Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geomatics Engineering Or Surveying

Engineering Mechanics

Hydrology

Transportation Engineering

Strength of Materials Or Solid Mechanics

Reinforced Cement Concrete

Steel Structures

Irrigation

Environmental Engineering

Engineering Mathematics

Structural Analysis

Geotechnical Engineering

Fluid Mechanics and Hydraulic Machines

General Aptitude

1

Let z_{0} be a root of the quadratic equation, x^{2} + x + 1 = 0, If z = 3 + 6iz$$_0^{81}$$ $$-$$ 3iz$$_0^{93}$$, then arg z is equal to :

A

$${\pi \over 4}$$

B

$${\pi \over 6}$$

C

$${\pi \over 3}$$

D

0

1 + x + x^{2} = 0

x = $${{ - 1 \pm \sqrt {1 - 4} } \over 2} = {{ - 1 \pm i\sqrt 3 } \over 2}$$

z_{0} = w, w^{2}

Now

z = 3 + 6iz$$_0^{81}$$ $$-$$ 3iz$$_0^{93}$$

z = 3 + 6iw^{81} $$-$$ 3iw^{93} (w^{93} = w^{81} = 1)

$$ \Rightarrow $$ z = 3 + 3i

then arg(z) = tan^{$$-$$1}$$\left( {{3 \over 3}} \right)$$ = tan^{$$-$$1} (1) = $${\pi \over 4}$$

x = $${{ - 1 \pm \sqrt {1 - 4} } \over 2} = {{ - 1 \pm i\sqrt 3 } \over 2}$$

z

Now

z = 3 + 6iz$$_0^{81}$$ $$-$$ 3iz$$_0^{93}$$

z = 3 + 6iw

$$ \Rightarrow $$ z = 3 + 3i

then arg(z) = tan

2

Let z_{1} and z_{2} be any two non-zero complex numbers such that $$3\left| {{z_1}} \right| = 4\left| {{z_2}} \right|.$$ If $$z = {{3{z_1}} \over {2{z_2}}} + {{2{z_2}} \over {3{z_1}}}$$ then -

A

$${\rm I}m\left( z \right) = 0$$

B

$$\left| z \right| = \sqrt {{17 \over 2}} $$

C

$$\left| z \right| =$$ $${1 \over 2}\sqrt {9 + 16{{\cos }^2}\theta } $$

D

Re(z) $$=$$ 0

Given, $$3\left| {{z_1}} \right| = 4\left| {{z_2}} \right|$$

$$ \Rightarrow $$ $${{\left| {{z_1}} \right|} \over {\left| {{z_2}} \right|}} = {4 \over 3}$$

$$ \Rightarrow $$ $${{\left| {3{z_1}} \right|} \over {\left| {2{z_2}} \right|}} = {4 \over 3} \times {3 \over 2} = 2$$

As we know, for any compled number

$${{3{z_1}} \over {2{z_2}}} = {{\left| {3{z_1}} \right|} \over {\left| {2{z_2}} \right|}}$$(cos$$\theta $$ + i sin$$\theta $$)

= 2(cos$$\theta $$ + i sin$$\theta $$)

$$ \therefore $$ $${{2{z_2}} \over {3{z_1}}}$$ = $${1 \over {2\left( {\cos \theta + i\sin \theta } \right)}}$$

= $${1 \over {2\left( {\cos \theta + i\sin \theta } \right)}} \times {{\left( {\cos \theta - i\sin \theta } \right)} \over {\left( {\cos \theta - i\sin \theta } \right)}}$$

= $${\left( {{1 \over 2}\cos \theta - {i \over 2}\sin \theta } \right)}$$

Now, given

$$z = {{3{z_1}} \over {2{z_2}}} + {{2{z_2}} \over {3{z_1}}}$$

= 2(cos$$\theta $$ + i sin$$\theta $$) + $${\left( {{1 \over 2}\cos \theta - {i \over 2}\sin \theta } \right)}$$

= $${{5 \over 2}\cos \theta + {3 \over 2}i\sin \theta }$$

So, |z| = $$\sqrt {{{25} \over 4}{{\cos }^2}\theta + {9 \over 4}{{\sin }^2}\theta } $$

= $${1 \over 2}\sqrt {9 + 16{{\cos }^2}\theta } $$

z is neither purely real nor purely imaginary and |z| depends on $$\theta $$.

$$ \Rightarrow $$ $${{\left| {{z_1}} \right|} \over {\left| {{z_2}} \right|}} = {4 \over 3}$$

$$ \Rightarrow $$ $${{\left| {3{z_1}} \right|} \over {\left| {2{z_2}} \right|}} = {4 \over 3} \times {3 \over 2} = 2$$

As we know, for any compled number

$${{3{z_1}} \over {2{z_2}}} = {{\left| {3{z_1}} \right|} \over {\left| {2{z_2}} \right|}}$$(cos$$\theta $$ + i sin$$\theta $$)

= 2(cos$$\theta $$ + i sin$$\theta $$)

$$ \therefore $$ $${{2{z_2}} \over {3{z_1}}}$$ = $${1 \over {2\left( {\cos \theta + i\sin \theta } \right)}}$$

= $${1 \over {2\left( {\cos \theta + i\sin \theta } \right)}} \times {{\left( {\cos \theta - i\sin \theta } \right)} \over {\left( {\cos \theta - i\sin \theta } \right)}}$$

= $${\left( {{1 \over 2}\cos \theta - {i \over 2}\sin \theta } \right)}$$

Now, given

$$z = {{3{z_1}} \over {2{z_2}}} + {{2{z_2}} \over {3{z_1}}}$$

= 2(cos$$\theta $$ + i sin$$\theta $$) + $${\left( {{1 \over 2}\cos \theta - {i \over 2}\sin \theta } \right)}$$

= $${{5 \over 2}\cos \theta + {3 \over 2}i\sin \theta }$$

So, |z| = $$\sqrt {{{25} \over 4}{{\cos }^2}\theta + {9 \over 4}{{\sin }^2}\theta } $$

= $${1 \over 2}\sqrt {9 + 16{{\cos }^2}\theta } $$

z is neither purely real nor purely imaginary and |z| depends on $$\theta $$.

3

Let $$z = {\left( {{{\sqrt 3 } \over 2} + {i \over 2}} \right)^5} + {\left( {{{\sqrt 3 } \over 2} - {i \over 2}} \right)^5}.$$ If R(z) and 1(z) respectively denote the real and imaginary parts of z, then -

A

R(z) = $$-$$
3

B

R(z) < 0 and I(z) > 0

C

I(z) = 0

D

R(z) > 0 and I(z) > 0

$$z = {\left( {{{\sqrt 3 + i} \over 2}} \right)^5} + {\left( {{{\sqrt 3 - i} \over 2}} \right)^5}$$

$$z = {\left( {{e^{i\pi /6}}} \right)^5} + {\left( {{e^{ - i\pi /6}}} \right)^5}$$

$$ = {e^{i5\pi /6}} + {e^{ - i5\pi /6}}$$

$$ = \cos {{5\pi } \over 6} + i{{\sin 5\pi } \over 6} + \cos \left( {{{ - 5\pi } \over 6}} \right) + i\sin \left( {{{ - 5\pi } \over 6}} \right)$$

$$ = 2\cos {{5\pi } \over 6} < 0$$

$${\rm I}(z) = 0$$ and $${\mathop{\rm Re}\nolimits} (z) < 0$$

$$z = {\left( {{e^{i\pi /6}}} \right)^5} + {\left( {{e^{ - i\pi /6}}} \right)^5}$$

$$ = {e^{i5\pi /6}} + {e^{ - i5\pi /6}}$$

$$ = \cos {{5\pi } \over 6} + i{{\sin 5\pi } \over 6} + \cos \left( {{{ - 5\pi } \over 6}} \right) + i\sin \left( {{{ - 5\pi } \over 6}} \right)$$

$$ = 2\cos {{5\pi } \over 6} < 0$$

$${\rm I}(z) = 0$$ and $${\mathop{\rm Re}\nolimits} (z) < 0$$

4

Let $${\left( { - 2 - {1 \over 3}i} \right)^3} = {{x + iy} \over {27}}\left( {i = \sqrt { - 1} } \right),\,\,$$ where x and y are real numbers, then y $$-$$ x equals :

A

$$-$$ 85

B

85

C

$$-$$ 91

D

91

$${\left( { - 2 - {i \over 3}} \right)^3} = - {{{{\left( {6 + i} \right)}^3}} \over {27}}$$

$$ = {{ - 198 - 107i} \over {27}} = {{x + iy} \over {27}}$$

Hence, $$y - x = 198 - 107 = 91$$

$$ = {{ - 198 - 107i} \over {27}} = {{x + iy} \over {27}}$$

Hence, $$y - x = 198 - 107 = 91$$

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