JEE Mathematics Complex Numbers MCQs Set 18

Answer: (c) \( \left|z-\dfrac1{1-a}\right|=\dfrac1{|1-a|} \)

Explanation:
1. \( z_k=\frac{1-a^k}{1-a} \), so \( z_k-\frac1{1-a}=-\frac{a^k}{1-a} \).
2. Taking modulus: \( \left|z_k-\frac1{1-a}\right|=\frac{|a|^k}{|1-a|}<\frac1{|1-a|} \) since \( |a|<1 \).
3. So all vertices lie within the circle of centre \( \frac1{1-a} \) and radius \( \frac1{|1-a|} \).

Teacher's Note:
1. Sum the geometric progression first.
2. The distance from the limit point 1/(1 − a) shrinks like |a|^k.

Answer: (b) Are similar

Explanation:
1. Rearranging, \( \frac{c-a}{b-a}=r=\frac{w-u}{v-u} \).
2. Equal ratios mean the triangles with vertices (a, b, c) and (u, v, w) have a common angle between two sides and sides in the same ratio.
3. Hence the triangles are similar (but not necessarily congruent).

Teacher's Note:
1. The ratio (c − a)/(b − a) encodes both a length ratio and an angle.
2. Equal complex ratios imply similar triangles with the same orientation.

Answer: (a) \( 3-\tfrac12i \) or \( 1-\tfrac32i \)

Explanation:
1. The diagonals of a rhombus bisect each other at right angles, and \( BD=2AC \) gives \( DM=\tfrac12BD=AC=2\,AM \).
2. So \( MA=\pm\tfrac i2(D-M) \) with \( D-M=-1+2i \): \( \tfrac i2(-1+2i)=-1-\tfrac i2 \).
3. Therefore \( A=M\pm\left(-1-\tfrac i2\right) \), giving \( 1-\tfrac32i \) or \( 3-\tfrac12i \).

Teacher's Note:
1. Rotating a half-diagonal by 90° gives the other half-diagonal direction.
2. The length ratio fixes the factor 1/2.

Answer: (b) \( a=b=2-\sqrt3 \)

Explanation:
1. Equal sides: \( |z_1|=|z_2| \) gives \( a^2+1=1+b^2 \), so \( a=b \) (both positive).
2. Then \( |z_1-z_2|^2=2(1-a)^2=|z_1|^2=a^2+1 \), so \( a^2-4a+1=0 \).
3. Thus \( a=2\pm\sqrt3 \); since \( 0<a<1 \), \( a=b=2-\sqrt3 \).

Teacher's Note:
1. Set two sides equal, then the third.
2. Reject the root greater than 1.

Answer: (a) The origin

Explanation:
1. \( z_2=\frac{-a-ib}{a^2+b^2}=-\frac{z_1}{|z_1|^2} \).
2. So \( z_2 \) is a negative real multiple of \( z_1 \): the two points lie on opposite sides of the origin on the same line through it.
3. The join of \( z_1 \) and \( z_2 \) therefore passes through the origin.

Teacher's Note:
1. Rationalise 1/(−a + ib).
2. A real multiple (here negative) means the points are collinear with the origin.

Answer: (d) 0

Explanation:
1. \( z_1 \) is the point dividing the segment joining \( z_3 \) and \( z_2 \) in a real ratio (the section formula).
2. So A lies on the line BC and its distance from the line is 0.

Teacher's Note:
1. A real weighted average of two points lies on the line through them.
2. Compare with the section formula (mx2 + nx1)/(m + n).

Answer: (b) \( (z_1-z_2)^2=2(z_1-z_3)(z_3-z_2) \)

Explanation:
1. At C, \( z_2-z_3=\pm i(z_1-z_3) \) (equal legs, perpendicular).
2. Choose coordinates \( z_3=0,\ z_1=1,\ z_2=\pm i \) and test the options.
3. For \( z_2=i \): \( (z_1-z_2)^2=(1-i)^2=-2i \) and \( 2(z_1-z_3)(z_3-z_2)=2(1)(-i)=-2i \). The same holds for \( z_2=-i \). Only option (b) works.

Teacher's Note:
1. Place C at the origin and A at 1 to test the options.
2. The condition is symmetric under swapping the roles of the two legs.

Answer: (d) \( \tfrac\pi6 \)

Explanation:
1. \( \sqrt3-i=2e^{-i\pi/6} \), so \( (\sqrt3-i)^{25}=2^{25}e^{-25i\pi/6} \).
2. The reciprocal has argument \( \tfrac{25\pi}6=4\pi+\tfrac\pi6\equiv\tfrac\pi6 \).

Teacher's Note:
1. Reciprocal negates the argument.
2. Reduce 25π/6 modulo 2π.

Answer: (d) \( -12+9i \)

Explanation:
1. A rotation through 180° (either sense) multiplies by \( -1 \).
2. Stretching three times multiplies by 3: \( -3(4-3i)=-12+9i \).

Teacher's Note:
1. Rotation by 180° is multiplication by −1.
2. Stretch factor multiplies the modulus.

Answer: (b) \( \tfrac{-15}2+10i \)

Explanation:
1. Multiplying by \( -2.5 \): \( -2.5(3-4i)=-7.5+10i \).

Teacher's Note:
1. Rotate by 180° (× −1) and then scale.
2. −2.5 × 3 = −7.5, −2.5 × (−4) = 10.

Answer: (d) \( |z_1+z_2|\ge|z_1|-|z_2| \)

Explanation:
1. By the triangle inequality, \( |z_1+z_2|\ge\big||z_1|-|z_2|\big|\ge|z_1|-|z_2| \).
2. The other options reverse the standard inequalities.

Teacher's Note:
1. The reverse triangle inequality gives the lower bound.
2. Test with z1 = 1, z2 = i to eliminate (a) and (b).

Answer: (b) 2

Explanation:
1. The vertices are \( z,\ iz,\ z+iz \). The sides from \( z+iz \) are \( iz \) and \( z \), of equal length \( |z| \) and at right angles.
2. Area \( =\tfrac12|z|^2=2 \), so \( |z|=2 \) (a modulus cannot be negative).

Teacher's Note:
1. Draw the three points: they form a right isosceles triangle.
2. A modulus is always non-negative.

Answer: (a) Collinear

Explanation:
1. \( B-A=2-6i \) and \( C-A=-4+12i=-2(2-6i) \).
2. Since \( C-A \) is a real multiple of \( B-A \), the points are collinear.

Teacher's Note:
1. Collinear points have (z3 − z1)/(z2 − z1) real.
2. Compute the ratio directly.

Answer: (d) 5

Explanation:
1. Collinearity means \( \frac{z_3-z_1}{z_2-z_1} \) is real.
2. Here \( z_2-z_1=-3+2i \) and \( z_3-z_1=-1+\left(\tfrac a3-1\right)i \).
3. The slopes must agree: \( \frac{b}{-1}=\frac{2}{-3} \) with \( b=\tfrac a3-1 \), so \( b=\tfrac23 \) and \( a=3\left(1+\tfrac23\right)=5 \).

Teacher's Note:
1. Equal slopes between the points give the same condition.
2. Three points are collinear if the slopes of two pairs agree.

Answer: (b) \( \tfrac12|z|^2\sin\alpha \)

Explanation:
1. Two sides from the origin have length \( |z| \) each, with angle \( \alpha \) between them.
2. Area \( =\tfrac12|z|\cdot|z|\sin\alpha=\tfrac12|z|^2\sin\alpha \).

Teacher's Note:
1. Use area = ½ ab sinC.
2. The angle between z and z e^(iα) is α.

Answer: (c) 2

Explanation:
1. Test \( z=i \): \( -i-i+2i=0 \), so \( z-i \) is a factor.
2. Dividing: \( z^3+iz^2+2i=(z-i)(z^2+2iz-2) \), and the quadratic has roots \( z=-i\pm1 \).
3. The vertices are \( i,\ 1-i,\ -1-i \): base 2 (the two lower points) and height 2, so the area is \( \tfrac12\times2\times2=2 \).

Teacher's Note:
1. Look for a root by inspection before dividing.
2. Two vertices share the same imaginary part, so the base is horizontal.

Answer: (a) The real axis

Explanation:
1. The condition says \( |z-5i|=|z+5i| \): z is equidistant from \( 5i \) and \( -5i \).
2. That is the perpendicular bisector of the segment joining them, the real axis.

Teacher's Note:
1. Equal distances from two points give the perpendicular bisector.
2. The bisector of 5i and −5i passes through the origin along the x-axis.

Answer: (d) \( x \)-axis

Explanation:
1. The condition says z is equidistant from \( -\tfrac i2 \) and \( \tfrac i2 \).
2. Expanding, \( x^2+(y+\tfrac12)^2=x^2+(y-\tfrac12)^2 \) gives \( 2y=0 \), i.e. \( y=0 \): the x-axis.

Teacher's Note:
1. Equal distance from two points is their perpendicular bisector.
2. The bisector of ±i/2 is the real axis.

Answer: (b) A straight line

Explanation:
1. With \( z=x+iy \), \( iz+1=(1-y)+ix \).
2. The imaginary part of the quotient is \( \frac{2y(1-y)-x(2x+1)}{(1-y)^2+x^2} \).
3. Setting it equal to \( -2 \): \( 2y-2y^2-2x^2-x=-2(1-y)^2-2x^2 \) which simplifies to \( x+2y=2 \), a straight line.

Teacher's Note:
1. The quadratic terms cancel after clearing the denominator.
2. Check with the point (2, 0): the quotient is 5/(1 + 2i) = 1 − 2i, whose imaginary part is −2.

Answer: (a) \( 2x+4y-5=0 \)

Explanation:
1. The points are \( 2-i \) and \( 3+i \). Squaring: \( (x-2)^2+(y+1)^2=(x-3)^2+(y-1)^2 \).
2. Expanding: \( -4x+4+2y+1=-6x+9-2y+1 \), so \( 2x+4y-5=0 \).

Teacher's Note:
1. Square both sides and cancel x² and y².
2. Check that the midpoint (5/2, 0) satisfies the answer.

Answer: (d) \( x-y+1=0 \)

Explanation:
1. Here \( z-2-3i=(x-2)+i(y-3) \) with amplitude \( \tfrac\pi4 \), so \( y-3=x-2 \) with both parts positive.
2. This gives \( x-y+1=0 \) (a ray from \( 2+3i \)).

Teacher's Note:
1. Amplitude π/4 means slope tan(π/4) = 1.
2. The locus is a half-line, lying on the line x − y + 1 = 0.

Answer: (b) A square

Explanation:
1. With \( z=x+iy \): \( |z+\bar z|=2|x| \) and \( |z-\bar z|=2|y| \).
2. So \( |x|+|y|=1 \), whose graph is a square with vertices \( (\pm1,0),\ (0,\pm1) \).

Teacher's Note:
1. Use z + conj(z) = 2x and z − conj(z) = 2iy.
2. |x| + |y| = 1 is a square (diamond) with diagonals along the axes.

Answer: (a) A straight line

Explanation:
1. The points are \( -1+i \) and \( 1-i \).
2. The condition says z is equidistant from them, so z lies on their perpendicular bisector: the line \( y=x \).

Teacher's Note:
1. Rewrite as |z − (−1 + i)| = |z − (1 − i)|.
2. Equal distances from two fixed points give a line.

Answer: (c) Ellipse

Explanation:
1. The sum of distances from the fixed points \( 4i \) and \( -4i \) is the constant 10.
2. Since \( 10>8=|4i-(-4i)| \), the locus is an ellipse with foci \( \pm4i \).

Teacher's Note:
1. Constant sum of distances to two foci defines an ellipse.
2. The constant must exceed the distance between the foci.