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If Z Square 1 Z 2 1 Then Z Lies On

Solution: Given |z2 – 1| = |z|2 + 1 Put z = x+iy |( x+iy)2 –1| = |x+iy|2 + 1 |( x+iy)2 –1| = |x+iy|2 + 1 Taking modulus and squaring... View Article

If Z X Plus Iy And Z 2 I Z 3 I Then Locus Of Z Is

Solution: Given |z - 2 + i| = |z - 3 - i| z = x + iy |x+iy - 2 + i| = |x+iy - 3 - i| |(x-2)+(y+1)i| = |(x-3)+(y-1)i| Taking modulus and... View Article

If Arg Z A Pi By 4 Where A Belongs To R Then The Locus Of Z Belongs

Solution: Given arg (z-a) = π/4 arg(x+iy-a) = π/4 arg((x-a)+iy) = π/4 tan -1(y/x-a) = π/4 y/(x-a) = tan π/4 y/(x-a) = 1 y = x-a x-y= a,... View Article

The Locus Of Z Given By Mod Z 1 By Z I 1 Is

Solution: Given |(z-1)/(z-i)| = 1 |(z-1)|= |(z-i)| Put z = x+iy |(x+iy-1)|= |(x+iy-i)| |(x-1)+iy|= |(x+(y-1)i)| Taking modulus and... View Article

The Conjugate Of A Complex Number Is 1 By I 1 Then That Complex Number Is

Solution: Let z = 1/(i-1) = 1/(-1+i) Conjugate of z = 1/(-1-i) = -1/(1+i) Hence option (4) is the answer.... View Article

A Point Z Moves On Argand Diagram In Such A Way That Z 3i 2 Then Its Locus Will Be

Solution: Given |z - 3i| = 2 Put z = x+iy |x+iy - 3i| = 2 |x+(y-3)i| = 2 Taking modulus and squaring both sides, we get x2+(y-3)2 = 4... View Article

If Z Lambda Plus 3i Root5 Lambda 2 Then The Locus Of Z Is A

Solution: Given z = (λ+3)+i√(5-λ2) z = x+iy Comparing we get x = (λ+3) => x-3 = λ y = √(5-λ2) y2 = (5-λ2) y2+λ2 = 5 Put λ = x-3... View Article

If Z1 Root 2z 1 Then The Locus Described By The Point Z In The Argand Diagram Is A

Solution: Given |z+1| = √2|z-1| Put z = x+iy |x+iy+1| = √2|x+iy-1| |(x+1)+iy| = √2|(x-1)+iy| √[(x+1)2+y2] = √2√[(x-1)2+y2] Squaring both... View Article

If Z1 1 Plus I Z2 Equal 2 Plus 3i And Z3 Equal Ai By 3 Where I 1 Are Collinear Then The Value Of A Is

Solution: Given z1 = 1 + i, z2 = –2 + 3i and z3 = ai/3 are collinear. z1 = 1 + i => (x1, y1) = (1,1) z2 = –2 + 3i =>(x2, y2) = (-2,3)... View Article

If Z 2 By Z 3 Equal 2 Represents A Circle Then Its Radius Is Equal To

Solution: |z - 2|/|z - 3| = 2 |z - 2| = 2|z - 3| Put z = x+iy |x+iy - 2| = 2|x+iy- 3| √((x-2)2+y2) = 2√((x-3)2+y2) x2-4x+4+y2 =... View Article

Poq Is A Straight Line Through Of Origin O P And Q Represent The Complex Numbers Respectively And

Solution: Let POQ is the straight line. O is the origin P = a+ib Q = c+id O = 0+i0 Since OP = OQ, O is the mid point. a+ib = c+id The... View Article

In The Argand Plane The Vector Is Turned In The Clockwise Sense Through 180 Degree And Stretched Three Times

Solution: z = 4 – 3i | z| = √(16+9) = 5 Consider z1 be the new vector obtained by rotating z in the clockwise sense through 1800. So z1 =... View Article

If Z 2 Then The Points Representing The Complex Numbers 1 Plus 5z Will Lie On A

Solution: Given |z| = 2 Consider z1 = x+iy lies on locus represented by -1+5z z1 = -1+5z z1+1 = 5z Take modulus on both sides | z1+1| =... View Article

The Equation Z Bar Z A Bar Z Bar A Z Plus B Equal 0 B Belongs To R Represents A Circle If

Solution: |z+a|2 = |a2-b|, represents a circe if |a2|>b. Hence option (2) is the answer.... View Article

Length Of The Line Segment Joining The Points 1 I And 2 Plus 3i Is

Solution: Given points are –1 –i and 2 + 3i . Let A be (-1,-1) and B( be (2,3) Distance between AB = √((2+1)2+(3+1)2) (distance formula) =... View Article

If Z1 And Z2 Be The Nth Roots Of Unity Which Subtend Right Angle At The Origin Then N Must Be Of The Form

Solution: Given z1 and z2 be the nth roots of unity which subtend right angle at the origin. arg (z1/z2) = π/2 So z1/z2 = cos π/2 + i sin... View Article

If Argz Less Than 0 Then Arg Z Arg Z Equal

Solution: Let z = r (cos θ+i sin θ) arg (z) = θ < 0 -z = -r(cos θ+i sin θ) = r(cos (π+θ) +i sin (π+θ)) arg (-z) = π+θ arg (-z)-arg (z) =... View Article

The Value Of A B Omega C Omega B C Omega A Omega A B Omega C Omega C A Omega B Omega Will Be

Solution: (a+bω+cω2)/(b+cω+aω2) + (a+bω+cω2)/(c+aω+bω2) = [(a+bω+cω2)/(b+cω+aω2)]+(a+bω+cω2)ω/(c+aω+bω2)ω = [(a+bω+cω2)(1+ω)]/(b+cω+aω2)... View Article

The Value Of Cos 30 Plus I Sin 30 By Cos 60 I Sin 60 Is

Solution: (cos 300+i sin 300)/(cos 600-i sin 600) = [(√3/2)+i (½) ]/[(1/2)-(√3/2)i] = (√3+i)/(1-i√3) Multiply numerator and denominator with... View Article

If Z 25 By Z 1 5 Then Find The Value Of Z

Solution: Let z = x+iy |z| = √(x2+y2) …(i) Given |(z-25)/(z-1)| = 5 |(z-25)| = 5|(z-1)| |(x+iy-25)| = 5|(x+iy-1)| Squaring both sides,... View Article

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