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1

AIEEE 2002

MCQ (Single Correct Answer)
The locus of the centre of a circle which touches the circle $$\left| {z - {z_1}} \right| = a$$ and$$\left| {z - {z_2}} \right| = b\,$$ externally

($$z,\,{z_1}\,\& \,{z_2}\,$$ are complex numbers) will be
A
an ellipse
B
a hyperbola
C
a circle
D
none of these

Explanation

Let the circle be $$\left| {z - {z_3}} \right| = r.$$

Then according to given conditions

$$\left| {{z_3} - {z_1}} \right| = r + a$$ (Shown in the image)

and $$\left| {{z_3} - {z_2}} \right| = r + b.$$ (Shown in the image)

Eliminating $$r,$$ we get

$$\left| {{z_3} - {z_1}} \right| - \left| {{z_3} - {z_2}} \right| = a - b.$$

$$\therefore$$ Locus of center $${z_3}$$ is

$$\left| {z - {z_1}} \right| - \left| {z - {z_2}} \right| = a - b$$ = constant.

Definition of hyperbola says, when difference of distance between two points is constant from a particular point then that particular point will lie on a hyperbola.

Here distance of z1 from z3 is = $$r + a$$ and distance of z2 from z3 is = $$r + b$$

Now their difference = ($$r + a$$) - ($$r + b$$) = $$a - b$$ = a constant

$$\therefore$$ Locus of z3 is a hyperbola.
2

AIEEE 2002

MCQ (Single Correct Answer)
If $$\left| {z - 4} \right| < \left| {z - 2} \right|$$, its solution is given by
A
$${\mathop{\rm Re}\nolimits} (z) > 0$$
B
$${\mathop{\rm Re}\nolimits} (z) < 0$$
C
$${\mathop{\rm Re}\nolimits} (z) > 3$$
D
$${\mathop{\rm Re}\nolimits} (z) > 2$$

Explanation

Given $$\left| {z - 4} \right| < \left| {z - 2} \right|$$

Let $$\,\,\,z = x + iy$$

$$\Rightarrow \left| {\left. {\left( {x - 4} \right) + iy} \right)} \right| < \left| {\left( {x - 2} \right) + iy} \right|$$

$$\Rightarrow {\left( {x - 4} \right)^2} + {y^2} < {\left( {x - 2} \right)^2} + {y^2}$$

$$\Rightarrow {x^2} - 8x + 16 < {x^2} - 4x + 4$$

$$\Rightarrow 12 < 4x$$

$$\Rightarrow x > 3$$

$$\Rightarrow {\mathop{\rm Re}\nolimits} \left( z \right) > 3$$
3

AIEEE 2002

MCQ (Single Correct Answer)
z and w are two nonzero complex numbers such that $$\,\left| z \right| = \left| w \right|$$ and Arg z + Arg w =$$\pi$$ then z equals
A
$$\overline \omega$$
B
$$- \overline \omega$$
C
$$\omega$$
D
$$- \omega$$

Explanation

Let $$\left| z \right| = \left| \omega \right| = r$$

$$\therefore$$ $$z = r{e^{i\theta }},\omega = r{e^{i\phi }}$$

where $$\,\,\theta + \phi = \pi .$$

$$\therefore$$ $$z = r{e^{i\left( {\pi - \phi } \right)}} = r{e^{i\pi }}.$$ $${e^{ - i\phi }} = - r{e^{ - i\phi }} = - \overline \omega .$$

[as $$\,\,\,\,\overline \omega = r{e^{ - i\phi }}$$ ]

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