Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

If $$\,\omega = {z \over {z - {1 \over 3}i}}\,$$ and $$\left| \omega \right| = 1$$, then $$z$$ lies on

A

an ellipse

B

a circle

C

a straight line

D

a parabola

Given $$\,\omega = {z \over {z - {1 \over 3}i}}\,$$ and $$\left| \omega \right| = 1$$

$$\therefore$$ $${{\left| z \right|} \over {\left| {z - {1 \over {\sqrt 3 }}i} \right|}} = \left| \omega \right|$$

$$ \Rightarrow $$ $${{\left| z \right|} \over {\left| {z - {1 \over {\sqrt 3 }}i} \right|}} = 1$$

$$ \Rightarrow $$ $$\left| z \right| = \left| {z - {1 \over {\sqrt 3 }}i} \right|$$ ..........equation (1)

$$\left| z \right|$$ represent distance of $$z$$ from point (0, 0) and

$$\left| {z - {1 \over {\sqrt 3 }}i} \right|$$ represent distance of $$z$$ from point $$\left( {0,{1 \over {\sqrt 3 }}} \right)$$.

According to the equation (1) the distance of $$z$$ from point (0, 0) and $$\left( {0,{1 \over {\sqrt 3 }}} \right)$$ is equal. Only if z is on a straight line then it will be equal distance from the both the points.

$$\therefore$$ $${{\left| z \right|} \over {\left| {z - {1 \over {\sqrt 3 }}i} \right|}} = \left| \omega \right|$$

$$ \Rightarrow $$ $${{\left| z \right|} \over {\left| {z - {1 \over {\sqrt 3 }}i} \right|}} = 1$$

$$ \Rightarrow $$ $$\left| z \right| = \left| {z - {1 \over {\sqrt 3 }}i} \right|$$ ..........equation (1)

$$\left| z \right|$$ represent distance of $$z$$ from point (0, 0) and

$$\left| {z - {1 \over {\sqrt 3 }}i} \right|$$ represent distance of $$z$$ from point $$\left( {0,{1 \over {\sqrt 3 }}} \right)$$.

According to the equation (1) the distance of $$z$$ from point (0, 0) and $$\left( {0,{1 \over {\sqrt 3 }}} \right)$$ is equal. Only if z is on a straight line then it will be equal distance from the both the points.

2

MCQ (Single Correct Answer)

If $$\,\left| {{z^2} - 1} \right| = {\left| z \right|^2} + 1$$, then z lies on

A

an ellipse

B

the imaginary axis

C

a circle

D

the real axis

Given $$\,\left| {{z^2} - 1} \right| = {\left| z \right|^2} + 1$$,

By squaring both sides we get,

$${\left| {{z^2} - 1} \right|^2}$$ = $${\left( {{{\left| z \right|}^2} + 1} \right)^2}$$

$$ \Rightarrow $$ $$\left( {{z^2} - 1} \right)$$$$\overline {\left( {{z^2} - 1} \right)} $$ = $${\left( {{{\left| z \right|}^2} + 1} \right)^2}$$ [ as $${{{\left| z \right|}^2}}$$ = $$z\overline z $$ ]

$$ \Rightarrow $$ $$\left( {{z^2} - 1} \right)$$$$\left( {{{\left( {\overline z } \right)}^2} - 1} \right)$$ = $${\left( {{{\left| z \right|}^2} + 1} \right)^2}$$

$$ \Rightarrow $$ $${\left( {z\overline z } \right)^2}$$ $$-$$ $${{z^2}}$$ $$-$$$${{{\left( {\overline z } \right)}^2}}$$ $$+$$ 1 = $${\left| z \right|^4}$$ $$+$$ 2$${{{\left| z \right|}^2}}$$ $$+$$ 1

$$ \Rightarrow $$ $${\left| z \right|^4}$$ $$-$$ $${{z^2}}$$ $$-$$$${{{\left( {\overline z } \right)}^2}}$$ $$+$$ 1 = $${\left| z \right|^4}$$ $$+$$ 2$${{{\left| z \right|}^2}}$$ $$+$$ 1

$$ \Rightarrow $$ $${{z^2}}$$ $$+$$$${{{\left( {\overline z } \right)}^2}}$$ $$+$$ 2$${z\overline z }$$ = 0

$$ \Rightarrow $$ $${\left( {z + \overline z } \right)^2}$$ = 0

$$ \Rightarrow $$ $${z + \overline z }$$ = 0

$$ \Rightarrow $$ $$z$$ = $$-$$ $${\overline z }$$

If $$z$$ = x + iy

then $${\overline z }$$ = x - iy

$$\therefore$$ x + iy = - (x - iy)

$$ \Rightarrow $$ x + iy = - x + iy

$$ \Rightarrow $$ x = 0

$$\therefore$$ z is purely imaginary.

So, it is lie on the imaginary axis.

By squaring both sides we get,

$${\left| {{z^2} - 1} \right|^2}$$ = $${\left( {{{\left| z \right|}^2} + 1} \right)^2}$$

$$ \Rightarrow $$ $$\left( {{z^2} - 1} \right)$$$$\overline {\left( {{z^2} - 1} \right)} $$ = $${\left( {{{\left| z \right|}^2} + 1} \right)^2}$$ [ as $${{{\left| z \right|}^2}}$$ = $$z\overline z $$ ]

$$ \Rightarrow $$ $$\left( {{z^2} - 1} \right)$$$$\left( {{{\left( {\overline z } \right)}^2} - 1} \right)$$ = $${\left( {{{\left| z \right|}^2} + 1} \right)^2}$$

$$ \Rightarrow $$ $${\left( {z\overline z } \right)^2}$$ $$-$$ $${{z^2}}$$ $$-$$$${{{\left( {\overline z } \right)}^2}}$$ $$+$$ 1 = $${\left| z \right|^4}$$ $$+$$ 2$${{{\left| z \right|}^2}}$$ $$+$$ 1

$$ \Rightarrow $$ $${\left| z \right|^4}$$ $$-$$ $${{z^2}}$$ $$-$$$${{{\left( {\overline z } \right)}^2}}$$ $$+$$ 1 = $${\left| z \right|^4}$$ $$+$$ 2$${{{\left| z \right|}^2}}$$ $$+$$ 1

$$ \Rightarrow $$ $${{z^2}}$$ $$+$$$${{{\left( {\overline z } \right)}^2}}$$ $$+$$ 2$${z\overline z }$$ = 0

$$ \Rightarrow $$ $${\left( {z + \overline z } \right)^2}$$ = 0

$$ \Rightarrow $$ $${z + \overline z }$$ = 0

$$ \Rightarrow $$ $$z$$ = $$-$$ $${\overline z }$$

If $$z$$ = x + iy

then $${\overline z }$$ = x - iy

$$\therefore$$ x + iy = - (x - iy)

$$ \Rightarrow $$ x + iy = - x + iy

$$ \Rightarrow $$ x = 0

$$\therefore$$ z is purely imaginary.

So, it is lie on the imaginary axis.

3

MCQ (Single Correct Answer)

If $$z = x - iy$$ and $${z^{{1 \over 3}}} = p + iq$$, then

$${{\left( {{x \over p} + {y \over q}} \right)} \over {\left( {{p^2} + {q^2}} \right)}}$$ is equal to

$${{\left( {{x \over p} + {y \over q}} \right)} \over {\left( {{p^2} + {q^2}} \right)}}$$ is equal to

A

- 2

B

- 1

C

2

D

1

Given $${z^{{1 \over 3}}} = p + iq$$

$$ \Rightarrow $$ z = (p + iq)^{3}

= p^{3} + (iq)^{3} +3p(iq)(p + iq)

= p^{3} - iq^{3} +3ip^{2}q - 3pq^{2}

= p(p^{2} - 3q^{2}) - iq(q^{2} - 3p^{2})

Given that $$z = x - iy$$

$$\therefore$$ $$x - iy$$ = p(p^{2} - 3q^{2}) - iq(q^{2} - 3p^{2})

By comparing both sides we get,

$${x \over p} = {p^2} - 3{q^2}$$ and $${y \over q} = {q^2} - 3{p^2}$$

$$\therefore$$ $${{\left( {{x \over p} + {y \over q}} \right)} \over {\left( {{p^2} + {q^2}} \right)}}$$

= $${{{p^2} - 3{q^2} + {q^2} - 3{p^2}} \over {{p^2} + {q^2}}}$$

= $${{ - 2{q^2} - 2{p^2}} \over {{p^2} + {q^2}}}$$

= $${{ - 2\left( {{q^2} + {p^2}} \right)} \over {{p^2} + {q^2}}}$$

= $$-2$$

$$ \Rightarrow $$ z = (p + iq)

= p

= p

= p(p

Given that $$z = x - iy$$

$$\therefore$$ $$x - iy$$ = p(p

By comparing both sides we get,

$${x \over p} = {p^2} - 3{q^2}$$ and $${y \over q} = {q^2} - 3{p^2}$$

$$\therefore$$ $${{\left( {{x \over p} + {y \over q}} \right)} \over {\left( {{p^2} + {q^2}} \right)}}$$

= $${{{p^2} - 3{q^2} + {q^2} - 3{p^2}} \over {{p^2} + {q^2}}}$$

= $${{ - 2{q^2} - 2{p^2}} \over {{p^2} + {q^2}}}$$

= $${{ - 2\left( {{q^2} + {p^2}} \right)} \over {{p^2} + {q^2}}}$$

= $$-2$$

4

MCQ (Single Correct Answer)

Let z and w be complex numbers such that $$\overline z + i\overline w = 0$$ and arg zw = $$\pi $$. Then arg z equals

A

$${{5\pi } \over 4}$$

B

$${{\pi } \over 2}$$

C

$${{3\pi } \over 4}$$

D

$${{\pi } \over 4}$$

Given $$\overline z + i\overline w = 0$$

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

$$ \Rightarrow \overline{\overline z} = - \overline {i\overline w } $$

$$ \Rightarrow \overline{\overline z} = - \overline i \overline{\overline w} $$

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

$$ \Rightarrow z = - \left( { - i} \right)w$$

$$ \Rightarrow z = iw$$

Now given that Arg(zw) = $$\pi $$

$$ \Rightarrow $$ Arg(z$$ \times $$$${z \over i}$$) = $$\pi $$

$$ \Rightarrow $$ Arg(z^{2}) - Arg(i) = $$\pi $$

$$ \Rightarrow $$ 2Arg(z) - $${\pi \over 2}$$ = $$\pi $$

[ $$i$$ complex number represent (0, 1) point on imaginary axis and Arg($$i$$) means the angle made by the point (0, 1) with real axis which is $${\pi \over 2}$$]

$$ \Rightarrow $$ 2Arg(z) = $${{3\pi } \over 2}$$

$$ \Rightarrow $$ Arg(z) = $${{3\pi } \over 4}$$

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

$$ \Rightarrow \overline{\overline z} = - \overline {i\overline w } $$

$$ \Rightarrow \overline{\overline z} = - \overline i \overline{\overline w} $$

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

$$ \Rightarrow z = - \left( { - i} \right)w$$

$$ \Rightarrow z = iw$$

Now given that Arg(zw) = $$\pi $$

$$ \Rightarrow $$ Arg(z$$ \times $$$${z \over i}$$) = $$\pi $$

$$ \Rightarrow $$ Arg(z

$$ \Rightarrow $$ 2Arg(z) - $${\pi \over 2}$$ = $$\pi $$

[ $$i$$ complex number represent (0, 1) point on imaginary axis and Arg($$i$$) means the angle made by the point (0, 1) with real axis which is $${\pi \over 2}$$]

$$ \Rightarrow $$ 2Arg(z) = $${{3\pi } \over 2}$$

$$ \Rightarrow $$ Arg(z) = $${{3\pi } \over 4}$$

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Complex Numbers

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