1
Numerical

### JEE Main 2021 (Online) 1st September Evening Shift

If for the complex numbers z satisfying | z $-$ 2 $-$ 2i | $\le$ 1, the maximum value of | 3iz + 6 | is attained at a + ib, then a + b is equal to ______________.

Correct Answer is 5

## Explanation

| z $-$ 2 $-$ 2i | $\le$ 1

| x + iy $-$ 2 $-$ 2i | $\le$ 1

|(x $-$ 2) + i(y $-$ 2)| $\le$ 1

(x $-$ 2)2 + (y $-$ 2)2 $\le$ 1

| 3iz + 6 |max at a + ib

$\left| {3i} \right|\left| {z + {6 \over {3i}}} \right|$

$3{\left| {z - 2i} \right|_{\max }}$

From figure maximum distance at 3 + 2i

a + ib = 3 + 2i = a + b = 3 + 2 = 5 Ans.
2
Numerical

### JEE Main 2021 (Online) 31st August Morning Shift

A point z moves in the complex plane such that $\arg \left( {{{z - 2} \over {z + 2}}} \right) = {\pi \over 4}$, then the minimum value of ${\left| {z - 9\sqrt 2 - 2i} \right|^2}$ is equal to _______________.

Correct Answer is 98

## Explanation

Let $z = x + iy$

$\arg \left( {{{x - 2 + iy} \over {x + 2 + iy}}} \right) = {\pi \over 4}$

$\arg (x - 2 + iy) - \arg (x + 2 + iy) = {\pi \over 4}$

${\tan ^{ - 1}}\left( {{y \over {x - 2}}} \right) - {\tan ^{ - 1}}\left( {{y \over {x + 2}}} \right) = {\pi \over 4}$

${{{y \over {x - 2}} - {y \over {x + 2}}} \over {1 + \left( {{y \over {x - 2}}} \right).\left( {{y \over {x + 2}}} \right)}} = \tan {\pi \over 4} = 1$

${{xy + 2y - xy + 2y} \over {{x^2} - 4 + {y^2}}} = 1$

$4y = {x^2} - 4 + {y^2}$

${x^2} + {y^2} - 4y - 4 = 0$

locus is a circle with center (0, 2) & radius = $2\sqrt 2$

min. value = ${(AP)^2} = {(OP - OA)^2}$

$= {\left( {9\sqrt 2 - 2\sqrt 2 } \right)^2}$

$= {\left( {7\sqrt 2 } \right)^2} = 98$
3
Numerical

### JEE Main 2021 (Online) 26th August Evening Shift

The least positive integer n such that ${{{{(2i)}^n}} \over {{{(1 - i)}^{n - 2}}}},i = \sqrt { - 1}$ is a positive integer, is ___________.

Correct Answer is 6

## Explanation

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

$= {{{{(2i)}^{{{n + 2} \over 2}}}} \over {{{( - 1)}^{{{n - 2} \over 2}}}}} = {{{2^{{{n + 2} \over 2}}};{i^{{{n + 2} \over 2}}}} \over {{{( - 1)}^{{{n - 2} \over 2}}}}}$

This is positive integer for n = 6
4
Numerical

### JEE Main 2021 (Online) 26th August Morning Shift

Let $z = {{1 - i\sqrt 3 } \over 2}$, $i = \sqrt { - 1}$. Then the value of $21 + {\left( {z + {1 \over z}} \right)^3} + {\left( {{z^2} + {1 \over {{z^2}}}} \right)^3} + {\left( {{z^3} + {1 \over {{z^3}}}} \right)^3} + .... + {\left( {{z^{21}} + {1 \over {{z^{21}}}}} \right)^3}$ is ______________.

Correct Answer is 13

## Explanation

$z = {{1 - \sqrt {3i} } \over 2} = {e^{ - i{\pi \over 3}}}$

${z^r} + {1 \over {{z^r}}} = 2\cos \left( { - {\pi \over 3}} \right)r = 2\cos {{r\pi } \over 3}$

$\Rightarrow 21 + \sum\limits_{r = 1}^{21} {{{\left( {{z^r} + {1 \over {{z^r}}}} \right)}^3} = 8\left( {{{\cos }^3}{{r\pi } \over 3}} \right) = 2\left( {\cos r\pi + 3\cos {{r\pi } \over 3}} \right)}$

$\Rightarrow 21 + {\left( {z + {1 \over 2}} \right)^3} + {\left( {{z^2} + {1 \over {{z^2}}}} \right)^3} + ....{\left( {{z^{21}} + {1 \over {{z^{21}}}}} \right)^3}$

$= 21 + \sum\limits_{r = 1}^{21} {{{\left( {{z^r} + {1 \over {{z^r}}}} \right)}^3}}$

$= 21 + \sum\limits_{r = 1}^{21} {\left( {2\cos r\pi + 6\cos {{r\pi } \over 3}} \right)}$

$= 21 - 2 - 6$

$= 13$

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