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If 2x 3 5i Then The Value Of 2x 3 2x 2 7x 72 Is

Solution: Given 2x = 3+5i x = (3+5i)/2 x3 = (27+135i-225-125i)/8 = (-198+10i)/8 x2 = (9-25+30i)/4 x2 = (-16+30i)/4 2x3... View Article

If Z Is A Complex Number Such That The Imaginary Part Of Z Is Non Zero And A Z 2 Z 1 Is Real Then A Cannot Take The Value

Solution: Given a = z2+z+1 z2+z+1−a = 0 If solution is not real then b2-4ac<0. 1-4(1-a)<0 1-4+4a <0 -3+4a<0 4a<3... View Article

Argument Of Complex Number 1 3i 2 I Is

Solution: Let z = (-1-3i)/(2+i) = (-1-3i)(2-i)/(2+i)(2-i) = (-2-6i+i-3)/5 = (-5-5i)/5 = (-1-i) Put z = r cos θ+i r... View Article

If Z 1 Z 2 Z 1 Z Then Find Arg Z 1 Z 2

Solution: Consider z1 = x1+iy1 z2 = x2+iy2 We have |Z1 + Z2| = |Z1| + |Z2| √((x1+x2)2+(y1+y2)2 = √(x12+y12)+√(x22+y22)... View Article

The Product Of All Values Of Cos Alpha I Sin Alpha 3 Is

Solution: (cosα + i sinα)3/5 = cos(3α/5)+i sin (3α/5) Product = (cos(3α/5)+i sin (3α/5))(cos (2π+3α)/5)+i sin(2π+3α)/5))×(cos (4π+3α)/5)+i sin (4π+3α)/5)(cos (6π+3α)/5)+i... View Article

The Imaginary Part Of 1 I 2 I 2i 1

Solution: (1+i)2/i(2i-1) = (1+i)2/(-2-i) = (1+i)2 (-2+i)/(-2-i)(-2+i) = (1+i)2 (-2+i)/(4–1) = (1+i)2 (-2+i)/5 = (1+2i-1)(-2+i)/5 = 2i(-2+i)/5 = (-4i-2)/5 Imaginary... View Article

The Point 4 1 Undergoes The Following Three Transformations Successively

Solution: When the point (4,1) undergoes reflection, it becomes (1,4) After translation through a distance of 2 units (1+4)+(2,0) =... View Article

If Z 8 Z 8 16 Where Z Is A Complex Number Then The Point Z Will Lie On

Solution: Given |z + 8| + |z – 8|= 16 Let z = x+iy |x+iy+8|+|x-iy-8| = 16 √(x+8)2+y2 = 16-√((x-8)2+y2)... View Article

If Magnitude Of A Complex Number 4 3i Is Tripled And Is Rotated Anticlockwise By An Angle Pi Then Resulting Complex Number Would Be

Solution: |4-3i| = √(42+32) = √25 = 5 When tripled, magnitude = 3(5) = 15 (eiπ )3(4-3i) = (cos π+i... View Article

If Z X Iy Then Area Of The Triangle Whose Vertices Are Points Z Iz Z Iz Is

Solution: Let A = z, B = iz, C = z+iz |AB| = |iz-z| |BC| = z |AC| = iz... View Article

The Locus Of The Point Z X Iy Satisfying The Equation Z 1 1 1 Is Given By

Solution: Given |(z-1)/(z+1)| = 1 z = x+iy Put z in given equation |x+iy-1)/(x+iy+1)| = 1 √((x-1)2+y2) = √((x+1)2+y2) Squaring... View Article

If Im Z 1 2z 1 4 Then Locus Of Z Is

Solution: Given (z-1)/(2z+1) = 4 (z-1)/(2z+1) = (x+iy-1)/(2(x+iy)+1) = ((x-1)+iy)/(2x+1)+2iy Multiply numerator and denominator with (2x+1)-2iy) = ((x-1)+iy)((2x+1)-2iy)/((2x+1)+2iy)(2x+1-2iy) = (x-1)(2x+1)+2y2+iy(-2x+2+2x+1)/((2x+1)2+4y2)... View Article

If Z 2 1 Z 2 1 Then Z Lies On

Solution: Given |z2 -1| = |z|2 + 1, |z2 +(-1)| = |z|2 + |-1| Arg (z2) = arg (-1) 2... View Article

If The Complex Numbers Z1 Z2 Z3 Are In An Ap Then They Lie On

Solution: Since z1, z2, z3 are in AP z2 = (z1+z3)/2 z2 is the midpoint of z1 and z3. So... View Article

The Region Of The Argand Diagram Defined By Mod Z 1 Plus Mod Z Plus 1 Less Than Equal To 4 Is An

Solution: Given region defined by |z – 1|+ |z + 1| ≤ 4. ..(1) For an ellipse PS+PS’ = 2a... View Article

The Point In The Set Z Belongs C Arg Z 2 By Z 6i Equal Pi By 2 Where C Denotes The Set Of All Complex Number Lie On The Curve Which Is

Solution: Given arg ((z-2)/(z-6i) = π/2 Arg (z-2)-arg(z-6i) = π/2 z = x+iy arg ((x-2)+iy) – arg(x+(y-6)i = π/2 =... View Article

Among The Complex Number Z Satisfying Condition Mod Z Plus 1 I Less Than Or Equal 1 The Number Having The Least Positive Argument Is

Solution: Given condition | z +1 – i| ≤ 1 | z – (-1+i)| ≤ 1 This will represent a... View Article

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