1

### 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 ${\mathop{\rm Re}\nolimits} (z) < 0$
2

### 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, $y - x = 198 - 107 = 91$
3

### JEE Main 2019 (Online) 11th January Evening Slot

Let z be a complex number such that |z| + z = 3 + i (where i = $\sqrt { - 1}$). Then |z| is equal to
A
${{\sqrt {34} } \over 3}$
B
${5 \over 3}$
C
${5 \over 4}$
D
${{\sqrt {41} } \over 4}$

## Explanation

$\left| z \right| + z = 3 + i$

$z = 3 - \left| z \right| + i$

Let  $3 - \left| z \right| = a \Rightarrow \left| z \right| = \left( {3 - a} \right)$

$\Rightarrow z = a + i \Rightarrow \left| z \right| = \sqrt {{a^2} + 1}$

$\Rightarrow 9 + {a^2} - 6a = {a^2} + 1 \Rightarrow a = {8 \over 6} = {4 \over 3}$

$\Rightarrow \left| z \right| = 3 - {4 \over 3} = {5 \over 3}$
4

### JEE Main 2019 (Online) 12th January Morning Slot

If ${{z - \alpha } \over {z + \alpha }}\left( {\alpha \in R} \right)$ is a purely imaginary number and | z | = 2, then a value of $\alpha$ is :
A
${1 \over 2}$
B
$\sqrt 2$
C
2
D
1

## Explanation

${{z - \alpha } \over {z + \alpha }} + {{\overline z - \alpha } \over {\overline z + \alpha }} = 0$

$z\overline z + z\alpha - \alpha \overline z - {\alpha ^2} + z\overline z - z\alpha + \overline z \alpha - {\alpha ^2} = 0$

${\left| z \right|^2} = {\alpha ^2},$  $a = \pm 2$