discriminant = (a+c+2b+2d)2–12 (ac+2bd) Thus sum of roots = – 2a = Ex.5 Sol. or x2 + (a+ b)x + ab = ( b–a) x + ( b– a) c or x2 + 2ax + ab + ca – bc = 0 Since product of the roots = 0 +  2  2  1 (  1) 2 + (  1) 2 , (  1) 2  (1  c) (  1) 2  (1  c) Putting value of 1 – c = ( + 1) ( +1)  1  1 = +     1  1 = 1   Find the quadratic equation whose one root 1 is 2 5 1 Given root = = 5– 2 2 5 = Ex.6 Sol. So the other root = – 5 – 2. Then sum of the roots = – 4, product of the roots =–1 Hence the equation is x2 + 4x – 1 = 0 Ex.7 Sol. = [(a+2d)–(c+2b)]2+8 Sol.  2  2  1 2  2  2  c   2  c Here the equation is x2 – p(x + 1) – c = 0     = p,  = – (p + c)  ( + 1) ( + 1) = 1 – c Now given expression = –12(ac+2bd) Ex.4 2bc bc If  are the roots of x2–p(x+1)–c=0 then find = [(a+2d)–(c+2b)]2+ 4 (a+2d)(c+2b) (c–b)(d–a)>0. 1 1 1 For the equation – = , if the x  b x  c xa product of roots is zero. find the sum of roots 1 1 1 – = xa xb xc 1 ba = xc x2  (b  a) x  ab bc bc Ex.8 Sol. Find the value of m for which one of the roots of x2 – 3x + 2m = 0 is double of one of the roots of x2 – x + m = 0 Let  be the root of x2 – x + m = 0 and 2  be the root of x2 – 3x + 2m = 0. Then,  2 –  + m = 0 and 4  2 – 6  + 2m = 0 1  2 = =   m2 = – 2m 2  2 m  4m  m = 0, m = – 2 If roots of the equation x2 + ax + 25 = 0 are in the ratio of 2 : 3 then find the value of a Here k = 2/3 so from the condition (k  1) 2 b2 = k ac a2 ( 2 / 3  1) 2 = 25 2/3  25 3 a2  = 9 2 25  25 a2 = 6 25 Page # 7 COMPLEX NUMBER AND QUADRATIC EQUATION a2 =  Ex.9 Sol.  25 25  25 a = 6 6 Ex.13 If z= If x = 2 + 3 then find the value of x3 – 7x2 + 13 x – 12 x = 2 + 3 x – 2 = 3  x2 + 4– 4x =3  x2 – 4x = – 1  x2 – 4x+ 1 = 0 Now we can write the given equation as x3 –7x2 + 13 x – 12 = x(x2 – 4x + 1) – 3x2 + 12 x – 12 = x(x2 – 4x + 1) –3(x2 – 4x + 1) – 9 Now putting the value of x2 – 4x + 1 = 0 = x (0) – 3(0) – 9 = – 9 Ex.10 Find Sol. 2 2 × =  3 = 2 6 ( i ) 2i × Ex.14 If = – Sol. AB = ( 3  1) 2  (1  2) 2 = BC = ( 2  3) 2  ( 3  1) 2 = CD = (2  2) 2  ( 2  3) 2 = DA = (1  2)2  ( 2  2) 2 = 17 17 AC = ( 2  1) 2  ( 3  2) 2 = 34 and BD = ( 2  3) 2  ( 2  1) 2 = 34 AB = BC = CD = DA and AC = BD ABCD is a square |z| = (1/ 2)  i 2 (1/ 2)  1    tan   i  sin  cos  2 2   1  2i sin 2 = 4 2 2 4 – i =  ,  5 5 5 5 is purely imaginary then     tan   i  sin  cos  1  2i sin 2 2 2   = ×   1  2i sin 1  2i sin 2 2         tan   2 sin  sin  cos   i sin  cos  2 tan  sin  2 2 2  2 2 2  1  4 sin2 2 tan  – 2 sin  2     sin  cos  = 0 2 2  sin  – (1 – cos ) – sin  = 0 cos   1  cos    – (1 – cos ) = 0  sin    cos    (1 – cos ) (tan  – 1) = 0 cos  = 1   = 2n and  tan  = 1   = n + 4 Ex.15 The complex numbers sin x + i cos 2x and cos x – i sin 2x are conjugate to each other then find x Sol. sin x + i cos 2x = cos x + i sin 2x  tan x = 1 and tan 2x = 1  n   x = n  + and x = + 4 2 8 7   3   5  9  , , , , ,.....  x   ..., 4 4 4 4 4  RS T 17 17 2 Multiply above and below by conjugate of denominator and put real part equal to zero. 6 the vertices of any quadrilateral are = 1+ 2i, B = –3 + i, C = –2 – 3i , and = 2 – 2i, show that it is a square = (1, 2), B = (–3, 1) = (–2,–3), D = (2, –2) z then find the value of z–1 find the general value of  = 3i Diagonal   z–1 = 3 Ex.11 If x be real , find the relation between a 1  ix and b , when = a – ib 1  ix 1  ix Sol.  = a – ib 1  ix on taking modulus; we get 1  ix | a – ib | = 1  ix | 1  ix | 1  ix  a2  b2 = = | 1  ix | = 1 1  ix  a2 + b 2 = 1 Ex.12 If A D Sol. A C Sol. FG 1 , 1IJ , H2 K RS T    ...,  UV W UV W 7   3   5  9  , , , , ,..... 8 8 8 8 8 there is no common value of x. Ex.16 If z1,z2 are any two complex numbers and a, b are any two real numbers, then prove that |az1–bz2|2+|bz1+az2|2 = (a2+b2)(|z1|2+|z2|2) Sol. Expression = (az1–bz2) (az1  bz2 ) +(bz1 + az2) (bz1  az 2 ) = (az1– bz2)(a z1 – b z2 ) Page # 8 COMPLEX NUMBER AND QUADRATIC EQUATION + (bz1 + az2) (b z1 + a z2 ) = a2 |z1 |2 + b2 | z2 |2 + b2 |z1|2 + a2 |z2|2 = (a2 + b2) (|z1 |2 + | z2 |2) Ex.17 Find the amplitude of 1–cos – i sin  Sol. Let z = 1 – cos  – i sin  r(cos  + i sin  ) sin   tan   = – 1 cos   2 sin( / 2)cos( / 2) = = – cot ( / 2) 2 sin 2 ( / 2) = – tan tan   = Ex.21 Find the Square root of – 8 – 6i Sol. Let  amp(z) =   – 2 2  – 8 – 6i = a2 – b2 + 2iab  a2 – b2 = – 8 ...[1] 2ab = – 6  ab = – 3 ...[2]  FG IJ H K    FG    .........IJ H2 2 2 K 2 3 FG z  z IJ  H z  z K = 4 , then find |z–7–9i| F z z I  If z = x + iy, then amp G z  z J = H K 4 1 number such that amp a2 + b2 = 10 from equation So, 2 x 2  y 2  14 x  18 y  130 = 3 2. {cos(  / 3)  i sin(  / 3)} ( 3  i) i1 Here |z| z = = |cos (  / 3 )  i sin(  / 3 )|| 3  i| = |i  1| = ( /3)} + amp ( 3 + i) – amp (– 1 + i)    11  = – + – = – 4 3 6 12 Therefore 11 11 cos   i sin  z = 2 12 12 FG H IJ K IJ K FG H  k 1 2k 2k  i cos  7 7  2k 2k   i cos  sin  7 7   2k 2k    i sin  = – i  cos 7 7   6 2 2 Again amp(z) = amp {cos( /3) – i sin RS FG T H 6   sin  2 IJ UV KW (1 – 3i) Ex.23 Find the value of Sol. Ex.20 Find the polar form of complex number  8  6i = ± x y – = (a2 – 3b2) – (b2 – 3a2) a b = 4(a2 – b2)  k = 4  ...(1) (from 1) (2) and (3) Ex.22 If z = x + iy, z1/3 = a – ib and x y – = k(a2 – b2), then find k a b Sol. Here x + iy = (a – ib)3 = (a3 – 3ab2) + i (–3a2b + b3)  x = a3 – 3ab2 , y = b3 – 3a2b 2  x2 + y2 – 14x – 18y + 112 = 0 Now |z – 7 – 9i| ...[3] a = 1, b = –3 1 =  8  6i = ± (a + ib) = (–8)2 + (–6)2 = 64 + 36 = 100 Ex.19 If z1 = 10+6i, z2 = 4+6i and z is a complex Sol. = (a2 + b2)2 = (a2 – b2)2 + 4a2b2 Ex.18 If xn = cos (/2n) + i sin (/2n) , then find x1x2 x3........... Sol. x1x2 x3 .......     2  3  ......... = cos 2 2 2 Sol. 2 FG    IJ H 2 2K F  I tan GH  JK 2 2 or + i sin RScos FG  11  2IJ  i sin FG  11  2IJ UV T H 12 K H 12 K W Rcos FG 13 IJ  i sin FG 13 IJ UV 2 S T H 12 K H 12 K W =    sin k 1 = – ie 2ki 7 2k 2k  i cos  7 7  4πi  2πi  = – i e 7 + e 7 + ......6 terms   12πi  2i  1 - e 7  = – ie 7  ( e 2i  1) 2πi  1 - e 7    = – i  2πi   e 7 - 1 = i  2πi  1- e 7   