Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (Single Correct Answer)

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

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

Given quadratic equation,

$${Z^2} + aZ + b = 0$$

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 -

$$Z_1^2$$ + $$Z_2^2$$ + $$Z_3^2$$ = $${Z_1}{Z_2}$$ + $${Z_2}{Z_3}$$ + $${Z_3}{Z_1}$$" ]

$${Z^2} + aZ + b = 0$$

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.

[

$$Z_1^2$$ + $$Z_2^2$$ + $$Z_3^2$$ = $${Z_1}{Z_2}$$ + $${Z_2}{Z_3}$$ + $${Z_3}{Z_1}$$ ]

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.

[

"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

MCQ (Single Correct Answer)

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

$$i$$

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)$$ \times $$$$i$$ = $$-i$$

$$\left| {z\omega } \right| = 1$$

$$ \Rightarrow $$ $$\left| z \right|\left| \omega \right|$$ = 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 $$z$$ = $$-\overline z $$ ]

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

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

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

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

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

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

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

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

($$z,\,{z_1}\,\& \,{z_2}\,$$ are complex numbers) will be

A

an ellipse

B

a hyperbola

C

a circle

D

none of these

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 z_{1} from z_{3} is = $$r + a$$ and distance of z_{2} from z_{3} is = $$r + b$$

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

$$\therefore$$ Locus of z_{3} is a hyperbola.

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 z

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

$$\therefore$$ Locus of z

4

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

$${\mathop{\rm Re}\nolimits} (z) > 2$$

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

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

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

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

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

$$ \Rightarrow 12 < 4x$$

$$ \Rightarrow x > 3$$

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

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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Complex Numbers

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