1

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) = ${\pi \over 4}$
2

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 $\theta$.
3

JEE Main 2019 (Online) 10th January Evening Slot

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

Explanation

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

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

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

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

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

${\rm I}(z) = 0$ and ${\mathop{\rm Re}\nolimits} (z) < 0$
4

JEE Main 2019 (Online) 11th January Morning Slot

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

Explanation

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

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

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