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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 +... 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) (x2+3x2)2(x2+3x+1) = [(ω-1)2+3(ω-1)2][(ω-1)2+3(ω-1)+1] = (ω2+ω-2)2(ω2+ω-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... 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 θ)... 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... 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 Comparing we get A =... 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... 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... 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 | = 23(25)/100 = 23/4... 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.... 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 ((3+3i+1-i)x-6i-2)+(6-9i+2i+3)y+3i-1)/(9-(-1) = i (4x+2ix)+(9y-7iy)-3i-3 = 10i (4x+9y-3)+(2x-7y-13)i = 0... 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... 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... View Article

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