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 circle with centre (-1,1) and radius is unity. This... View Article
In The Argand Plane The Distinct Roots Of 1 Plus Z Plus Z Power 3 Plus Z Power 4 Equal 0 Represent Vertices Of Solution: 1+ z+ z3 + z4 = 0 => (1+z)+z3(1+z) = 0 => (1+z3)(1+z) = 0 z = -1 z3 = -1 z = -1, - ω, -ω2 are cube roots of... View Article
[latex]The \: Value \: of (\frac{-1}{2}) + \left ( \sqrt{\frac{3i}{2}}\right )^{1000} is[/latex] (1) (2) (3) (4) Solution: (-1+√(3i))/2 = ω We know ω3n = 1 (-½)+√(3i)/2)1000 = ω1000 = ω3(333) ω = ω Hence option (4) is the answer.... View Article
If Root X Plus Iy A Plus Ib Then X Iy Is Equal To Solution: Given √(x+iy) = ±(a+ib) Squaring both sides (x+iy) = (a+ib)2 = a2+2aib-b2 = a2-b2+2aib Comparing real and imaginary... View Article
One Of The Values Of 1 Plus I Root 2 Power 2 By 3 Is Solution: Let Z = ((1+i)/√2)2/3 = ((1/√2)+(i/√2))2/3 Z = (cos (π/4) + i sin (π/4))2/3 Z = cos (8k+1)π/6 + i sin... View Article
If X 3 Plus I Root 3 2 Is A Complex Number Then The Value Of X Power 2 Plus 3x Power 2 Power 2 X 2 3x Plus 1 Is Solution: Given x = (-3+i√3)/2 = (-1+i√3)/2-(1/1) = ω-1 [since ω = (-1+i√3)/2] (x2+3x2)2(x2+3x+1)... View Article
The Real Part Of 1 Cos Theta 2 I Sin Theta 1 Is Solution: (1-cos θ+2i sin θ)-1 = (2 sin2 θ/2 + 2i sin θ)-1 = (2 sin2θ/2 + 2i sin θ/2 cos θ/2)-1 = (2... View Article
If M1 M2 M3 And M4 Respectively Denote The Moduli Of The Complex Numbers 1 Plus 4i 3 I 1 I And 2 3i Then The Correct One Among The Following Is Solution: Given m1 = |1+4i| = √(1+16) = √17 m2 = |3+i| = √(9+1) = √10 m3 = |i-1| =... View Article
If 3 2 Cos Theta I Sin Theta A Ib Then A 2 2 B 2 Is Solution: Given 3/(2+ cos θ+i sin θ) = a+ib (3/(2+ cos θ+i sin θ))(2+cos θ)-i sin θ)/(2+cos θ)-i sin... View Article
The Locus Of Z Satisfying The Inequality Z 2i 2z I 1 Where Z X Iy Is Solution: Given |(z+2i)/(2z+i)|<1 z = x+iy So |x+iy+2i |/ |2(x+iy)+i | <1 |x+i(y+2) |/ |2x+(2y+1)i | <1 |x+i(y+2) |/<|2x+(2y+1)i... View Article
If Z1 And Z2 Two Non Zero Complex Numbers Such That Z1 Z2 Z1 Z2 Then Arg Z1 Z2 Is Equal To Solution: |Z1 + Z2|= |Z1||Z2| This implies z1 and z2 are parallel. Hence the angle between them is 0. Therefore arg(z1/z2) =... View Article
If Omega Not Equal 1 Is A Cube Root Of Unity And 1 Plus Omega 7 A B Omega Then A And B Are Respectively Solution: Given (1+ω)7 = A+Bω (1+ω)7 = (1+ω)(1+ω)6 = (1+ω)(ω2)6 = 1+ω A+Bω = 1+ω... View Article
If A B C And U V W Are Complex Numbers Representing The Vertices Of Two Triangles Such That C 1 R A Rb And W 1 R U Rv Where R Is A Complex Number Then The Two Triangles Solution: Consider that the triangles b are ΔABC and ΔUWV. Vertices of ΔABC given by a,b,c and ΔUWV given by u,v,w. AB... View Article
If Z 2 I 2 3 2 I 3 I 4 Then Amplitude Of Z Is Solution: Given z(2- i2√3)2 = i(√3 + i)4 Let ω be the cube root of unity, ω3 = 1 and 1+ω+ω2 = 0 ω =... View Article
If Z 3+I 3 3i 4 2 8 6i 2 Then Z Is Equal To Solution: Given Z = (√3+i)3(3i+4)2/(8+6i)2 |Z| = |(√3+i)3(3i+4)2/(8+6i)2 | = |√3+i| 3| 3i+4| 2/| 8+6i)2 | =... View Article
Let Z1 Z2 And Z1 Z2 If Z1 Has Positive Real Part And Z2 Has Negative Imaginary Part Then Z2 Z1 Z1 Z2 May Be Solution: Given z1 ≠ z2 and |z1| = |z2| The numerator is purely imaginary and the denominator is purely real. Im(z) and Re(z) are real... View Article
The Values Of X And Y Satisfying The Equation 1i X 2i 3 I2 3iy I 3 I Equal I Are Solution: Given (((1+i)x-2i)/(3+i))+((2-3i)y+i))/(3-i)) = i ((1+i)(3-i)x-2i(3-i) + (2-3i)(3+i)y+ i(3+i))/(3+i)(3-i) = i... View Article
The Smallest Positive Integer N For Which 1 I 2n 1 I 2n Is Solution: Given (1+ i)2n = (1 - i)2n [(1+i)/(1-i)]2n = 1 Multiply numerator and denominator with (1+i) We get [(1+i)(1+i)/(1-i)(1+i)]2n = 1... View Article
Check The Solution The Points Z 1 Z 2 Z 3 Z 4 In The Complex Plane Are The Vertices Of A Parallelogram Taken In Order If And Only If Solution: Since z1, z2, z3, z4 are the vertices of the parallelogram, the midpoint of the diagonals are same. (z1+z3)/2 = (z2+z4)/2 =>... View Article
If Z X Iy And Omega Equals 1 Iz Z I Then Omega 1 Implies That In The Complex Plane Solution: z = x+iy and ω = (1-iz)/(z-i) Given |ω| = 1 |ω| = |(1-iz)/(z-i)| = 1 |ω|2 = |(1-iz)/(z-i)|2 ... View Article
