1

### JEE Main 2019 (Online) 9th January Morning Slot

Let $\alpha$ and $\beta$ be two roots of the equation x2 + 2x + 2 = 0 , then $\alpha ^{15}$ + $\beta ^{15}$ is equal to :
A
-256
B
512
C
-512
D
256

## Explanation

Given equation,

x2 + 2x + 2 = 0

$\therefore$  x = ${{ - 2 \pm \sqrt {4 - 4.1.2} } \over {2.1}}$

x = $-$ 1 $\pm$ i

$\therefore$  $\alpha$ = $-$ 1 + i

and $\beta$ = $-$ 1 $-$ i

Note :

x + iy = r (cos$\theta$ + isin$\theta$)

$\therefore$  (x + iy)n = rn (cosn$\theta$ + isinn$\theta$)

$\therefore$  $-$ 1 + i = $\sqrt 2$ [cos${{3\pi } \over 4}$ + isin${{3\pi } \over 4}$ ]

$\Rightarrow$  ($-$ 1 + i)15 = ${\left( {\sqrt 2 } \right)^{15}}$ [cos$\left( {{{15.3\pi } \over 4}} \right) + i\sin \left( {{{15.3\pi } \over 4}} \right)$]

And $-$1 $-$ i = $\sqrt 2$ $\left[ {\cos \left( { - {{3\pi } \over 4}} \right) + i\sin \left( { - {{3\pi } \over 4}} \right)} \right]$

= $\sqrt 2 \left[ {\cos {{3\pi } \over 4} - \sin {{3\pi } \over 4}} \right]$

$\therefore$  ($-$1 $-$ i)15 = ${\left( {\sqrt 2 } \right)^{15}}\left[ {\cos \left( {{{15.3\pi } \over 4}} \right) - i\sin \left( {{{15.3\pi } \over 4}} \right)} \right]$

Now

$\alpha$15 + $\beta$15

= ($-$1 + i)15 + ($-$ 1 $-$ i)15

=  ${\left( {\sqrt 2 } \right)^{15}}$ $\left[ {2\cos \left( {{{15.3\pi } \over 4}} \right)} \right]$

= ${\left( {\sqrt 2 } \right)^{15}}\left[ {2\cos \left( {11\pi + {\pi \over 4}} \right)} \right]$

= ${\left( {\sqrt 2 } \right)^{15}}\left[ {2\left( { - \cos {\pi \over 4}} \right)} \right]$

= ${\left( {\sqrt 2 } \right)^{15}} \times 2 \times - {1 \over {\sqrt 2 }}$

= $- {\left( {\sqrt 2 } \right)^{14}}.2$

= $-$ 27 $\times$ 2

= $-$ 28

= $-$ 256
2

### JEE Main 2019 (Online) 9th January Morning Slot

Let
A = $\left\{ {\theta \in \left( { - {\pi \over 2},\pi } \right):{{3 + 2i\sin \theta } \over {1 - 2i\sin \theta }}is\,purely\,imaginary} \right\}$
. Then the sum of the elements in A is :
A
${5\pi \over 6}$
B
$\pi$
C
${3\pi \over 4}$
D
${{2\pi } \over 3}$

## Explanation

Given complex number,

${{3 + 2i\sin \theta } \over {1 - 2i\sin \theta }}$

$= {{\left( {3 + 2i\sin \theta } \right)\left( {1 + 2i\sin \theta } \right)} \over {1 + 4{{\sin }^2}\theta }}$

$= {{3 + 6i\sin \theta + 2i\sin \theta - 4{{\sin }^2}\theta } \over {1 + 4{{\sin }^2}\theta }}$

$= {{\left( {3 - 4{{\sin }^2}\theta } \right) + i\left( {8\sin \theta } \right)} \over {1 + 4{{\sin }^2}\theta }}$

As complex number is purely imaginary, So real part of this complex number is zero.

$\therefore$   ${{3 - 4{{\sin }^2}\theta } \over {1 + 4{{\sin }^2}\theta }}$ = 0

$\Rightarrow$   $3 - 4{\sin ^2}\theta = 0$

$\Rightarrow$   $\sin \theta = \pm {{\sqrt 3 } \over 2}$

as   $\theta$ $\in$ $\left( { - {\pi \over 2},\pi } \right)$

$\therefore$   $\theta$ $=$ $-$ ${\pi \over 3},{\pi \over 3},{{2\pi } \over 3}$

$\therefore$   Sum of those values of A is

$= - {\pi \over 3} + {\pi \over 3} + {{2\pi } \over 3}$

$= {{2\pi } \over 3}$
3

### JEE Main 2019 (Online) 9th January Evening Slot

Let z0 be a root of the quadratic equation, x2 + 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

## Explanation

1 + x + x2 = 0

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

z0 = w, w2

Now

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

z = 3 + 6iw81 $-$ 3iw93      (w93 = w81 = 1)

$\Rightarrow$  z = 3 + 3i

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

### JEE Main 2019 (Online) 10th January Morning Slot

Let z1 and z2 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

## Explanation

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 $\theta$.