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If 1 Omega And Omega 2 Are The Cube Roots Of 1 Then 1 Plus Omega Equals

Solution: Use ω3 = 1 and (1+ω) = -ω2, 1+ω2 = ω (1+ω)(1+ω2)(1+ω4)(1+ω8) = (1+ω)( -ω2)(1+ ω)(1+ ω2) = (1+ω)2( -ω2)-ω2 = (-ω2)2ω2 = ω6... View Article

If W Not Equal 1 Is Cube Root Of Unity And 1 W2 N Equal 1 W 4 N Then Smallest Possible Value Of N Is

Solution: We know 1+w+w2 = 0 1+w2 = -w 1+w4 = 1+w3w = 1+w = -w2 (1+w2)n/(1+w4)n = (-w/-w2)n = (1/w)n For n = 3, wn = 1. Hence... View Article

Let Alpha Beta Be Real And Z Be A Complex Number If Z Alpha Z Beta Equal Has Two Distinct Roots On The Line Then It Is Necessary That

Solution: z2 + αz + β = 0 Let z1 and z2 be the roots. z1 = x+iy z2 = x-iy Sum of roots z1 +z2 = -α z1z2 = β ..(i) Given roots lie on... View Article

1 Plus I 2 Ai 2 3i 3 Plusi Equal X Plus Iy X Y Belongs To R And X Equal Y Then A Equal

Solution: (1+i)(2+ai)+(2+3i)(3+i) = x+iy (2+2i+ai-a+6+9i+2i-3 = x+iy 5-a+(13+a)i = x+iy Equate real and imaginary parts 5-a = x 13+a = y... View Article

The Number Of Complex Numbers Z Such That Z 1 Equal Z Plus 1 Equal Z I Equal

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

If 1plus2i By 2 Plus I R Cos Theta Plus I Sin Theta Then R And Theta Equal

Solution: Given (1+2i)/(2+i) = r (cos θ+i sin θ) (1+2i)(2-i)/(2+i)(2-i) = r (cos θ+i sin θ) (2+4i-i+2)/(4+1) = r (cos θ+i sin θ) (4+3i)/5 =... View Article

If Z 4 By Z Equal 2 Then The Maximum Value Of Z Is 1

Solution: We know |a+b| ≤ |a|+|b| Given |z-4/z| = 2 |z| ≤ |z -(4/z)|+|(4/z)| |z| ≤ 2+|(4/z)| |z|- 2+|(4/z)| ≤ 0 [|z|2-2|z|-4] / |z| ≤ 0... View Article

If W Z Z 1 By 3 I And W 1 Then Z Lies On

Solution: Given w = z/(z-⅓)i and |w| = 1 w = 3z/(3z-i) |w| = 3|z|/|(3z-i)| 3|z| = |(3z-i)| Put z = x+iy |3(x+iy)| = |(3x+i(3y-1)|... View Article

Let Zw Be Two Complex Numbers Such That Bar Z I Bar W 0 And Arg Zw Pi Then Arg Z Equal

Solution: z = iw w = -iz arg (zw) = π arg (iz2) = π arg (-i)+2 arg(z) = π -(π/2) + 2 arg(z) = π Since arg (-i) = -(π/2) arg(z) =... View Article

If Z X Iy And Z 1 By 3 P Plus Iq Then X P Plus Y Q By P 2 Q2 Equal

Solution: Given z = x-iy and z1/3 = p+iq x-iy = (p+iq)3 x-iy = p3-3pq2+i(3p2q-q3) Equating real and imaginary part x = p3-3pq2 x/p =... View Article

If 1 Plus I X 2i By 3 I Plus 2 3i Y Plus I By 3 I I Then X Y Equal

Solution: Given ((1+i)x-2i)/(3+i) +((2-3i)y+i)/(3-i) = i [((1+i)x-2i)(3-i)/(3+i)(3-i)]+ [((2-3i)y+i)(3+i)/(3-i)(3+i)] = i... View Article

The Expression Of Complex Number 1 1 Cos Theta I Sin Theta In The Form A Plus Ib Is

Solution: 1/(1+cos θ-i sin θ) = 1(1+cos θ+i sin θ)/(1+cos θ-i sin θ) (1+cos θ+i sin θ) = (1+cos θ+i sin θ)/(1+cos θ)2+sin2θ = (1+cos θ+i sin... View Article

If The Cube Roots Of Unity Are 1 Omega Omega Square Then 1 Omega Omega2 Equal

Solution: We know sum of three cube roots of unity equals zero. 1+ω+ω2 = 0 (1+-½-√3i/2 + -½+√3i/2 = 0) Hence option (2) is the answer.... View Article

If Z Equal 4 3i 5 3i Then Z 1 Equal

Solution: Z = (4+3i)/(5-3i) Z-1 = 1/Z = (5-3i)/(4+3i) Multiply numerator and denominator with (4-3i) = (5-3i)(4-3i)/(4+3i)(4-3i) =... View Article

I Power 1 I Power 2 Plus Equal

Solution: i1+i2+i3+i4+...+i100 i1+i2+i3+i4= i-1+i+1 = 0 Powers of i follow a cyclicity of 4. So i1+i2+i3+i4+...+i100 = 0 Hence option... View Article

The Complex Numbers Z1 Z2 And Z3 Satisfying Z1 Z3 Z2 Z3 1 Root 3 I 2 Are The Vertices Of A Triangle Which Is

Solution: (z1-z3)/(z2-z3) = (1- √3i)/2 Multiply numerator and denominator with (1+√3i) = (1- √3i)(1+√3i)/2(1+√3i) = 1+1(3)/(2(1+√3i) =... View Article

Evaluate I Power 19 1 I Power 25 2

Solution: (i19+(1/i)25)2 = (i16 i3+1/i24 i)2 (since i4 = 1, i2 = -1) = ( -i+1/ i)2 = (-i+i3)2 = (-i-i)2 = (-2i)2 = 4 (-1) =... View Article

If Z X Iy X Y Belongs To R And 3x 3x Yi 4 6i Then Z

Solution: 3x+(3x-y)i = 4-6i Equate real part, we get 3x = 4 x = 4/3 Equate imaginary part, we get 3x-y = -6 3(4/3)-y = -6 4-y = -6 y =... View Article

Let Z And Omega Be Two Non Zero Complex Numbers Such That Z Omega And Arg Z Arg Omega Pi Then Z Equals

Solution: z = | z |(cos θ+ i sin θ) Where θ = arg z If θ1 = arg ω So θ = π- θ1 z = | ω | (cos (π- θ1)+ i sin (π- θ1 ) = | ω | (-cos θ1... View Article

If Z Not Equal 1 And Z 2 Z 1 Is Real Then The Point Represented By The Complex Number Z Lies

Solution: Let z = x+iy z2/(z - 1) = (x+iy)2/(x+iy-1) = (x2-y2+2xiy)/(x-1)+iy Multiply numerator and denominator with (x-1)-iy =... View Article

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