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 ______________.
Your Input ________

Answer

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 _______________.
Your Input ________

Answer

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 ___________.
Your Input ________

Answer

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 ______________.
Your Input ________

Answer

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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