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In Which Thermodynamic Process Work Done Is Minimum

In Isochoric process, the work done is always zero. Isochoric process is a thermodynamic process that takes place in a constant volume.... View Article

Find The Value Of The Given Determinant

The value of the determinant (1) 0 (2) 1 (3) 2 (4) 3 Solution: The value of the determinant (1) 0 (2) 1 (3) 2 (4) 3... View Article

If Omega Not Equal 1 Is A Cube Root Of Unity And 1 Plus Omega Power 7 Then A And B Are Respectively

Solution: We know 1+ω+ω2 = 0 1+ω= -ω2 (1+ω)7 = (-ω2)7 = -ω14 = -ω2 = 1+ω Comparing with A + Bω We get A = 1 and B = 1 Hence... View Article

If Cos Theta I Sin Theta Cos 2 Theta I Sin 2 Theta Cos N Theta Then The Value Of Theta Is

Solution: Given (cos θ + i sinθ )(cos 2θ + i sin 2θ)...(cos nθ + i sin nθ) = 1 eiθe2iθe3iθ….einθ = 1 ei(θ+2θ+3θ+...nθ) = 1 ein(n+1)θ/2 =... View Article

In Which Quadrant Of The Complex Plane The Point 1 Plus 2i By 1 I Lies

Solution: (1+2i)/(1-i) = (1+2i)(1+i)/(1-i)(1+i) = (1+2i+i-2)/(1+1) = (-1+3i)/2, lies in second quadrant. Hence option (3) is the answer.... View Article

If A Equal Cos Theta Plus I Sin Theta Then 1 Plus A By 1 A Is Equal To

Solution: Let a = cos θ + i sin θ (1+a)/(1-a) = (1+ cos θ + i sin θ)/(1-cos θ - i sin θ) Multiply numerator and denominator with (1-cos θ+ i... View Article

The Equation Z 5i Z 5i 12 Where Z X Iy Represents An

Solution: |z – 5i| /|z + 5i| = 12 |z – 5i| = 12 |z + 5i| Put z = x+iy | x+iy– 5i| = 12 |x+iy + 5i| | x+(y-5)i| = 12 |x+(y+5)i| Taking... View Article

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

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