1

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

If |z $-$ 3 + 2i| $\le$ 4 then the difference between the greatest value and the least value of |z| is :
A
$2\sqrt {13}$
B
8
C
4 + $\sqrt {13}$
D
$\sqrt {13}$

## Explanation

$\left| {z - \left( {3 - 2i} \right)} \right| \le 4$ represents a circle whose center is (3, $-$2) and radius = 4.

$\left| z \right|$ = $\left| z -0\right|$  represents the distance of point 'z' from origin (0, 0)

Suppose RS is the normal of the circle passing through origin 'O' and G is its center (3, $-$2).

Here, OR is the least distance

and OS is in the greatest distance

OR = RG $-$ OG and OS = OG + GS . . . . .(1)

As, RG = GS = 4

OG = $\sqrt {{3^2} + \left( { - 2{)^2}} \right)}$ = $\sqrt {9 + 4}$ = $\sqrt {13}$

From (1), OR = 4 $-$ $\sqrt {13}$ and OS = 4 + $\sqrt {13}$

So, required difference = $\left( {4 + \sqrt {13} } \right)$ $-$ $\left( {4 - \sqrt {13} } \right)$

= $\sqrt {13} + \sqrt {13}$ = $2\sqrt {13}$
2

### JEE Main 2018 (Online) 16th April Morning Slot

The least positive integer n for which ${\left( {{{1 + i\sqrt 3 } \over {1 - i\sqrt 3 }}} \right)^n} = 1,$ is :
A
2
B
3
C
5
D
6

## Explanation

${\left( {{{1 + i\sqrt 3 } \over {1 - i\sqrt 3 }}} \right)^n} = 1$

$\Rightarrow \,\,\,\,\,{\left( {{{{{1 + i\sqrt 3 } \over 2}} \over {{{1 - i\sqrt 3 } \over 2}}}} \right)^n} = 1$

We know, $\omega$ = $-$ ${{1 - i\sqrt 3 } \over 2}$

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

$\Rightarrow$ $\,\,\,\,$ ${\left( {{{ - {\omega ^2}} \over { - \omega }}} \right)^n} = 1$

$\Rightarrow$ ($\omega$)n = 1 = $\omega$3

$\Rightarrow$ $\,\,\,\,$ n = 3
3

### 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
4

### 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

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

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