1

### JEE Main 2016 (Online) 9th April Morning Slot

The point represented by 2 + i in the Argand plane moves 1 unit eastwards, then 2 units northwards and finally from there $2\sqrt 2$ units in the south-westwardsdirection. Then its new position in the Argand plane is at the point represented by :
A
2 + 2i
B
1 + i
C
$-$1 $-$ i
D
$-$2 $-$2i

## Explanation

Here,

z $-$ (3 + 3i) = $2\sqrt 2$ (cos($-$135o) + i sin ($-$ 135o))

= $2\sqrt 2$ ($-$ ${1 \over {\sqrt 2 }}$ $-$${i \over {\sqrt 2 }}$)

= $-$ 2 $-$ 2i

$\Rightarrow$   z = 3 + 3 i $-$ 2 $-$ 2 i = 1 + i

Note :

Polar form of a complex number :

z = r (cos$\theta$ + i sin$\theta$)

Here r = modulus of z and $\theta$ argument of z.
2

### JEE Main 2017 (Offline)

Let $\omega$ be a complex number such that 2$\omega$ + 1 = z where z = $\sqrt {-3}$. If

$\left| {\matrix{ 1 & 1 & 1 \cr 1 & { - {\omega ^2} - 1} & {{\omega ^2}} \cr 1 & {{\omega ^2}} & {{\omega ^7}} \cr } } \right| = 3k$,

then k is equal to
A
z
B
-1
C
1
D
-z

## Explanation

Given 2$\omega$ + 1 = z;

z = $\sqrt 3 i$

$\Rightarrow$ $\omega = {{\sqrt 3 i - 1} \over 2}$

$\Rightarrow$ As $\omega$ is complex cube root of unity.

${\omega ^3} = 1$

$1 + \omega + {\omega ^2} = 0$

$\left| {\matrix{ 1 & 1 & 1 \cr 1 & { - {\omega ^2} - 1} & {{\omega ^2}} \cr 1 & {{\omega ^2}} & {{\omega ^7}} \cr } } \right| = 3k$

$\Rightarrow$ $\left| {\matrix{ 1 & 1 & 1 \cr 1 & \omega & {{\omega ^2}} \cr 1 & {{\omega ^2}} & \omega \cr } } \right| = 3k$

Applying R1 $\to$ R1 + R2 + R3

$\Rightarrow$ $\left| {\matrix{ 3 & 0 & 0 \cr 1 & \omega & {{\omega ^2}} \cr 1 & {{\omega ^2}} & \omega \cr } } \right| = 3k$

$\Rightarrow$ $3\left( {{\omega ^2} - {\omega ^4}} \right) = 3k$

$\Rightarrow$ $\left( {{\omega ^2} - \omega } \right) = k$

$\therefore$ $k = \left( {{{ - 1 - \sqrt 3 i} \over 2}} \right) - \left( {{{ - 1 + \sqrt 3 i} \over 2}} \right)$

$\Rightarrow$ k = ${ - \sqrt 3 i}$ = -z
3

### JEE Main 2018 (Offline)

If $\alpha ,\beta \in C$ are the distinct roots of the equation
x2 - x + 1 = 0, then ${\alpha ^{101}} + {\beta ^{107}}$ is equal to
A
2
B
-1
C
0
D
1

## Explanation

Given equation,

x2 $-$ x + 1 = 0

Roots of this equation

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

$\therefore\,\,\,$ $\propto \, = \,{{1 + \sqrt 3 \,i} \over 2}$

and $\beta = \,{{1 - \sqrt 3 \,i} \over 2}$

We know;

$\omega = {{ - 1 + \sqrt 3 \,i} \over 2} = - \left( {{{1 - \sqrt 3 \,i} \over 2}} \right) = - \beta$

and ${\omega ^2} = {{ - 1 - \sqrt 3 \,i} \over 2} = - \left( {{{1 + \sqrt 3 \,i} \over 2}} \right) = - \propto$

$\therefore\,\,\,$ $\propto \, = - {\omega ^2}$ and $\beta \, = \, - \omega$

$\therefore\,\,\,$ ${ \propto ^{101}} + {\beta ^{107}}$

$= {\left( { - {\omega ^2}} \right)^{101}} + {\left( { - \omega } \right)^{107}}$

$= {\left( { - 1} \right)^{101}}.{\left( {{\omega ^2}} \right)^{101}} + {\left( { - 1} \right)^{107}}.{\left( \omega \right)^{107}}$

$= - 1.{\left( {{\omega ^2}} \right)^{101}} - {\omega ^{107}}$

$= - \left( {{\omega ^{202}} + {\omega ^{107}}} \right)$

$= - \left( {{\omega ^{3.67}}.\omega + {\omega ^{3.35}}.{\omega ^2}} \right)$

$= - \left( {\omega + {\omega ^2}} \right)\,\,\,$ [ as $\,\,\,$ ${\omega ^{3n}} = 1$]

$= - \left( { - 1} \right)$ $\,\,\,\,\,\,$ [as $\,\,\,$ $1 + \omega + {\omega ^2} = 0$ ]

$= 1$
4

### JEE Main 2018 (Online) 15th April Morning Slot

The set of all $\alpha$ $\in$ R, for which w = ${{1 + \left( {1 - 8\alpha } \right)z} \over {1 - z}}$ is purely imaginary number, for all z $\in$ C satisfying |z| = 1 and Re z $\ne$ 1, is :
A
an empty set
B
{0}
C
$\left\{ {0,{1 \over 4}, - {1 \over 4}} \right\}$
D
equal to R

## Explanation

As w = ${{1 + \left( {1 - 8\alpha } \right)z} \over {1 - z}}$, w is purely imaginary

$\therefore w$ + $\bar w$ = 0

$\Rightarrow$ ${{1 + \left( {1 - 8\alpha } \right)z} \over {1 - z}}$ + ${{1 + \left( {1 - 8\alpha } \right)\bar z} \over {1 - \bar z}}$ = 0

$\Rightarrow$ [1 + (1 - 8$\alpha$)][1 - $\bar z$] + [1 + ( 1 - 8$\alpha$)][1 - z] = 0

$\Rightarrow$ 2 - (z + $\bar z$) + (1 - 8$\alpha$)(z + $\bar z$) - 2(1 - 8$\alpha$) = 0

$\Rightarrow$ 2 - (z + $\bar z$) + (z + $\bar z$) - 8$\alpha$(z + $\bar z$) - 2 + 16$\alpha$ = 0

$\Rightarrow$ 16$\alpha$ = 8$\alpha$(z + $\bar z)$

z + $\bar z$ = 2 or $\alpha$ = 0

but z + $\bar z$ = 2 is not possible as Re(Z) $\ne$ 1

$\therefore$ $\alpha$ = 0

$\therefore$ $\alpha$ $\in$ {0}