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If Z 1 A B Ic And A2 B2 C2 1 Then 1 Iz 1 Iz Is Equal To

Solution: Given z(1+a) = b+ic z = (b+ic)/(1+a) (1+iz)/(1-iz) = (1+i(b+ic)/(1+a))/(1-i(b+ic)/(1+a)) = (1+a-c+ib)/(1+a+c-ib) =... View Article

If Sigma K 0 To 100 Ik X Iy Then The Values Of X And Y Are

Solution: ∑k=0100 ik = x+iy 1+i+i2+i3+i4+....i100 = x+iy This is a G.P with common ratio i and first term 1. => 1(1-i101)/(1-i) = x+iy... View Article

If 1 I 1 Plus I 100 A Ib Then

Solution: ((1-i)/(1+i)) = ((1-i)(1-i)/(1+i)(1-i)) = (1-2i-1)/(1+1) = -2i/2 = -i ((1-i)/(1+i))100 = -i100 = (i4)25 = 1 Comparing... View Article

The Multiplicative Inverse Of A Number Is The Number Itself Then Its Initial Value Is

Solution: Check the options. If we consider, -1, the multiplicative inverse is 1/-1 = -1 Hence option (2) is the answer.... View Article

If The Imaginary Part Of Is Zero Where A Is A Real Number Then The Value Of A Is Equal To

Solution: (2+i)/(ai-1) = (2+i)(-1-ai)/(ai-1)(-1-ai) = (-2-i-2ai+a)/(1+a2) = (-2+a)+i(-2a-1)/(1+a2) Given imaginary part is zero. So -2a-1... View Article

If Z 1 I Then The Multiplicative Inverse Of Z 2 Is

Solution: Given z = 1+i z2 = (1+i)2 = 1+2i-1 = 2i Multiplicative inverse = 1/2i = -i/2 Hence option (3) is the answer.... View Article

X Iy 1 3 A Ib Then X A Y B Is Equal To

(1) 4(a2+b2) (2) 4(a2-b2) (3) 4(b2-a2) (4) None of these Solution: Given (x+iy)1/3 = a+ib (x+iy) = (a+ib)3 x+iy = a3-ib3+3a2bi-3ab2... View Article

If Z Not Equal To Zero Is A Complex Number Then

(1) Re (z) = 0 => Im (z2) = 0 (2) Re (z2) = 0 => Im (z2) = 0 (3) Re (z) = 0 => Re (z2) = 0 (4) None of these Solution: Let z =... View Article

The Real Values Of X And Y For Which The Equation Is X Iy 2 3i 4i Is Satisfied Are

Solution: Given (x+iy)(2-3i) = 4+i (x+iy)(2-3i) = 2x+2iy-3ix+3y = 2x+3y+i(2y-3x) Comparing real and imaginary parts 2x+3y = 4 -3x+2y = 1... View Article

If A2 B2 1 Then 1 B Ia 1 B Ia Is Equal To

Solution: (1+b+ia)/(1+b-ia) Multiply numerator and denominator with (1+b+ia) = (1+b+ia)(1+b+ia)/(1+b-ia)(1+b+ia) = (1+b+ia)2/((1+b)2+a2) =... View Article

Root 5 12i Root 5 12i Root 5 12i Root 5 12i Is Equal To

Solution: [√(5+12i)+ √(5-12i)]/[√(5+12i)- √(5-12i)] Multiply numerator and denominator with √(5+12i)+√(5-12i)... View Article

A Ib Greater Than C Id Can Be Explained Only When

Solution: Since inequality in complex numbers is not defined, a+ib> c+id is possible if b = 0 and d = 0 Hence option (2) is the answer.... View Article

The Imaginary Part Of 1 I 2 I Is

Solution: (1+i)2/(2-i) = (1+2i-1)/(2-i) = 2i/(2-i) = 2i(2+i)/(2-i)(2+i) = (4i-2)/(4+1) = (4i-2)/5 Imaginary part is 4/5. Hence option... View Article

If Z And Z Are Complex Numbers Such That Z Then Z Is Equal To

Solution: Let z = x+iy z’ = x-iy zz’ = (x+iy)(x-iy) = x2+ixy-ixy+y2 = x2+y2 Given z.z’ = z x2+y2 = x+iy Since imaginary part is... View Article

Re 1 I 2 3 I Is Equal To

Solution: (1-i)2/(3-i) = (1-2i-1)/(3-i) = -2i/(3-i) Multiply numerator and denominator with (3+i) = -2i(3+i)/(3-i)(3+i) = (-6i+2)/(9+1) =... View Article

1 By 1 2i Plus 3 By 1 I3 Is Equal To

Solution: [1/(1-2i)]+[3/(1+i)](3+4i)/(2-4i) = (1+i+3(1-2i))(3+4i)/(1-2i)(1+i)(2-4i) = (4-5i)(3+4i)/(3-i)(2-4i) =... View Article

Additive Inverse Of 1 I Is

Solution: Additive inverse of x+iy is -(x+iy). Additive inverse of 1-i is -(1-i) = -1+i Hence option (3) is the answer.... View Article

If Z1 And Z2 Be Two Complex Number Then Re Z1 Z2 Is Equal To

(1) Re (z1) Re (z2) (2) Re (z1) Im (z2) (3) Im (z1) Re (z2) (4) None of these Solution: Let z1 = x+iy Re z1 = x Im z1 = y z2 = a+ib Re... View Article

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

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

Solving 3 2yi 9x 7i Where I2 1 For X And Y Real We Get

Solution: Given 3-2yi = 9x-7i Comparing real and imaginary parts 9x = 3 x = ½ = 0.5 2y = 7 y = 7/2 = 3.5 Hence option (1) is the answer.... View Article

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