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1

### AIEEE 2003

Let $${Z_1}$$ and $${Z_2}$$ be two roots of the equation $${Z^2} + aZ + b = 0$$, Z being complex. Further , assume that the origin, $${Z_1}$$ and $${Z_2}$$ form an equilateral triangle. Then
A
$${a^2} = 4b$$
B
$${a^2} = b$$
C
$${a^2} = 2b$$
D
$${a^2} = 3b$$

## Explanation

$${Z^2} + aZ + b = 0$$
and two roots are $${Z_1}$$ and $${Z_2}$$.

$$\therefore$$ $${Z_1}$$ + $${Z_2}$$ = $$-a$$ and $${Z_1}$$$${Z_2}$$ = $$b$$

Question says,
There are three complex numbers:
1. Origin (0)
2. $${Z_1}$$
3. $${Z_2}$$
and they form an equilateral triangle. So They are the vertices of the triangle.

[ Important Point : If $${Z_1}$$, $${Z_2}$$ and $${Z_3}$$ are the vertices of an equilateral triangle then -
$$Z_1^2$$ + $$Z_2^2$$ + $$Z_3^2$$ = $${Z_1}{Z_2}$$ + $${Z_2}{Z_3}$$ + $${Z_3}{Z_1}$$ ]

In this question,
$${Z_1}$$ = 0, $${Z_2}$$ = $${Z_1}$$ and $${Z_3}$$ = $${Z_2}$$

By putting those values in the equation we get,

$${0^2}$$ + $$Z_1^2$$ + $$Z_2^2$$ = $$0$$ + $${Z_1}{Z_2}$$ + 0

$$\Rightarrow$$ $$Z_1^2$$ + $$Z_2^2$$ = $${Z_1}{Z_2}$$

$$\Rightarrow$$ $$Z_1^2$$ + $$Z_2^2$$ = $$b$$ [ as $${Z_1}$$$${Z_2}$$ = $$b$$ ]

$$\Rightarrow$$ $${\left( {{Z_1} + {Z_2}} \right)^2}$$ - $$2{Z_1}{Z_2}$$ = $$b$$

$$\Rightarrow$$ $${\left( {{Z_1} + {Z_2}} \right)^2}$$ - $$2b$$ = $$b$$

$$\Rightarrow$$ $${\left( {{Z_1} + {Z_2}} \right)^2}$$ = $$3b$$

$$\Rightarrow$$ $${\left( { - a} \right)^2}$$ = $$3b$$

$$\Rightarrow$$ $${a^2}$$ = $$3b$$

So Option (D) is correct.

[ Note : This question is asked to check if you know the following formula -

"If $${Z_1}$$, $${Z_2}$$ and $${Z_3}$$ are the vertices of an equilateral triangle then -
$$Z_1^2$$ + $$Z_2^2$$ + $$Z_3^2$$ = $${Z_1}{Z_2}$$ + $${Z_2}{Z_3}$$ + $${Z_3}{Z_1}$$" ]
2

### AIEEE 2003

If $$z$$ and $$\omega$$ are two non-zero complex numbers such that $$\left| {z\omega } \right| = 1$$ and $$Arg(z) - Arg(\omega ) = {\pi \over 2},$$ then $$\,\overline {z\,} \omega$$ is equal to
A
$$- i$$
B
1
C
- 1
D
$$i$$

## Explanation

Given that,
$$\left| {z\omega } \right| = 1$$
$$\Rightarrow$$ $$\left| z \right|\left| \omega \right|$$ = 1
$$\Rightarrow$$ $$\left| z \right|$$ = $${1 \over {\left| \omega \right|}}$$

and $$Arg(z) - Arg(\omega ) = {\pi \over 2}$$
$$\Rightarrow$$ $$Arg\left( {{z \over \omega }} \right)$$ $$= {\pi \over 2}$$

When argument of a complex number is $${\pi \over 2}$$, it means it is making an angle of $${\pi \over 2}$$ with the real axis in the counterclockwise, so it is along the imaginary axis and positive side of imaginary axis.

So, $${{z \over \omega }}$$ is a purely imaginary number that means there is no real part in this complex number.

So we can assume,
$${{z \over \omega }}$$ = $$ki$$

$$\Rightarrow$$ $${\left| {{z \over \omega }} \right|}$$ = $$\left| {ki} \right|$$

$$\Rightarrow$$ $${\left| {{z \over \omega }} \right|}$$ = $$\left| k \right|\left| i \right|$$

$$\Rightarrow$$ $${\left| {{z \over \omega }} \right|}$$ = $$k$$       [ as $$\left| i \right|$$ = 1 ]

$$\Rightarrow$$ $$\left| z \right|$$$$\times$$$${1 \over {\left| \omega \right|}}$$ = $$k$$

$$\Rightarrow$$ $$\left| z \right|$$$$\times$$ $$\left| z \right|$$ = $$k$$       [ as $${1 \over {\left| \omega \right|}}$$ = $$\left| z \right|$$ ]

$$\Rightarrow$$ $${\left| z \right|^2}$$ = $$k$$

$$\Rightarrow$$ $$\left| z \right|$$ = $$\sqrt k$$

$$\therefore$$ $$\left| \omega \right|$$ = $${1 \over {\sqrt k }}$$

As $${{z \over \omega }}$$ is imaginary so we can write,

$${{z \over \omega }}$$ = $$- {{\overline z } \over {\overline \omega }}$$
[ When $$z$$ is imaginary then $$z$$ = $$-\overline z$$ ]

$$\Rightarrow$$ $$\overline z \omega$$ = $$- z\overline \omega$$

$$\Rightarrow$$ $$\overline z \omega$$ = $$-{{z \over \omega }}$$.$$\overline \omega$$.$$\omega$$

$$\Rightarrow$$ $$\overline z \omega$$ = $$-{{z \over \omega }}$$.$${\left| \omega \right|^2}$$

$$\Rightarrow$$ $$\overline z \omega$$ = $$-ki$$.$${\left( {{1 \over {\sqrt k }}} \right)^2}$$

$$\Rightarrow$$ $$\overline z \omega$$ = $$-ki$$.$${1 \over k}$$

$$\Rightarrow$$ $$\overline z \omega$$ = $$-i$$

Method 2 :

Given that,
$$\left| {z\omega } \right| = 1$$
$$\Rightarrow$$ $$\left| z \right|\left| \omega \right|$$ = 1
$$\Rightarrow$$ $$\left| z \right|$$ = $${1 \over {\left| \omega \right|}}$$

and $$Arg(z) - Arg(\omega ) = {\pi \over 2}$$
$$\Rightarrow$$ $$Arg\left( {{z \over \omega }} \right)$$ $$= {\pi \over 2}$$

When argument of a complex number is $${\pi \over 2}$$, it means it is making an angle of $${\pi \over 2}$$ with the real axis in the counterclockwise, so it is along the imaginary axis and positive side of imaginary axis.

So, $${{z \over \omega }}$$ is a purely imaginary number that means there is no real part in this complex number.

So we can assume,
$${{z \over \omega }}$$ = $$ki$$

$$\Rightarrow$$ $${\left| {{z \over \omega }} \right|}$$ = $$\left| {ki} \right|$$

$$\Rightarrow$$ $${\left| {{z \over \omega }} \right|}$$ = $$\left| k \right|\left| i \right|$$

$$\Rightarrow$$ $${\left| {{z \over \omega }} \right|}$$ = $$k$$       [ as $$\left| i \right|$$ = 1 ]

$$\Rightarrow$$ $$\left| z \right|$$$$\times$$$${1 \over {\left| \omega \right|}}$$ = $$k$$

$$\Rightarrow$$ $$\left| z \right|$$$$\times$$ $$\left| z \right|$$ = $$k$$       [ as $${1 \over {\left| \omega \right|}}$$ = $$\left| z \right|$$ ]

$$\Rightarrow$$ $${\left| z \right|^2}$$ = $$k$$

$$\Rightarrow$$ $$\left| z \right|$$ = $$\sqrt k$$

$$\therefore$$ $$\left| \omega \right|$$ = $${1 \over {\sqrt k }}$$

(1) Magnitude of $$\overline z \omega$$

= $$\left| {\overline z } \right|\left| \omega \right|$$

= $$\left| z \right|\left| \omega \right|$$ [ as $$\left| z \right|$$ = $$\left| {\overline z } \right|$$ ]

= $$\sqrt k$$.$${{1 \over {\sqrt k }}}$$

= 1

$$\therefore$$ The distance from the origin of $${\overline z \omega }$$ is 1.

(2) Argument of $${\overline z \omega }$$ = $$Arg\left( {\overline z \omega } \right)$$

= $$Arg\left( {\overline z } \right) + Arg\left( \omega \right)$$

= $$-Arg\left( z \right) + Arg\left( \omega \right)$$

= $$- \left( {Arg\left( z \right) - Arg\left( \omega \right)} \right)$$

= $$- {\pi \over 2}$$

$$\therefore$$ $${\overline z \omega }$$ is at (0, -1) on the negative side of imaginary axis and making an angle of $${\pi \over 2}$$ clockwise.

$$\therefore$$ $${\overline z \omega }$$ = 0 + (-1)$$\times$$$$i$$ = $$-i$$
3

### AIEEE 2002

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.
4

### AIEEE 2002

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

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

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

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

$$\Rightarrow 12 < 4x$$

$$\Rightarrow x > 3$$

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

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