Complex Numbers

lGonmilltuunm Algebra of GomplexNumbers,modulus and argument,triangularinequality, cube roots of unity.

A numberin theformof a + ib,wherea, b arerealnumbers andi = J-t is calleda complexnumber. A complexnumbercan also be definedas an orderedpair of real numbersa and b and may be writtenas (a, b), wherethe first numberdenotesthe real part and the secondnumberdenotesthe imaginarypart.lf z = a + ib, then the real partof z is denotedby Re (z) and the imaginarypart by lm(z).A complexnumberis saidto be purelyrealif lm(z)= 0, and is saidto be purely imaginary if Re(z)= 0. Thecomplexnumber0 = 0 + i0 is bothpurelyrealandpurelyimaginary. Two complexnumbersare said to be equalif and only if their real partsand imaginarypartsare se p a r a telyequal i.e.a+ib=c+idimplie s a = c a n d b = d . Ho we v e r, t h e re is n o o t h e rre la t i o n betweencomplexnumbersthatis of the typea + ib < (or >) c + id.

Remark: r

Clea rl y i 2 = -1 ,i 3 = i 2 .i = -i ,

i a = 1.In

general ,i 4n= 1, 14n+-i1,i 4n* 2 = -1 , i 4n+ 3--i

foran

integern.

A complexnumberz = x + iy, writtenas an orderedpair (x, y), can be represented by a pointP whoseCartesiancoordinates are (x, y) referredto axes OX and OY, usuallycalledthe real and the imaginaryaxes.The planeof OX and Oy is calledthe Arganddiagramor thecomplexplane.SincetheoriginO lieson both OX and OY, the corresponding complexnumberz = 0 is bothpurelyrealandpurelyimaginary.

Modulus and Argument of a Complex Number: we definemodulusof the complexnumberz = x + iy to be the realnumber ^'*y' by lzl.lt maybe notedthatlzl > 0 andlzl = 0 wouldimplythatz = 0.

and denote it

lf z = x + iy, thenangle0 givenby tan0 = I is saidto be the argument or amplitude of the complex X

nu m b e r zandisdenotedbyarg(z)ora mpln (zc).a s e o f x = 0 (wh e re y + o ), a rg (z )= + n l2 o r-n l 2 ']llltCc

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upony> 0ory< 0 and t h e c o mp le xn u mb e ris c a llepdu re ly ima g in alfryy'= g ( w h e r e d e p e n ding upon x > 0 orx < 0 and the complexnumberis calledpurely x+ 0), thenarg(z)= 0 orn depending of thecomplexnumber0 is notdefined' real.Theargument the We can definethe argumentof a complexnumberalso as any value of the 0 which satisfies

e= ofequations system "o" l;fr;,

sin0= ffix2 +Y'

then The argumentof a complexnumberis not unique.lf 0 is a argumentof a complexnumber, the satisfying of 0 value The of n. 2nn + 0 (n integer)is also argumentof z for variousvalues inequality - n < 0 < n is calledtheprincipal vatueof the argument. = = words Fromfigure1, we canseethatoP = Jx2 + y2 = lzl and lf 0 ZPoM, thentanO y/x' In other pointzfrom the originand arg(z)is the anglewhichOP fzl is the lengthof Op i.e.the distanceof makeswiththe Positivex-axis.

Trigonometric(or Potar)form of a ComplexNumber: LetOP = r, thenx = r cos0, andY = r sine + z=x + iy = r cos0 + ir sine = r ( cos0 + i sin0 ). Thisis knownas Trigonometric of 0' (or polar)formof a ComplexNumber.Herewe shouldtakethe principalvalue Forgeneralvaluesof theargument (wheren is an integer) z=rlcos ( 2nn+ 0) + isin ( 2nn+ 0)l Studentsshouldnotethatsometimescos0 + i sin 0 is, in short,writtenas cis(0)' e' Eulerrsformula:cose + isin 0 = ei

Remark: r

Methodof findingthe principalvalueof the argumentof a complexnumberz

Stepl: Findtan0=

= x + iy

tvl and this givesthe valueof 0 in the first

FI

quadrant. Step ||: Find the quadrantin which z lies,with the he|pof sign of X x and y co-ordinates. Then argument of z wilf be 0, n - e, e - n, and -0 lll: Step accordingas z lies in the first second,third or fourth quadrant

lllustration1. For z = tb - i, rinatheprincipalvaluearg(z). Solution: Herex = {3, y= -1

1l + tano= l- -=1 ? 0 = ; I ./3 |

I

b

valueof atgz= -L . (Sincez liesin thefourthquadrant) + principal

ffit-+

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RSM-91 1-P3-MA-Complex Number

Unimodular Complex Nu mher: A complexnumberz for which lzl = is'said to be complexnumber.Sincelzl 1, z lies on a unimodular circleof radius1 unitandcentre(0, 0). lI lzl= 1 + z= cos0 + i sin0, + 1lz= (cos0 + i sinefl = cos0 - i sinO.

Algebraic Operations with Complex Numbers:

I

(a + ib) + (c+id)= (a + c) + i ( b+d) Addition (a + ib) - (c+id)= (a - c) + i ( b - d) Subtraction Multiplication (a + ib) (c+id;= (ac - bd) + i ( ad + bc)

I

Division

I I

a+ib c. id _ (ac + bd)

(when at least one of c and d is non-zero)

_ (bc - ad) c2 +d2 c2 +d2

Geometricalmeaning of Addition and Subtraction: lel z1 =Xr + iYr and z2 = *, * iy2 be two complex numbersrepresentedby the points P1(x1,y1) and Pz(xz,yz)respectively. By definition21+22 shouldbe represented by the pointP(x1+xr,yt+ yz).Thispoint P is the vertexwhich completesthe parallelogram OP1PP2 with the line segmentsjoiningthe origin withPr and Pzas the adjacentsides = lz1+zzl=OP.

P(x1+x,.yt+yz)

P'z('xz,-Yz\ Also by definitionz,r- zz shouldbe represented by the point(xr- xz,yr - yz).Thispointis the vertexQ whichcompletesthe parallelogram OPjaPl with the linesegmentsjoiningthe originwith Pr and P'2 (wherethe pointP'2repr€sents -22;the point-22 c?n be obtainedby producingthe directedline PzO by fengthlz2)as the adjacentsides= lzt - z2l= OQ = PzPr.

Remarks: r

r

In any triangle,sum of any two sidesis greaterthanthe thirdsideand differenceof any two sldesis l'essthanthe thirdside;we have = 0. (i) hereequalityholdswhen arg(4122) l4l+lz2l>-14+zzl; (ii) llzrl)zrll3lzt- z2l;hereequalityholdswhen arg(4lz) = 0. In the parallelogram OP1PP2, the sum of the squaresof its sidesis equalto the sum of the squaresof itsdiagonals ; i.e. OP2+ P2P?= OPI +P,P2+PPl +P2O2 +l z, + zz 12+ | zt - zz f = 2( z'r12+ | =, lt ).

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GeometricalMeaningof Productand Division: isin01),z2-- r2(cos0z + isinOz)be complexnumbersrepresentedby Q1and Qz' Letzr= 11(cos01+

Constructionfor the Point Representingthe Productzlz2:

(i)

unity,so that OL Let L be the pointon OX whichrepresents = 1. Drawthe triangleOQzPdirectlysimilarto the triangle OLQrmakingZLOQr= 7Q2OPand ZOLQr= ZOQzPthe produclz1z2 ThenpointP represents Explanation: Dueto similartriangles oP =t? oP - oQz + op = rrrz . that is t,r 1 OQ, OL Also ZQzOP = /,LOQI = 0r 3 ZLOP = 0t + 0z Sincezp2 =r1r2{cos(01+ 0z)+ isin (01+ 0z)} , P representsz1z2'

Constructionfor the Point Representingthe Quotient4lz2:

(ai)

Drawthe triangleOQrPdirectlysimilarto the triangle oQ2L. the quotientzrlzz. ThenP represents Explanation: rt oP oQn oP " = = ctlon' Fromthe rastconstru oe2 o L k= 1 l- l

t-

|

Op = ]-11= li1-l also IQ,oL = 0r ond IQaOP= 02 lzzl lzzl :. IPOL = 0r - 0z

- or)\= ? - o,) * isin(0., OP(z)= 1(cos(01 .

22

l2 t

by P =1t . + numberrepresented z2

Remark: t

Snd lf z1= 11(Cos0r + isin 01), and Zz=r2(cos02+ isin 0z),thenZjzz= r.,r,ei(e'*e')

l t -tt ."(et-',). H enc e|l =' .'t= r l= \ r z : lz llzt r la n dH= ! =f2H, / r l+ 0 . 22

t I

12

e2l

lzrl

arg(z'12)= 0r + 0z= arg(zi + ary(7,2) ( z,\ argl a | = 0r -02= arg(21)-arg(22) \zz )

Square Rootof a ComPIexNumber: Letzl =xr + iyr be the givencomplexnumberand we haveto obtainits squareroot. Let x + iy = (xr + iyr)%= x' -f + 2ixy= Xr + iYr

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sX1 = x'-y'andy1 =2xy = ztl+ \.

> x2-l

v = ,1Yl

2

lf y1>0=x=+

x'- 4=^ , 4x'

| v rl

F;t;,

\l

2,

Y=+

I zrl-x, 2

(Re(2,)* |., l)

2 lfyl 0 .

r

Now let us considera circledescribedon a line.segment AB, (A(21),B(22))as its diameter. Let P(z) be any point on the circle.As the angle in the semicircleis nl2. IAPB = nl2 ( =n- z) .^ :+ afg -z) \zz

=

A(zr)

B(zz)

t-" t-=t * :=L =o is purelyimaginary = z-zz z -z z Z- Zz

+ (z - .r)(Z -2r) + (z - zr) (2 -21) = 0 r

Let z1 and z2 be two given complex numbers and z be any , ,:jLl= complexnumbersuch that,arg | o , *her" ae(0, ru). \z-zz) Then 'z' would lie on an arc of segmentof a circle oo 2122, containingangle cr.Clearlyif cr,e (0, nl2),2 would lie on the major arc (excluding the points 21 and z) and if a e (n12,7t).'z' would lie on the minor arc(excludingthe pointsz1 and z2).

Note: The sign of o determinesthe side of zl 22otl which the segmentlies. Thus a is positivein fig. (1) and negativein fig. (2)

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fis.( 1) A(zz) A(zz)

B(zr)

B(zr)

ris.(2)

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Let ABCD be a cyclic quadrilateral such that A(21),B(22),C(23)and D(z+)lie on a circle' Clearly

r

D(a

lA+lC=n (zo -2,,\

(=, -."\

*sl#,)+ars\-:---.

-

)=

n-

-=,\(._z_:z)=* _ aro ?o -21)\24 -4 \zz 24 -z't

>

)

.22

-23

zz-z't z+-zl

A(zt

is purely real.

ThuspointsA (zr),8(2 , ), C(z a )a n d D(z + )(t a k e n in o rd e r)wo u |d b e c o n c y c | i c i f (=o- =,)(., - zs) ,^ -Pur€lYr€al' (zz-z'r)\z+-zs) 'o given point: Equation of tangent to a given circle at a tangentat A(zr)'Let us takeany Let lz - zsl=r be the givencircleano we haveto obtainthe pointP(z)on thetangentlineat A(21)' Clearly IPAB= nl2

=-='' ispurelvimasinary. I=,n,, = iaa z o-z ' r -=.

' ng( ''\=o

-

)

'-='' * !:-1- =g zo -z'r zo - 4

(z- zi (4 -A) * G -4)Go - 4) = o

,. B1a) I I l

', az

=0 =(4 -2r)*2(=o -2 . )+ z ' , 4 -z t Z o+ 2 , 2 ,-4 z o -ztzo -Zzo =0 + z(4 -Z)*2(.o -zr)+2lz',12 = r, equationof the tangenl al z = Zt would be, ln particularif given circle is lzl =2r2 . A +zt=zlzl2

.

andz is an given complexnumbers ftlz-=.|=^ (^ e R*,r + 1),where21 end22 are lz-zzl arbitrarycomplexnumberthenz wouldlie on a circle'

b llal - lz2ll . lzt+ z2l?= (zt + zr)(2t + 2z)=lz.,l2+ ltl'*.,72+ c arg(z4z) = arg(4) + ary(22) . arg (z1lz2)= arg(z) - arg(22)

zzZt

De Moivre's Theorem :lf z= r(cos 0 + isin 0), and n is a positiveinteger,then 2kn+0 2krc+01 1t^ .,^f " . =0,1,2,.....n_1. 2"" =t"" lcos-+isin-"-l,k LNNJ The nth Root of Unity Let xbe the nth rootof unity.Then xn =1=cos2kn+isin2kn 2kn 2kr -1. 3 X = cos-"" + isin-'"" k = 0,1,2,....,n nn

(wherek isan integer)

Let cr= cos2-!+ isin?I.tnen the nth roots of unity are crt nn = - 1),i.e.the nth rootsof unityare 1,a,a2,.....an-1 (t 0,1,2,....,n . sum of the Roots 1+a+

n-1 4 n o'*....+on-1=1--o =0 =)I"or2kn=0

1-aftnft n

n .r and lrin?E=O

Thusthe sumof the rootsof unityis zero. Productof the Roots 1,a.a2.....crn-1=(-1)"(-1)= C1)n*, Concept of Rotation : lf 21,22,zs, erathe threeverticesof a triangleABC describedin the counterclockwise sense.then z q- z t O Q, = _(COSc[ -__:_________.j_

zz-zt

OP'

+ tStnc)

CA

BA

,^ l z,-znl i^ e'" = :___l____-l-r. e'u

lzz-ztl Note that arg (zs- z.) - arg(22- zi = o is the anglethrough which OP must be rotatedin the anti-clock-wise directionso that it becomesparalleltoOQ.

,,.,P*,

GeometricalApplications: Condition for Collinearity lf there are three real numbers(otherthan0) l, m and n such that lz1+ fizz* 1123 = Q and l+ m + n = 0 then complexnumberszl,22etldz2willbe collinear. Equation of a Straight Line . Equation of a straight line with the help of rotation formula: Let A(21) and B(22)be any two points lying on any line and we have to obtainthe equation of this line.For this purposelet us take any point C(z) on this line / . Z-2" {z-zt\ ^ or r-; z-2, '==---:-. slnceargl-l=u zz-zt Zz-Zt \zz-zt,) jrlltcc

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E=

Generalequationof the line: (1)we get za+-za+b= 0 ' Fromequation and ib = Z z 2 -2 . 2 r,b e R wheree=i(4-l) plane' Thisis the generalequationof a linein the complex Slopeof a givenline: ne(a) =lI za+-za+a = 0 is thegivenlinethenitsslope t("') b = 0 is zd +-za+ l' = 0 Equationof a lineparallelto the line za +-za+ (wherel, is a realnumber)' - _7a+ i1.= 0 . to the line za +-za+b= 0 is za Equationof a lineperpendicular (wherel, is a realnumber)' Equationof PerpendicularBisector: bisector' Letthe line'L',be it's perpendicular considera linesegmentjoiningA(21)and B(22)' = PB' lf P(z)be any pointon the'L',we havePA =lz-z2l =lz-zl = + z(2, - a\ +z(2, - zt)+ 2.4 - zr2, o Distanceof a given point from a given line: givenpointbe z" thendistanceof z" fromthis Letthe givenlinebe za+-za+b=0 , andthe lz^a+z^a+ol lineis--r:#' zPl .l

Equationof a Circle its centreat zsis lz-zsl= r' Equationof a circleof radiusr and having _ 12' = _^) = f2> fi. + az +62+b = 0, where_ ? = 7o andb zo4 > |z_zo|,= f * ( z - zi (z plane' the generalequationof a circle in the complex It represents oE quationofacircledes c rib e d o n a |in e s e g me n t A B , (A (2 1 ), 8 (2 2 ))a s d i a m e t e r i s (z-z)(2-2r)*(t- zr ) (7 -4 )= o ' oLetzlA fidzzbetwogiv e n c o mp |e x n u mb e rs a n d z b e a a n y c o mp |e x n u m b e r s u c h t h a t , (lieon thearcof a circle' - \ arol'-" l=a,where a e (0,n).Then'z'will \z-zz ) Then o

circle' that A(21)'B(22)'c(4) and D(a) lie on a Let ABCD be a cyclic quadrilateralsuch (zo -2.).(zr-zt) is purelyreat. (zr-2.)(zo-2")

Some lmportant Resultto Remember z. 22'zs ts -^--,^-. numbersz1' points representedby complex^..*hara r The trianglewhose verticesare the

e q u i l a te razz-zg l i fl ++-zs-zt .= +=0i'2't-72 €' ' itzl+z?' + z!:zlzz+ 2223+zsz1' C|z-21|+|z_:,12|=l,,representsane||ipseif|21_22| 3, thendeterminethe leastvalueof l,tzl* -l and the corresponding

Solufion;

rzr>3 vt-ttas l'.:l-l'',-*l= I'o-#l=

Let f(x) = 1-

L forx > 3 X

f'(x)=1*Iro I

I

II I

I

I

xfunction+ f(x).1nat x =3 increasing is + f(x)

l -* 1 l -=l ' - -, 1l = - 1 -= 8 . - "3-5

l' 21,''llzllal,o*

3'

I ffi"

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RSM-91 1-P3-MA-Complex Number

To find z: Let z = r eio,where r = 3.

=lr"''* 1"-''l=l9.o.e* 9,r,nrl =I rnen lz* 1l r 3 zl t I

-

I

I

13

|

/ro\2 ,^ /g)2 .,^ / a )2 / t o )2 l -l c o s -U + l -l s l n -0 = l -l = l -l cos'e =l,9)' cos2o+ cos' e= o \.3/ \.3i \31 \3/ \3/

+ 0=

= - 3i.

2'2

Problem 6. ABCD is a rhombus.lts diagonalsAC and BD intersectat the point M and satisfyBD=2AC. ItspointsDandMrepres e n t t h e c o mp le x n u mb e rs l+ ia n d 2 -ire s p e c t iv e l y . F i n d t h e by A. complexnumberrepresented Solution: LetA be (x,y) lt is giventhatBD = 2AC+ MD = 2AM to AM AlsoDM is perpendicular > (1-2)'+ (1+1)2= 4 l$-2)2+ (y+1)21...(1)

Y * 1 . 1 * l= - 1 + 2 (y+ 1 ) = x- 2 " n ox-2 1-2 s + 1 )2 =ll4 W ithx -2=2(y + 1),(1)giv e (y =y= -112,-312= x=3, 1 + A representsz = 3 - i12,or 1-3i12 Alternative Sol. MD = 2AMandAM IDM i.e. ZAMD= nl2

= J:-(2:)-=

(1+i)-(2-i)

3Y MD

"tT

= +it2> z-(2- i1= 1

iu'+ 2i\ > z= 3-il2or 1-3i12.

Problem 7. Provethe followinginequalitiesgeometricallyand analytically: tl

zl @ l_Z_4 0 ) (C) lztl = lZ t Solution:

ztzt =12.,l'2.,-1= arg(zi|)= ary(4) = arg(22) = zz= kz',-11k> 01 7 ' 2.= Hence(A)is the correctanswer. Problem 2. ff a > 0, a e R, z= a + 2i andzlzl - az + 1 = 0 then (B) z is always a negativereal number (A) z is always a positiverealnumber (D) sucha complexz doesnot exist. (C) z is purely imaginary number

Solution: parts,we get imaginary andcomparing Puttingz = a+ 2i in thegivenequation . a2+ 4 = a2,whichis notpossible Hence(D) is the correctanswer. Problem 3. tf a is the angtewhicheach sideof a regularpolygonof n srdessubfendsat its centre,then 1 + cosa+ cos2a+ cos3a... + cos(n-l)ais equalto (B) 0 (A) n (D) noneof these (c) 1 Solution: n-1

n-l

r=0

r=0

irl

=0 = = f cosro nef e; sumofthen rootsof unity Hence(B) is the correctanswer. Problem4. lf the imaginary paft of the expression ' -,1* "t' o" zero, then locusof z is eo' z -1 (B) a parabola (A) a straight line parallelto x-axis (D) none of these (C) a circle of radius 1

Solution: Let U=

z-1 eoi

7

et' z -1

+_ - =- .

1 u

Giventhat imaginaryPartof u +

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RSM-91 1-P3-MA-Complex Number

f" *1)-(o *1)=o ( u /\. u/ (u-u).H: o + t,-ol(''-*) =t t

-

,t-#=o =lul2=1

.t2

= 1 = lz-11= 1,whichisacircle. = l=l le''l Hence(C) is the correctanswer. Problem 5. lf 4 , 22, 23 are the vertices of an equilateraltriangleinscribedin the circlelz | = t then the area of the triangle having zt, - 22, 23, as ifs yerticesis lo

1411' ,2

@9 4

,'.r3J3

(D) noneof these

l"/

4

B (zz)

Solution: DE= BD- BE = 2*312=112 1 Areaof AACD= DE T AC 2

= 1,1 Ja =€ .

224 Hence(B) is the correctanswer.

D(-zz)

Problem 6.

fieon Thecomplexnumbersz = x + iy which satisfythe eouationlzriLl=1 ' lz+ 1il

(A) the x-axis (C) a circlepassingthroughthe origin

(B) the straightliney - 5 (D) Noneof these

Solutlon: lz - 5il=1 =lz-sil =lz + 5 il l-l lz + 5il + z wouldlie on the rightbisectorof the line segmentconnectingthe points5i and - 5i Thusz wouldlie on thex-axis. Hence(A) is the correctanswer. Problem 7. the regiongivenby Theinequalitylz-l l< f z-2f represents (B) Re (z) . O (A) Re (z) > 0 > (D) Noneof these (C) Re (z) 2 Solution: lz - 4l < | z-21+ | z- 4l' < | z - 2l' + ( z-Q Q -\ < ( z-2)(z -2) + z2 - 4(z+2) +16< z2 - 2 (z + 2 ) + 4 G|lltGC Ltd., FIITJEE Hoase,2g-+ Kalu 5arai, garvapriya vihaD New Delhl -tto

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RSM-9 1 1-P3-MA-ComPlex Number

36 (z) > 3 + 2(z +2) > 12 = 4Re(z) > 12+ Re

Hence(D) is the correctanswer' Problem 8. =ft' the equationof a circle,wherezF 2+3i' z2=4+3i lf the equation lz-zl2 +lz-2212 represents of k is are the extremitiesof a diameter'thenthe value (B) 4 (A) 1/ 4 (D) Noneof these (c)

2

Sofufion; As zr and 22 ate the extremitiesof the diameter'

lz- =rt'*V'lr)ii-=V',- it'*

k = lz,- zrl'= | 2 +3i- 4 - 3if = l2l2= 4

-Hence(B) is the correctanswer' Problem9. tf lzl = min {lz - llt,lz + 1V'then (A )lz+zl=1 2 (C)lz * zl-- 1

(B )z + 2 = 1 (D) noneof these

Solufion; + - - 1l for R. (z)> 0 lzl = min{lz - 11,lz 1l]r lz B utlzl= lz -1 1=+R" (4 = ; ' = + 1l' for R" (z) 0 a n dr= 7 , 2 , 3 , . . . , n -2 .

Solution: Gi v e n z n -' + t z n -z+ z n -3+ .....+ z+ 1 = 0 = (z-1)(= n-, 4 2 n = 1 = e i 2 n ' (re N )

* zn-2 + ....+ z+ 1) (z* 1)

i2nn

=z=

g n i2x

=+Therootsareen,€ Zt

i4n

i6r

i(2n-2)n

n, € n,.....,g

n

i2nn

whichisaG.p.withcommonratioeT.

i2n

-1 n Also a t' = d" , whichis a constant. Hence(C) and (D)are the correctanswers.

Problem 12. lf f(x) and g(x) are two polynomialssuch that the potynomial h (x )= x f f )+ f

divisibleby t' +x +1, then (A) f(1) =s(1)

g ( x 6 )i s

(B) f(1) = - s( 1)

(C) f(1) = g(1) *0

(D) f(l) = -g(1) *0

tolutian: Rootsof x2+x +1 = 0 are complexcuberootsof unity, so h(w)= h(w2)=0 =wf(1) + # g(1) = 0 and # (l) + wg (1 )= 0 = (1) = g(J)= 0. Hence(A) and (B)arethe correctanswers. Prablem 13. One vertex of the triangleof maximum area that can be inscribed in the curue lz - 2 il =2,is 2 +2i, remainingveriices is / are (A) -1+i( 2 +Ji 1 (B)-1- i( 2 +Ji 1 _1+ (c) i( 2_ $ ) ( D) - 1- i( 2- Ji) Solutlon: Clearlythe inscriedtriangleis equilateral.

+

z z -z o= zt-zo

z z -z o=

"'?, zt -zo

"-'?

+ zz= -1 +i(2* Jd ) andz3= -1 + i(2- Jg I Hence(A)and (C) arethe correctanswers. Prablem 14.

jllltcc

Zz

Ltd., FrrlJEE Hoase,2g-A, Kalu Safai, Sauapfiya vihar, New Dethi -tto

z1(2+2i)

076, ph 26575949, 26569492 Fax 265tgg42

F

suchthatz2* az* a' = 0, then numberand'a'be a realparameter Let,z,be a comptex (B) Iocus of z is a circle (A) tocusof z is a pairof straightlines

=t+ (c)arg(z)

(D)lzl=lal .

Solution: '"-'r,r'is nonreal root of cube unity) ) + az+ a2= 0 = z= ar, ar" ( where or arg(a)+ arg(o2) arg(a)+ arg(CI) = locusof z is a pairof straightlinesand arg(z)= + arg(z)= tT

3

+ lzl = lal' also,lzl= lallrrrlor lal lr''r2l Hence(A),(C) and (D) arethe correctanswers' Problem15. Roofsof the equationf -1 = 0' n el, (B) Iie on a circle' (A) are collinear non-collinear' (D) (C) form a regularPolYgonof unit circumradius. are Solution:

complex where* = cos?+isin?I. The distanceof the clearly,rootsare 1, a", ..2,...on-1, pointslie on a circle' by theserootsfromoriginis 1 i'e' all these numbersrepresented =Theyarenon-co|linear+Theyformaregularpo|ygonofunitcircum.radius. Hence(B),(C) and (D) arethe correctanswers'

ReadthefollowingcomprehensioncarefullyandanswerthefollowingProblems(1t[-16 definedbelow LetA, B, C be threesetsof complexnumbersas A = {z: lmz> 1}

B = l r:l z -2- il= 3 ) - i) z ) = J 2'\ c ={=:R e((1

Problem16. Thenumberof elementsin the setA n B n C is @)1 (A )o @)* (C)2 Solution: Y = 1-intheArgandplane' i = Setof pointson and abovethe^line + (y - 1)2= !2' (x 2)' g = S"t of pointson tne circte = C = Re(1-i)z=Re ((1- i) (x + iy ) + x + y J ' Hencein.l'e n C) = hasonlyone pointof intersection' Hence(B) is the correctanswer' Problem17. +lz - s - il2 tiesbetween Letz be anYPointin A n B n C' Then' lz + 1- il2 (B)30and34 (A) 25 and 29 (D)40and44 (C) 35 and 39 Solution: ttttot

I

RSM-911-P3-MA-Complex Number

39

ThepointsC 1,1) and(5, 1)aretheextremities of a diameter of thegivencircle. Hencelz + 1 -il2 + lz- 5 - i; 2= 3 6 . Hence(C) is the correctanswer. Problem18. Let z be any point in A n B n C and let w be any point satisfyinglw-2-ll 1, find all the complexnumbersz that satis$ the equation z+clz+11+i=0.

4.

in a circleof AssumethatAr(i = 1, 2,.....n) are the verticesof a regularpolygoninscribed radiusunity.Find

t I

I I

i

(i) (ii) 5.

|'*........* lR.',\|', In,n,|'* lA.,A, lA'n.| . In'n,I In'n.| .........

andminimumpossible valuesof lzl satisfying Findthecomplexnumberz withmaximum

(b)1..11= 3

r t"l lr*11= lzl 6.

7.

+ ... + cosgn= 2 where00,0r, + zn- 1cosQr Showthat all the rootsof the equationzncos0e e R lieoutsidethe drclelzl = 112. 02...0n 21,zz, zs era three non-zerocomplexnumberssuch that 22* 21,and a = lzil, b = lzrl, labcl r. , 2

6 = ;2.;.ffilb I lc

8.

c

arg al=0,thenshowthatary.z:= - I -

a

Ul

22

z::+l

\ z z -z t )

Prove linescut the circlelzl = r in pointsa, b, c and d respectively. Two differentnon-parallel thattheselinesmeetin the pointz givenoy. ={l!}{}f

L

lfnisanoddintegergre a t e rt h a n 3 , b u t n o t a mu lt ip le o f 3 . P ro v e t h a t x 3 + f + x i s a f a c t o r o f (x+1)n - xn-1.

10.

= X*1" whereA, B, c, ...,d, b, c,... andl. .* Showthat the equatio" *.*. are realnumbers,cannot havecomplexroots.

11 .

lf lz + -l = a, wherez is a complexnumber,find the leastand the greatestvaluesof lzl and lzl complexnumbers.Alsofind a for whichthe greatestand the leastvalues the corresponding complexnumbers. of lzl are equalandthe corresponding

I

JllltCC

1l

Ltd., FIITJEE House,2g-+ Kalu Sarai, Saruapdya mhar, ,lew Delhi -tlo

076, Ph 26515949, 26569493, Fax 26513942

r

--=

RSM-91 1-P3-MA-ComPlex Number

12.

Provethat,for integralvaluesof n > 1, all the rootsof the equation n,. nzn= 1 + z* z' + ...+znliewithinthe circlelrl' = n-1

13.

the complexnumberszt, zz, ..., 26 respectivelyare the lf pointsAr, Az,...,A'6representing if ze be the complexnumberrepresentingthe centroid and hexagon verticesof a regular thenprovethatzl+z|+....+zl =67f,. thehexagon

at points havingvertices trianste an equitaterat 14. consider ^(ft"'J cl1.i

tJ \/ g

thenfindthevalueol AP2+ BP2+ CP2. l. rt e1r1is anypointon itsincircle,

I

15(i). Findthevalueof fq4l' lP i-1 , / (ii).

t[#"*J

"zmicotl(o).

Findthe rootscommonto the equationsxu- xt + x' - 1 = 0, xa= 1'

talllcc

Ltd., FrIrtEE House,2g-A, Kalu Sarai, Saruapriya Vihar, New Delhi -tto

076' Ph 26515941 26569493' Fax 26573942

"ro

RSM-91 1-P3-MA-Complex Number

45

Objective : Level-| 1.

n,theargument Foranyinteger ofz =

,. ffi

2.

(A);

F);

Q);

D\

2:T 3

lf cr,F,y arethecuberootsof p, (p 3/a @) pl < 1t2 (D) noneof these (C) 112< lzl < 314

23.

then z-31}, wherez is a complexnumber, lf lz-21=min{lz-11,1 = (B) (A) Re(z) 172 Re(z)= 572 (c) Re (') . {:, :} t2' 2l

24.

( lzl2 +2lzl ^ \ lf a comprex numberx satisfiesl,cetrl-z

lfrF;fiJ.

pointrepresented by z is (A)lzl = 5 (C)lzl>1 25.

(D)noneor these

o,tnen locus/ region ofthe

(B) lzl < 5 (D)2 < lz 0 40.

41.

(B) l' (z) = g (D) Re (z) > 0, l'' (z) < 0

The locus of Z which lies in shaded regionis best representedby (A)z lz + 1l > 2, larg(z+ 1)l < nA (B)z lz - 1l > 2, larg(z- 1)l< nl4 (c) z lz + 1l 7=0andz5.to1l=1are ( A)o=f:i,z=i (B)c6=*1,2=i = (C) ro *i,z= +i (D) none ofthese

n''itee

LH,, FTfitEE Uou*,29-+

Katu gnl,

&napriya

til tl

Thenumberof complexnumberssatisfying lz+ 2l + lz-21= 8 and lz- l1+ lz+ 1l= 2is (A) 4 (B)2

(c)0

il

whar, Itew pethl -tto

ot6, m 265t5g4g

26569493, Fax 26513942

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i

il

RSM-9 1 1-P3-MA'GornPlex

44.

45.

Thegreatestpositiveargumentofcomplexnumbersatisfying|241=Re(z)is

(A);

@) +

p);

(D);

B, C of the tr4BC are e'0' tD' The complex number associated with the vertices A, of unity and cosO > Re(rrr))'tn:l respectively ( where rrr, rrr are the complex cube roots

of the the circumcircle complexnumberof the pointwhereanglebisectorof A meets is (B) -e" (A) - e't (D) rrr+ ro (C) roro

46.

lf k + lk + 221=1212(k e R-) then argumentof z is

(A) 0 (C) nt2

47.

'2nr withthe pointA, on Argandplane, (t < r < n) be the complexno associated Let.,, = ",? pointB is (2,0) thenthevalueof BA1'B& ' B& """ BA"is equalto

(A) n

(C) 2n 48.

49.

50.

(B) " (D) noneof these

[3]i,_i

(1 + i)x + (1 - i)y = lf x and y are complexnumbers,then the systemof equations 2ix+2y=1+ih a s (B)no solution (A)uniquesolution (D)noneof these (C) infinitenumberof solutions tzt = (1 - t)22,where21,z2 Statement1:Locusof p(z) is a straightlinein the equationz in an argandplane' the pointsA and B respectively because m : n is collinearwith the Statement2: A pointdividingthe two fixed pointin the ratio throughthe givenPoints. correctexplanation (A) Statement-1 is iiue, Statement-2 is True; Statement-2is a Statement-1 is NoT a correctexplanation is True;Statement-2 iril Et"i"*"nt -1 is True,Statement-2 Statement-1 -2 is False (C) Statement-1 is True,Statement -2 is True (OiStat"tent -1 is False,Statement valueof lz1-z2lis22' 1: lf lzrl= 12andlzz-F + ZJ6 i)l= 3; thenmaximum statement because joiningcentreof two.circles' statement2: Maximumvalueof lzt - zzloccursalongthe line is a correctexplanation (A) Statement-1 is irue, Statementl2 is True; Slatement-2 Statement-1 a correctexplanation (B) Statement-1 is True,statement-2 is True;statement-2is NoT Statement-1 (C) Statement-1 is True,Statement-2 is False -2 is True iOiSt"t"t"nt -1 is False,Statement

ffi*"ro""o+o*"pnr"ro",n"*

tttttot

51

RSM-91 1-P3-MA-Complex Number

Level- ll 1.

2.

lf lzl = lz2l andarg(21)+arce2)= nl2 , lhen (A)zp2is purelyreal (c) (4+7"'12is PurelYimaginary

by the complexnumber2 - i is rotatedaboutthe originthroughan The pointrepresented of thepointis anglel . Thenewposition z (A )1+2i (c)1-2i

3.

(B)zp2 is purelyimaginary = -nl2 + ary(7,2'1) (D) arg(21-1)

(B )-1 -2 i (D)2 + i

It z = -2 +2Jdt, then z2n+ 22nzn + 24"may be equal to

(B) 0 (D) noneof these

(A) 2" (c) 3.24" '. ., 5 \ .l - | ll n+sln -n12 -+cos

4.

5.

Thevalueof 169e\ (A ) 119-120i (c) 119+ 120i

13

tsi is (B ) -i(1 2 0+ 1 1 e i) (D) noneof these

o) then n l1f:=21= k, (,.1,22+ '

lz,,z+zrl (A) for k = 1 locusof z is a straightline (B) for k e {1,0} z lieson a circle (C) for k = 0 z represents a point joining2 ana -2 bisector of thelinesegment (D) for k = 1, z lieson the perpendicular

zj

21

6.

lf z1 = ar + ibr and z2 = a, * ib2are complexnumberssuch that lzi = 1, lzzl = 2 and t3Z lnd lr,2=2b1+ibzsatisfy 0, thenthe pairof complexnumbers(Dr= ?r + Re(2.122)= (A) lroll= 1 =0 (C) Re(rrr1rrr2)

7.

=2 (B)lcozl =2 (D) lm(c,r1ro2)

the complexnumber21on the curvelzl = 2, pairof tangentsare lf froma pointP representing = drawnto the curvelzl 1, meetingat pointQ(zz)and R(4), then 21+22+23 willlieon thecurvelzl = '1 (A)compfex number 3

n B\(!*l.1lf1*a*-1_l= ' ' 22 zs ) [zr z2 zs )\zt

*(7)=+

(c)

of APQRwill coincide (D)orthocentre andcircumcentre 8.

Zc

of thecomplexnumber -l s lf lz1+ zzl= l4l + lzrl,thenoneof thevalueof arguments z1 (A) 0 (C)2n

fl,ltcc

(B)n (D)3n

House/2g-+ Katu sani, saryaprlya whar, New Delhi -tto Ltd., FTTTJEE

oto Ph 26515949, 2656949'

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the verticesof a

za,laken in that order, represent 21,22,23, numbers lf thecomPlex then (B) lz1- z2l= lzz- zsl (A ) z 1 + 2 3 = z 2 t 74

3=9-is G) ' zz-z+

10.

11.

12.

(D) noneof these

puretyimaginary

, , , -z r1 and-2.,canbe markedon , 1 1iz n =',= !(Ji+ i) , t h e n2 1 i2 2' (B)semicircle (A) circle (D) 3/4of a circle (C)quartercircle (whereall are centredat origin) Complexnumberscan be consideredas vectors in case of (B) sum (A)modulus (D) argument (C) difference = by pointson the circlelzl = 1 and lzl let 2.1,z2 be two comPlexnumbersrepresented then respectivelY (B) min lzl -z2l = 1 (A)maxl2z1+7r1=4 I rl (D) noneof these

(c)lzz*-l< 3 | I '1

,

13.

14.

r

rrD

-rh Forn as a Positiveinteger'(t + J s ) + (1 -r/ 3 ) is (B) alwaYsirrational (A) alwaysrational (D) noneof these (C) alwaysinteger

lf z= x+ iythentheequation l?il=

a circlewhen m represents

(A) m = 1/2 (C)m=2 15.

Fourpointsinthecomp|exp|anearerepresentedbythecomp|exnumbersZ1,22,23,2 the verticesof a order such thatz1 | 23= zz + z; ' Then these points are

(A)parallelogram (C)square 16.

(c) - Jj

I

(B) rectangle (D) quadrilateral

equation lf iz3+ 22-z + i = 0 then complexnumberz satisfyingthe (A )z = i

17.

(B )m= t (D)m= 3

lf (1 + i)z= (1 - i)Z, then

(B )z =

t: a/l

(D)z = J_,

(A)z lieson a straightline

3n (B)for allsuchz+ 0, arg(z)= 4

(C)all suchz aregivenby z= t(1- i),t e R

(D)z mustbe PurelYreal n -t^O OtO Ph 2651594'

26569494 Fax 26513942

RSM-9 1 1-P3-MA-Complex Number

18.

53

Let two distinctcomplexnumbersz1 and z2 both satisfythe equationz+z=2lz_11

and

arg(zr-zr)=1, th"n

4 (A) z1 and z2lie on a parabola (c) lm(21 + zz) = 2

19.

(B) Re(21 + z2) = 2 1D)no such zl andz2 exist

lf z1 + 7, 123= A,z1 + Z2rD* Zylz = B,zt * Zz1p2 * Zg1a=C, whefe 1, a, a2 afe the cube fOots

of unity,then (A )A + B +C=3zr

(B )A B C= z 1 z 2 z 3

(Q z?+z1r+z1r=Jz,,zrz, 20.

(D)lAl,+lel,+lcl2=s(l=rl, *l=rl,*lrrlr)

lf z.randz2?raanytwocomplexnumbers satisfuing lzt + zzl,

lz, _z2lthen

(A)Re[1) > o

re) Re[:l) < o

(c) ,m (+),,

@-;.*n(+).;

Numerical Based: 1.

lf,f(x)and g(x)are two polynomials suchthat polynomial h(x)= xf(x3)* xrg(xu)is divisibleby x'+ x + 1, thenthevalueof g(1)is _

2.

lf thereflectionof thelineaz + a 2 = 0 in t h e re a la x is is 6 z + b Z = 0 , t h e n t h e v a lu e o tf h e part imaginary of thecomplexnumberw = 6-a is

Comprehension: Gomprehension - l Readthe followingwriteup carefully: In argandplanelzl representthe distanceof a pointzfrom the origin.In generallzt-zzlrepresent the distancebetweentwo points\ and22.Alsofor a generalmovingpointz in argandplane, if arg(z) = 0, then7 = 1zlei9, whereeio= cosO+ i sinO Now answerthe following questions(1 - 3): 1.

Theequationlz-zl+lz-z2l= 1 0if 2 1 = J + 4 ia n dz z = -3 -4 ire p re s e n t s (A) pointcircle (B)orderedpair(0, 0) (C)ellipse (D)noneof these

2.

llz - zrl - lz - zrll = t, wheret is a realparameter alwaysrepresents (A) ellipse (B) hyperbola (C) circle (D)noneof these

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Ltd,, FIIDEE House.2g-A, KaIu garar)

Uihar, New Dellri -170 OtO ph 265t994,

2656949Z Fax 268t3942

RSM-9 1 1-P3-MA-ComPlex

(-

| \l orzis thenlocus tflz- (3+ 2i)l=lzcoslt-"*'.,}1,

3.

(B) parabola (D) hyperbola

(A) circle (C) elliPse Comprehension- ll

Readthe following writeup carefully: (k = 0, 1,2, ...,n - 1),so thatforn odd Then nthrootsof unityarecosZlt*1rin?!n, n-1

2kn Zr xn- 1 = (x- 1 )I I[ " - cos- - r sr n k=1 \ n-1 Zr

T)(--"o'?!"*'''n4) 1

-*-n r2=(*,,r - - 1,r ) l][**i- "".+ .) o r xn r2

x = co s 20+ i s i n20'w e wr itins andhence

n-1

n-1

z

r 2kn) =zz sineflicos2e-cos:-J,(nodd)' sinno

q

k=1'

n-2 i, 2kn\ z sin20[[cosZe - cos-.1 ' = Z sinnO n even for Similarly

Nowanswerthefollowingquestions(4- 6): Thevalueot sinfsinf sinf isequalto 4.

(A) I

(c)+ t

5.

(D)

*

efsin' z0- sin2;Xt' ^' 0- sin2?) ' """kisequalto l fsi n 5 0 =ksin (B )1 6 (D)- 32

(A)- 16 (c) 32 6.

@+

I"or4.orf lf cos-cosTcos-

=r"o,*"o'{"o'f -uurTws

(A )-1

(c)2

g ''

thenk isequalto (B) 1 (D)-2

- lll Comprehension Readthe following writeup carefully:

(suchthat nonumbersi:1,22 ?lfid23respectively A, B, C are the pointsrepresenting 11":9Tpl.* = = on the complexplaceand lztl lz2l lzsl' "i#quarl utt"to'

RSM-911-P3-MA-Complex Number

55

Now answer the following questions (7 - 9):

7.

lf the altitudeat the vertexA of MBC meetsthe circumcircle againat P, then complex pointP is numberrepresenting (A1-zzzs z^

@) -z lL z3

p) -3zz

1o1-Q'

z2

* =t) z1

8.

Thelocusof a pointQ (z)whichtouches thecircumcircle of AABCandthe line z+Z-2=0 (giventhatlz.rl= lzzl= lz3l= 1) is (A ) (z -2)=a(z+2)2 ( e) ( z - 2) ' * a( z+2)=0 (C)A rg(z)=2nn,nel (D)A rg(z -1 ) = 2 n n ,n e I

9.

) ! 1 a n dlz . , l= lz z l= l4 l= (2.,)=t,A rg(2 2 -) ;=, A r g ( 2 3 = LetA rg rt

r P (z) b e a n y

pointon incircle of triangle ABC,thenAP2+ BP2+ CP2is equalto (A) 1 (B)3 (c) 5 (D)7

NlatchtheColumn Match the

(A)

List I lf 21,22,zs a,rathe verticesof an equilateraltrianglewith (i) +21+zZ) centroid zs,thevatueof (z? " ['6)

List * tl 0

lf the complexnumberz satisfythe equation(i - z) (1 + 2i) + (ii) ( -iz\(3 -4i) = 1 + Ti.the nz + 2 + z z is e o u atlo

7

lf lz - il

B(i). (ii). (iii). (iv). (v). (vi).

A |Ipointsto wa rd s t h e rig h t o f x = le x c e p t t h e p o in t (2 ' 0 ) = All pointsinsidethe circlelzl S Allreal and purelyimaginarynumber (1' 0) Exteriorto a circleof radius 10 withcentre Regionbet we e n t h e t wo c o n c e n t ric c irc |e s o f ra d iu s la n d 4 w i t h c e n t r e ( 2 , 0 ) . centre(0' 0) n"iion outsidethe circleof radiusll2wilh 1r

L

_nt2

$),, 11. _L(rriJd)2,.1(.t

12.

$i.nTa

14.

a+b+c=0

4 y 2 + 4 x + s e c22

13. 15.

=o'

lz2l>1

Level'll 3.

-"'x":12:!-i r: lf c = 1,2= -1 - i ; lf 1 < c < ' ! 2 ' r= --G , _ 1 )

5.

2n ,_\_ ttJs \a)z= ---;)/ \

11.

l zl ^'" =-- - 2

4(i).

; no sotutionforc, J2

(ii)'

n

(b )

+ -4 i, t i

*-!*^'li ,=r^,=4=,2= !l!I+=J,,,n ft * *u' '.= r(E --[ z ) - "no

trueforall reala > 0;z= !2i ' 5 14. 15(i). 1

Ju'lcc Ltd,, HtrtEe aour.,zffif,En"priv"

(ii).

1 ,-1

vihar, NewDethi-770 oto Ph 2

57

RSM-91 1-P3-MA*Complex Number

Ob"iective:

1. 4. 7. 10. 13. 16. 19. 22. 25. 28. 31. 34. 37. 40. 43. 46. 49.

A B

A

2. 5. B. 11. 14. 17. 20. 23. 26. 29. 32. 35. 38. 41. 44. 47. 50.

B ,C,D A ,B A ,B ,C,D A ,D A ,C A ,C,D A ,D

2. 5. 8. 11 14 17 20

c B

c D B A

c A

c A A A

c c

c

3. 6. 9. 12. 15. 18. 21. 24. 27. 30. 33. 36. 39. 42. 45. 48.

D A A

c D A

c B

c A B A

c D B A

A B A D

c B B B B D D A B

c D

c

Level- ll 1. 4. 7. 10. 13. 16. 19.

A,B A , B , C, D A,C A,B,C A,B,D A,B,C A,D

3. 6. 9. 12. 15. 18.

B,C A,B,G,D A,B,C A,B,C A,B,C A,C

Based Numerical 1.

2.

0

: Comprehension 1. 4.C 7.

D A

2. 5. 8.

D B D

(i) (iv)

(c) (c)

3. 6. 9.

B A

(D) (D)

(iv) (iii)

c

['latchttreCalumn 1. 2.

(A) (A)

(iii) (i)

(B) (B)

(ii) (ii)

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