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1Complex NumbersP3A- LEVEL – MATHEMATICS(NOTES)1. Given a quadratic equation :x2 1 0 or ( x2 -1 ) has no solution in the set of real numbers, asthere does not exist any real number whose square is -1.Here we introduce a number (symbol ) i -1or i2 -1and we may deducei3 -ii4 1Now solution of x2 1 0x2 -1x -1orx i2. We define a complex number z ( x i y)Example : ( 4 3 i ),, 7i;x,y ϵ Rand 0 are complex numbers.a) Given a complex number z (a i b)Then real part of z aor Re z aand Imaginary part of z borimg z bb) Example i) z ( 4 3 i) is a complex numberii) ( 0 i ) is pure real numberiii) 7 i (0 7i ) is pure imaginary numberand0 0 i0

23. a) Equal Complex numbersGiven Complex numbers u ( a i b ) and v ( c i d )Then ( a i b) ( c i d )a c real and imaginary partsb d are separately equalb) Inequality in complex numbers is not definedu ( a i b)u v,v (c id)or v u is not defined4. Algebra of Complex Numbersa)Addition of Complex Numbers:Given u ( a i b ) and v (c i d ); a , b , c and d ε Ru v ( (a c ) i (b d) )Example : ( 2 3 i ) ( 4 - 7 i ) ( 2 4 ) i ( 3 ( - 7 ) ) (6–4i)b) Subtraction of Complex Numbers:u – v ( a i b ) - (c i d ) (a–c) i (c–d)Example : ( 2 3 i ) - ( 4 - 7 i ) ( 2 - 4 ) i ( 3 - ( - 7 ) ) ( -2 10 i )c) Multiplication of two complex numbers :( a i b ) (c i d ) ( ac - cd ) i ( ad bc)Example : ( 2 3 i ) ( 4 - 7 i ) ( 2 x 4 – 3 (-7) ) i ( 2 (-7) 3 x 4 ) ( 8 21 ) i ( -14 12 ) (29 – 2i)

3Note : i) ( a i b )2 ( a2 - b 2 ) 2 a b i( 2 3 i )2 ( 4 – 9 ) 2x 2x3i ( - 5 12 i )ii) ( a i b) ( a – i b) a2 b 2( 2 3 i) ( 2 – 3 i ) 2 2 3 2 13iii) ( a – i b)2 ( a2 - b 2 ) – 2 abi(2 – 3 i)2 (4 – 9 ) – 2 x 2 x 3 ( - 5 – 12 i )d) Division of two Complex Numbers : x x (-5. Conjugate of a Complex Numbers :Given z ( x i y )x,y εRConjugate of z denoted by z* ( x – y i)Example : i) z 4 3 iii) z 2 - 5 iz* 4 – 3 iz* 2 5 i6. Properties of Conjugate Complex Numbers :Given z ( a i b ) and w ( c i d )thenz* a – i bi)( z*) * zii)z z* 2 Re (z)iii)z – z * 2 i Im(z)iv)z z*and w* c - idz is pure real ii )

4v)z z* 0vi)z.z * ( Re z ) 2 ( Imz )2 a2 b2orvii)z is pure Ima.zz* z 2( z w )* z* w*viii) ( z - w )* z* - w*ix)x)( z w )* z*w*(* ; w 0xi) A quadratic equation with real coefficients :ax2 bx c 0such thathas conjugate complex roots.b2 – 4ac 0Example : z2 4 z 13 0 has conjugate complex rootsi.e ( - 2 3 i ) and ( - 2 – 3 i )6. Geometric Representation of a Complex NumbersTo each complex numbers z ( x i y) there corresponds a uniqueordered pair ( a, b ) or a point A (a ,b ) on Argand diagramExample : Represent the following complex numbers on an Argand Diagram :i) z ( 4 3 i )v) z1 iii) w -2 3 ivi) z2 3iii) u ( - 3 – 2 i )iv) v ( 4 - i )

5Solution:i) z ( 4 3 i )ii) w ( -2 3 i )A ( 4,3 )iv) v ( 4 - i )B ( - 2, 3 )iii) u ( - 3 – 2 i )v) z1 iC ( - 3 , -2 )vi) z2 37. Modulus of a Complex Numbers :Z (a ib)A (a,b)YA ( a,b )zOaModulus of z z OA ( a2 b2 )Example :z ( 4 3i )NOTE :bX; z 0z ( 4 2 32) 5z a ib, z* ( a – i b)zz* ( a i b) ( a – ib ) a 2 b2 ( (a2 b2 ))2 z zz* z 228. Properties of Modulus of a Complex Numbersz , z1 , z2 ε Ci)z 0z 0 or ( Re z 0 and Ima z 0 )D (4, -1 )E ( 0,1)F (3, 0)

6ii)z z* -ziii) - z Re z z2iv)zz* zv)z1z2 z1 z2 vi) vii)and - z Ima z z; z2 0 viii) z2 z29. To find square roots of a complex numberLet ( x i y) is square root of ( a i b )( x i y )2 ( a b i ) (i)Or ( x2 - y 2 ) 2xyi ( a b i)Orx2 - y 2 a (ii)2xy b (iii)Taking modulus on both sides of (i) ( x i y )2 a i b x i y 2 (a2 b2 )Or x2 y 2 (a2 b2 ) (iv)Adding equation (ii) and (iv) we get,2 x2 (a (a2 b2 )x2 ( a ( a2 b2 ) p ( let )x p (v)Now subtract equation (ii) from (iv)( a i b)

72 y 2 (a2 b2 ) - ay2 ( (a2 b2 ) - a ) q ( let )y qNOTE : Case I : Now from equation (iii) if 2xy b 0Then req. Square roots ( p i q )Case II : If 2xy b 0Then req. Square roots ( p - i q )10. Important application of modulus of complex numbers :If Complex numbers z1 and z1 are represented by points A and Brespectively.i) AB z2 - z1Bz2A z1and z2 - z1 ABoand z1 OAii) Given z - z1 z - z2 represents the locus of z, which is the set ofpoints P equidistant from two given points A ‘z1’ and B ‘z2’Hence , Locus of z is the perpendicular bisector of ABiii) Equation of circleP ‘z’a) z –z 0 RRCZ0

8Locus of z is a circle :Centre at C ‘ z0 ‘ and radius CP Rb) Show that : Complex equation :z z* a z* a* z b 0 .(i) ;aεC,b ε R2represent a circle with centre at ‘-a ‘ and radius R ( a - b)Solution : from equation ( i )z z* a z* a* z - bAdd aa* on both the sidesz z* a z* a* z a a* - b a a*z* ( z a ) a* ( z a ) a a* – b(z a ) ( z* a* )(z a ) (z a )* a 2 - b a 2 - b z a 2 a 2 - b z a ( a 2 - b ) ----------------------------(ii)Comparing it with { z - a R } equation ( ii ) represent a circle withcentre at ‘ – a ’ and R ( a 2 - b )c) z – z1 k z – z1 ; kR , k 1 represent a circle , kP zA z1‘z1 ‘iv) AP(z)C z- z1 z- z2 z1- z2 B ‘z 2 ‘( or PA PB AB )B z2

9P lies on segment AB‘z1 ‘v) Az2P ‘z ‘B z- z1 - z- z2 z1- z2 P lies on segment BP11)orPA - PB ABArgument of a Complex Numbers :z ( a b i ) complex number is represented by point Aarg z AOX θtan θ θ tan-1 ( )-π θ πHere θ is principal arguments of z.YA (a,b)OCASE Ia 0 , b 0θbaNYA(a,b)θ tan-1 ( ) ααOCASE IIa 0 and b 0tan-1 αθ (π - α )z a ibXbaXYA(a,b)bNπ- ααa

10CASE IIIa 0 and b 0tan-1 αθ - (π - α )CASE IVa 0 and b 0tan-1 αθ - α12. Application of argument of Complex numbersi)arg z Locus of z is half line OPii) arg ( z - 1) Locus of z is half line APiii) arg ( z – 1 3 i ) or arg [ z- (1 – 3 i ) Locus of z is half line BPOZ-(π-α)O- αAX

11iv) arg ( z 2 – i ) or ( z – ( -2 i ) ) Locus of z is half line CP13. Polar form of a Complex NumbersYGiven a Complex number z x i y z ( x2 y2 ) r (let)And arg z tan -1 θThen Polar form of zz r (cos θ i sin θ )14. Multiplication of Complex Numbers in Polar form:Given complex numbersz1 r 1 ( cos θ1 i sin θ1 )and z2 r 2 ( cos θ2 i sin θ2 )Then ,z1 z2 r 1 r 2 [ cos (θ1 θ2) i sin(θ1 θ2) ]Note : i) z1 z2 z1 z2 r 1 r 2ii) arg (z1 z2 ) θ1 θ2 arg z1 arg z2 K ( 2 π)K 0 if - π ( θ1 θ2 ) π-1 if (θ1 θ2 ) π 1 if ( θ1 θ2 ) -π15. Division of Complex Numbers In Polar Form: [ cos (θ1 – θ2 ) i sin ( θ1 - θ2)]P(z) z rθOryN

12 ; arg () θ1 - θ2θ ( arg z1 - arg z2 ) K ( 2 π)K 0 if - π θ1 - θ2 π-1if (θ1 - θ2 ) π 1 if ( θ1 - θ2 ) -π16. Square root of a Complex Number in Polar FormGiven Complex number z r ( cos θ i sin θ )Let the square root of z is w p ( cos α i sin α ) (i)[ p( cos α i sin α ) ] 2 r ( cos θ i sin θ )Or p2 ( cos 2α i sin 2 α ) r ( cos θ i sin θ )p2 rp rand 2 α θα i) arg wα andor- π if θ 0and π if θ 017. Exponential form a complex Number :Given a Complex number in Polar form z r ( cos θ i sin θ )Then exponential form z re iθ