Complex Numbers
197 Questions
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 28th June Morning Shift
The number of elements in the set {z = a + ib $\in$ C : a, b $\in$ Z and 1 < | z $-$ 3 + 2i | < 4} is __________.
Correct Answer: 40
Explanation:
at line $y=-2$, we have $(5,-2)(6,-2)(1,-2)(0,-2)$ $\Rightarrow 4$ points
at line $y=-1$, we have $(4,-1)(5,-1)(6,-1)(2,-1)$ $(1,-1)(0,-1) \Rightarrow 6$ points
at line $y=0$, we have $(0,0)(1,0)(2,0)(3,0)(4,0)$ $(5,0)(6,0) \Rightarrow 7$ points
at line $y=1$, we have $(1,1),(2,1),(3,1),(4,1),(5,1)$ i.e. 5 points
symmetrically
at line $y=-5$, we have 5 points
at line $y=-4$, we have 7 points
at line $y=-3$, we have 6 points
So Total integral points $=2(5+7+6)+4$
$ =40 $
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 26th June Evening Shift
If ${z^2} + z + 1 = 0$, $z \in C$, then
$\left| {\sum\limits_{n = 1}^{15} {{{\left( {{z^n} + {{( - 1)}^n}{1 \over {{z^n}}}} \right)}^2}} } \right|$ is equal to _________.
Correct Answer: 2
Explanation:
$\because$ ${z^2} + z + 1 = 0$
$\Rightarrow$ $\omega$ or $\omega$2
$\because$ $\left| {\sum\limits_{n = 1}^{15} {{{\left( {{z^n} + {{( - 1)}^n}{1 \over {{z^n}}}} \right)}^2}} } \right|$
$ = \left| {\sum\limits_{n = 1}^{15} {{z^{2n}} + \sum\limits_{n = 1}^{15} {{z^{ - 2n}} + 2\,.\,\sum\limits_{n = 1}^{15} {{{( - 1)}^n}} } } } \right|$
$ = \left| {0 + 0 - 2} \right|$
$ = 2$
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 24th June Evening Shift
Let S = {z $\in$ C : |z $-$ 3| $\le$ 1 and z(4 + 3i) + $\overline z $(4 $-$ 3i) $\le$ 24}. If $\alpha$ + i$\beta$ is the point in S which is closest to 4i, then 25($\alpha$ + $\beta$) is equal to ___________.
Correct Answer: 80
Explanation:
Here $|z - 3| < 1$
$ \Rightarrow {(x - 3)^2} + {y^2} < 1$
and $z = (4 + 3i) + \overline z (4 - 3i) \le 24$
$ \Rightarrow 4x - 3y \le 12$
$\tan \theta = {4 \over 3}$
$\therefore$ Coordinate of $P = (3 - \cos \theta ,\sin \theta )$
$ = \left( {3 - {3 \over 5},{4 \over 5}} \right)$
$\therefore$ $\alpha + i\beta = {{12} \over 5} + {4 \over 5}i$
$\therefore$ $25(\alpha + \beta ) = 80$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 31st August Evening Shift
If z is a complex number such that ${{z - i} \over {z - 1}}$ is purely imaginary, then the minimum value of | z $-$ (3 + 3i) | is :
A.
$2\sqrt 2 - 1$
B.
$3\sqrt 2 $
C.
$6\sqrt 2 $
D.
$2\sqrt 2 $
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 27th August Morning Shift
If $S = \left\{ {z \in C:{{z - i} \over {z + 2i}} \in R} \right\}$, then :
A.
S contains exactly two elements
B.
S contains only one element
C.
S is a circle in the complex plane
D.
S is a straight line in the complex plane
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th August Evening Shift
If ${\left( {\sqrt 3 + i} \right)^{100}} = {2^{99}}(p + iq)$, then p and q are roots of the equation :
A.
${x^2} - \left( {\sqrt 3 - 1} \right)x - \sqrt 3 = 0$
B.
${x^2} + \left( {\sqrt 3 + 1} \right)x + \sqrt 3 = 0$
C.
${x^2} + \left( {\sqrt 3 - 1} \right)x - \sqrt 3 = 0$
D.
${x^2} - \left( {\sqrt 3 + 1} \right)x + \sqrt 3 = 0$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th August Morning Shift
The equation $\arg \left( {{{z - 1} \over {z + 1}}} \right) = {\pi \over 4}$ represents a circle with :
A.
centre at (0, $-$1) and radius $\sqrt 2 $
B.
centre at (0, 1) and radius $\sqrt 2 $
C.
centre (0, 0) and radius $\sqrt 2 $
D.
centre at (0, 1) and radius 2
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 27th July Evening Shift
Let C be the set of all complex numbers. Let
S1 = {z$\in$C : |z $-$ 2| $\le$ 1} and
S2 = {z$\in$C : z(1 + i) + $\overline z $(1 $-$ i) $\ge$ 4}.
Then, the maximum value of ${\left| {z - {5 \over 2}} \right|^2}$ for z$\in$S1 $\cap$ S2 is equal to :
S1 = {z$\in$C : |z $-$ 2| $\le$ 1} and
S2 = {z$\in$C : z(1 + i) + $\overline z $(1 $-$ i) $\ge$ 4}.
Then, the maximum value of ${\left| {z - {5 \over 2}} \right|^2}$ for z$\in$S1 $\cap$ S2 is equal to :
A.
${{3 + 2\sqrt 2 } \over 4}$
B.
${{5 + 2\sqrt 2 } \over 2}$
C.
${{3 + 2\sqrt 2 } \over 2}$
D.
${{5 + 2\sqrt 2 } \over 4}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 27th July Morning Shift
Let C be the set of all complex numbers. Let
${S_1} = \{ z \in C||z - 3 - 2i{|^2} = 8\} $
${S_2} = \{ z \in C|{\mathop{\rm Re}\nolimits} (z) \ge 5\} $ and
${S_3} = \{ z \in C||z - \overline z | \ge 8\} $.
Then the number of elements in ${S_1} \cap {S_2} \cap {S_3}$ is equal to :
${S_1} = \{ z \in C||z - 3 - 2i{|^2} = 8\} $
${S_2} = \{ z \in C|{\mathop{\rm Re}\nolimits} (z) \ge 5\} $ and
${S_3} = \{ z \in C||z - \overline z | \ge 8\} $.
Then the number of elements in ${S_1} \cap {S_2} \cap {S_3}$ is equal to :
A.
1
B.
0
C.
2
D.
Infinite
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 22th July Evening Shift
Let n denote the number of solutions of the equation z2 + 3$\overline z $ = 0, where z is a complex number. Then the value of $\sum\limits_{k = 0}^\infty {{1 \over {{n^k}}}} $ is equal to :
A.
1
B.
${4 \over 3}$
C.
${3 \over 2}$
D.
2
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 20th July Morning Shift
If z and $\omega$ are two complex numbers such that $\left| {z\omega } \right| = 1$ and $\arg (z) - \arg (\omega ) = {{3\pi } \over 2}$, then $\arg \left( {{{1 - 2\overline z \omega } \over {1 + 3\overline z \omega }}} \right)$ is :
(Here arg(z) denotes the principal argument of complex number z)
(Here arg(z) denotes the principal argument of complex number z)
A.
${\pi \over 4}$
B.
$ - {{3\pi } \over 4}$
C.
$ - {\pi \over 4}$
D.
${{3\pi } \over 4}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 18th March Evening Shift
Let a complex number be w = 1 $-$ ${\sqrt 3 }$i. Let another complex number z be such that |zw| = 1 and arg(z) $-$ arg(w) = ${\pi \over 2}$. Then the area of the triangle with vertices origin, z and w is equal to :
A.
4
B.
${1 \over 4}$
C.
2
D.
${1 \over 2}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 18th March Morning Shift
If the equation $a|z{|^2} + \overline {\overline \alpha z + \alpha \overline z } + d = 0$ represents a circle where a, d are real constants then which of the following condition is correct?
A.
|$\alpha$|2 $-$ ad $\ne$ 0
B.
|$\alpha$|2 $-$ ad > 0 and a$\in$R $-$ {0}
C.
|$\alpha$|2 $-$ ad $ \ge $ 0 and a$\in$R
D.
$\alpha$ = 0, a, d$\in$R+
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 17th March Evening Shift
Let S1, S2 and S3 be three sets defined as
S1 = {z$\in$C : |z $-$ 1| $ \le $ $\sqrt 2 $}
S2 = {z$\in$C : Re((1 $-$ i)z) $ \ge $ 1}
S3 = {z$\in$C : Im(z) $ \le $ 1}
Then the set S1 $\cap$ S2 $\cap$ S3 :
S1 = {z$\in$C : |z $-$ 1| $ \le $ $\sqrt 2 $}
S2 = {z$\in$C : Re((1 $-$ i)z) $ \ge $ 1}
S3 = {z$\in$C : Im(z) $ \le $ 1}
Then the set S1 $\cap$ S2 $\cap$ S3 :
A.
has exactly three elements
B.
is a singleton
C.
has infinitely many elements
D.
has exactly two elements
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 17th March Morning Shift
The area of the triangle with vertices A(z), B(iz) and C(z + iz) is :
A.
1
B.
${1 \over 2}$| z |2
C.
${1 \over 2}$| z + iz |2
D.
${1 \over 2}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 16th March Evening Shift
The least value of |z| where z is complex number which satisfies the inequality $\exp \left( {{{(|z| + 3)(|z| - 1)} \over {||z| + 1|}}{{\log }_e}2} \right) \ge {\log _{\sqrt 2 }}|5\sqrt 7 + 9i|,i = \sqrt { - 1} $, is equal to :
A.
8
B.
3
C.
2
D.
$\sqrt 5 $
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 16th March Morning Shift
Let a complex number z, |z| $\ne$ 1,
satisfy ${\log _{{1 \over {\sqrt 2 }}}}\left( {{{|z| + 11} \over {{{(|z| - 1)}^2}}}} \right) \le 2$. Then, the largest value of |z| is equal to ____________.
satisfy ${\log _{{1 \over {\sqrt 2 }}}}\left( {{{|z| + 11} \over {{{(|z| - 1)}^2}}}} \right) \le 2$. Then, the largest value of |z| is equal to ____________.
A.
5
B.
8
C.
6
D.
7
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Evening Shift
If $\alpha$, $\beta$ $\in$ R are such that 1 $-$ 2i (here i2 = $-$1) is a root of z2 + $\alpha$z + $\beta$ = 0, then ($\alpha$ $-$ $\beta$) is equal to :
A.
$-$7
B.
7
C.
3
D.
$-$3
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Morning Shift
Let the lines (2 $-$ i)z = (2 + i)$\overline z $ and (2 $+$ i)z + (i $-$ 2)$\overline z $ $-$ 4i = 0, (here i2 = $-$1) be normal to a circle C. If the line iz + $\overline z $ + 1 + i = 0 is tangent to this circle C, then its radius is :
A.
${3 \over {2\sqrt 2 }}$
B.
$3\sqrt 2 $
C.
${1 \over {2\sqrt 2 }}$
D.
${3 \over {\sqrt 2 }}$
2021
JEE Mains
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: 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.
| 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.
2021
JEE Mains
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: 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$
$\arg \left( {{{x - 2 + iy} \over {x + 2 + iy}}} \right) = {\pi \over 4}$
$\arg (x - 2 + iy) - \arg (x + 2 + iy) = {\pi \over 4}$
${\tan ^{ - 1}}\left( {{y \over {x - 2}}} \right) - {\tan ^{ - 1}}\left( {{y \over {x + 2}}} \right) = {\pi \over 4}$
${{{y \over {x - 2}} - {y \over {x + 2}}} \over {1 + \left( {{y \over {x - 2}}} \right).\left( {{y \over {x + 2}}} \right)}} = \tan {\pi \over 4} = 1$
${{xy + 2y - xy + 2y} \over {{x^2} - 4 + {y^2}}} = 1$
$4y = {x^2} - 4 + {y^2}$
${x^2} + {y^2} - 4y - 4 = 0$
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$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th August Evening Shift
Let z1 and z2 be two complex numbers such that $\arg ({z_1} - {z_2}) = {\pi \over 4}$ and z1, z2 satisfy the equation | z $-$ 3 | = Re(z). Then the imaginary part of z1 + z2 is equal to ___________.
Correct Answer: 6
Explanation:
Let z1 = x1 + iy ; z2 = x2 + iy2
z1 $-$ z2 = (x1 $-$ x2) + i(y1 $-$ y2)
$\therefore$ $\arg ({z_1} - {z_2}) = {\pi \over 4}$ $\Rightarrow$ ${\tan ^{ - 1}}\left( {{{{y_1} - {y_2}} \over {{x_1} - {x_2}}}} \right) = {\pi \over 4}$
${y_1} - {y_2} = {x_1} - {x_2}$ ....... (1)
$|{z_1} - 3|\, = {\mathop{\rm Re}\nolimits} ({z_1}) \Rightarrow {({x_1} - 3)^2} + {y_1}^2 = {x_1}^2$ .... (2)
$|{z_2} - 3|\, = {\mathop{\rm Re}\nolimits} ({z_2}) \Rightarrow {({x_2} - 3)^2} + {y_2}^2 = {x_2}^2$ .... (3)
sub (2) & (3)
${({x_1} - 3)^2} - {({x_2} - 3)^2} + {y_1}^2 - {y_2}^2 = {x_1}^2 - {x_2}^2$
$({x_1} - {x_2})({x_1} + {x_2} - 6) + ({y_1} - {y_2})({y_1} + {y_2})$
$ = ({x_1} - {x_2})({x_1} + {x_2})$
${x_1} + {x_2} - 6 + {y_1} + {y_2} = {x_1} + {x_2} \Rightarrow {y_1} + {y_2} = 6$
z1 $-$ z2 = (x1 $-$ x2) + i(y1 $-$ y2)
$\therefore$ $\arg ({z_1} - {z_2}) = {\pi \over 4}$ $\Rightarrow$ ${\tan ^{ - 1}}\left( {{{{y_1} - {y_2}} \over {{x_1} - {x_2}}}} \right) = {\pi \over 4}$
${y_1} - {y_2} = {x_1} - {x_2}$ ....... (1)
$|{z_1} - 3|\, = {\mathop{\rm Re}\nolimits} ({z_1}) \Rightarrow {({x_1} - 3)^2} + {y_1}^2 = {x_1}^2$ .... (2)
$|{z_2} - 3|\, = {\mathop{\rm Re}\nolimits} ({z_2}) \Rightarrow {({x_2} - 3)^2} + {y_2}^2 = {x_2}^2$ .... (3)
sub (2) & (3)
${({x_1} - 3)^2} - {({x_2} - 3)^2} + {y_1}^2 - {y_2}^2 = {x_1}^2 - {x_2}^2$
$({x_1} - {x_2})({x_1} + {x_2} - 6) + ({y_1} - {y_2})({y_1} + {y_2})$
$ = ({x_1} - {x_2})({x_1} + {x_2})$
${x_1} + {x_2} - 6 + {y_1} + {y_2} = {x_1} + {x_2} \Rightarrow {y_1} + {y_2} = 6$
2021
JEE Mains
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: 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
$ = {{{{(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
2021
JEE Mains
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: 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$
${z^r} + {1 \over {{z^r}}} = 2\cos \left( { - {\pi \over 3}} \right)r = 2\cos {{r\pi } \over 3}$
$ \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$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th July Evening Shift
If the real part of the complex number $z = {{3 + 2i\cos \theta } \over {1 - 3i\cos \theta }},\theta \in \left( {0,{\pi \over 2}} \right)$ is zero, then the value of sin23$\theta$ + cos2$\theta$ is equal to _______________.
Correct Answer: 1
Explanation:
Re $(z) = {{3 - 6{{\cos }^2}\theta } \over {1 + 9{{\cos }^2}\theta }} = 0$
$\Rightarrow$ $\theta$ = ${{\pi \over 4}}$
Hence, sin23$\theta$ + cos2$\theta$ = 1.
$\Rightarrow$ $\theta$ = ${{\pi \over 4}}$
Hence, sin23$\theta$ + cos2$\theta$ = 1.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th July Evening Shift
The equation of a circle is Re(z2) + 2(Im(z))2 + 2Re(z) = 0, where z = x + iy. A line which passes through the center of the given circle and the vertex of the parabola, x2 $-$ 6x $-$ y + 13 = 0, has y-intercept equal to ______________.
Correct Answer: 1
Explanation:
Equation of circle is (x2 $-$ y2) + 2y2 + 2x = 0
x2 + y2 + 2x = 0
Centre : ($-$1, 0)
Parabola : x2 $-$ 6x $-$ y + 13 = 0
(x $-$ 3)2 = y $-$ 4
Vertex : (3, 4)
Equation of line $ \equiv y - 0 = {{4 - 0} \over {3 + 1}}(x + 1)$
$y = x + 1$
y-intercept = 1
x2 + y2 + 2x = 0
Centre : ($-$1, 0)
Parabola : x2 $-$ 6x $-$ y + 13 = 0
(x $-$ 3)2 = y $-$ 4
Vertex : (3, 4)
Equation of line $ \equiv y - 0 = {{4 - 0} \over {3 + 1}}(x + 1)$
$y = x + 1$
y-intercept = 1
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th July Morning Shift
Let $S = \left\{ {n \in N\left| {{{\left( {\matrix{
0 & i \cr
1 & 0 \cr
} } \right)}^n}\left( {\matrix{
a & b \cr
c & d \cr
} } \right) = \left( {\matrix{
a & b \cr
c & d \cr
} } \right)\forall a,b,c,d \in R} \right.} \right\}$, where i = $\sqrt { - 1} $. Then the number of 2-digit numbers in the set S is _____________.
Correct Answer: 11
Explanation:
Let $X = \left( {\matrix{
a & b \cr
c & d \cr
} } \right)$ & $A = {\left( {\matrix{
0 & i \cr
1 & 0 \cr
} } \right)^n}$
$\Rightarrow$ AX = IX
$\Rightarrow$ A = I
$ \Rightarrow {\left( {\matrix{ 0 & i \cr 1 & 0 \cr } } \right)^n} = I$
$ \Rightarrow {A^8} = \left[ {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right]$
$\Rightarrow$ n is multiple of 8
So, number of 2 digit numbers in the set
S = 11 (16, 24, 32, .........., 96)
$\Rightarrow$ AX = IX
$\Rightarrow$ A = I
$ \Rightarrow {\left( {\matrix{ 0 & i \cr 1 & 0 \cr } } \right)^n} = I$
$ \Rightarrow {A^8} = \left[ {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right]$
$\Rightarrow$ n is multiple of 8
So, number of 2 digit numbers in the set
S = 11 (16, 24, 32, .........., 96)
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 18th March Morning Shift
Let z1, z2 be the roots of the equation z2 + az + 12 = 0 and z1, z2 form an equilateral triangle with origin. Then, the value of |a| is :
Correct Answer: 6
Explanation:
For equilateral triangle with vertices z1, z2 and z3,
$z_1^2 + z_2^2 + z_3^3 = {z_1}{z_2} + {z_2}{z_3} + {z_3}{z_1}$
Here one vertex z3 is 0
$ \therefore $ $z_1^2 + z_2^2 = {z_1}{z_2} + 0 + 0$
Given, z1, z2 are roots of ${z^2} + az + 12 = 0$
$ \therefore $ ${z_1} + {z_2} = - a$
${z_1}{z_2} = 12$
$ \therefore $ $z_1^2 + z_2^2 + 2{z_1}{z_2} = {z_1}{z_2} + 2{z_1}{z_2}$
$ \Rightarrow {({z_1} + {z_2})^2} = 3{z_1}{z_2}$
$ \Rightarrow {( - a)^2} = 3 \times 12$
$ \Rightarrow {a^2} = 36$
$ \Rightarrow a = \pm 6$
$ \Rightarrow |a|\, = 6$
$z_1^2 + z_2^2 + z_3^3 = {z_1}{z_2} + {z_2}{z_3} + {z_3}{z_1}$
Here one vertex z3 is 0
$ \therefore $ $z_1^2 + z_2^2 = {z_1}{z_2} + 0 + 0$
Given, z1, z2 are roots of ${z^2} + az + 12 = 0$
$ \therefore $ ${z_1} + {z_2} = - a$
${z_1}{z_2} = 12$
$ \therefore $ $z_1^2 + z_2^2 + 2{z_1}{z_2} = {z_1}{z_2} + 2{z_1}{z_2}$
$ \Rightarrow {({z_1} + {z_2})^2} = 3{z_1}{z_2}$
$ \Rightarrow {( - a)^2} = 3 \times 12$
$ \Rightarrow {a^2} = 36$
$ \Rightarrow a = \pm 6$
$ \Rightarrow |a|\, = 6$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Morning Shift
Let z and $\omega$ be two complex numbers such that $\omega = z\overline z - 2z + 2,\left| {{{z + i} \over {z - 3i}}} \right| = 1$ and Re($\omega$) has minimum value. Then, the minimum value of n $\in$ N for which $\omega$n is real, is equal to ______________.
Correct Answer: 4
Explanation:
Let z = x + iy
| z + i | = | z $-$ 3i |
$ \Rightarrow $ y = 1
Now
$\omega$ = x2 + y2 $-$ 2x $-$ 2iy + 2
$\omega$ = x2 + 1 $-$ 2x $-$ 2i + 2
Re($\omega$) = x2 $-$ 2x + 3
Re($\omega$) = (x $-$ 1)2 + 2
Re($\omega$)min at x = 1 $ \Rightarrow $ z = 1 + i
Now,
$\omega$ = 1 + 1 $-$ 2 $-$ 2i + 2
$\omega$ = 2(1 $-$ i) = 2$\sqrt 2 {e^{i\left( {{{ - \pi } \over 4}} \right)}}$
$\omega$n = 2$\sqrt 2 {e^{i\left( {{{ - n\pi } \over 4}} \right)}}$
If $\omega$n is real $ \Rightarrow $ n = 4
| z + i | = | z $-$ 3i |
$ \Rightarrow $ y = 1
Now
$\omega$ = x2 + y2 $-$ 2x $-$ 2iy + 2
$\omega$ = x2 + 1 $-$ 2x $-$ 2i + 2
Re($\omega$) = x2 $-$ 2x + 3
Re($\omega$) = (x $-$ 1)2 + 2
Re($\omega$)min at x = 1 $ \Rightarrow $ z = 1 + i
Now,
$\omega$ = 1 + 1 $-$ 2 $-$ 2i + 2
$\omega$ = 2(1 $-$ i) = 2$\sqrt 2 {e^{i\left( {{{ - \pi } \over 4}} \right)}}$
$\omega$n = 2$\sqrt 2 {e^{i\left( {{{ - n\pi } \over 4}} \right)}}$
If $\omega$n is real $ \Rightarrow $ n = 4
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Evening Shift
Let z be those complex numbers which satisfy
| z + 5 | $ \le $ 4 and z(1 + i) + $\overline z $(1 $-$ i) $ \ge $ $-$10, i = $\sqrt { - 1} $.
If the maximum value of | z + 1 |2 is $\alpha$ + $\beta$$\sqrt 2 $, then the value of ($\alpha$ + $\beta$) is ____________.
| z + 5 | $ \le $ 4 and z(1 + i) + $\overline z $(1 $-$ i) $ \ge $ $-$10, i = $\sqrt { - 1} $.
If the maximum value of | z + 1 |2 is $\alpha$ + $\beta$$\sqrt 2 $, then the value of ($\alpha$ + $\beta$) is ____________.
Correct Answer: 48
Explanation:
Let, z = x + iy
Given, z(1 + i) + $\overline z $ (1 $-$ i) $ \ge $ $-$ 10
$ \Rightarrow $ z + $\overline z $ + i (z $-$ $\overline z $) $ \ge $ $-$ 10
$ \Rightarrow $ 2x + i (2iy) $ \ge $ $-$ 10
$ \Rightarrow $ x + i2 y $ \ge $ $-$ 5
$ \Rightarrow $ x $-$ y $ \ge $ $-$ 5 ...... (1)
Also given, | z + 5 | $ \le $ 4
$ \Rightarrow $ | z $-$ ($-$5 + 0i) | $ \le $ 4 ...... (2)
It represents a circle whose center at ($-$ 5, 0) and radius 4. z is inside of the circle.
From (1) and (2) z is the shaded region of the diagram.
Now, | z + 1 | = | z $-$ ($-$1 + 0 i) | = distance of z from ($-$1, 0).
Clearly 'p' is the required position of 'z' when | z + 1 | is maximum.
$ \therefore $ P $ \equiv $ ($-$5 $-$ 4 cos45$^\circ$, 0 $-$ 4sin45$^\circ$) = ($-$5$-$2$\sqrt 2 $, $-$2$\sqrt 2 $)
$ \therefore $ (PQ)2|max = 32 + 16$\sqrt 2 $
$ \Rightarrow $ $\alpha$ = 32
$ \Rightarrow $ $\beta$ = 16
Thus, $\alpha$ + $\beta$ = 48
Given, z(1 + i) + $\overline z $ (1 $-$ i) $ \ge $ $-$ 10
$ \Rightarrow $ z + $\overline z $ + i (z $-$ $\overline z $) $ \ge $ $-$ 10
$ \Rightarrow $ 2x + i (2iy) $ \ge $ $-$ 10
$ \Rightarrow $ x + i2 y $ \ge $ $-$ 5
$ \Rightarrow $ x $-$ y $ \ge $ $-$ 5 ...... (1)
Also given, | z + 5 | $ \le $ 4
$ \Rightarrow $ | z $-$ ($-$5 + 0i) | $ \le $ 4 ...... (2)
It represents a circle whose center at ($-$ 5, 0) and radius 4. z is inside of the circle.
From (1) and (2) z is the shaded region of the diagram.
Now, | z + 1 | = | z $-$ ($-$1 + 0 i) | = distance of z from ($-$1, 0).
Clearly 'p' is the required position of 'z' when | z + 1 | is maximum.
$ \therefore $ P $ \equiv $ ($-$5 $-$ 4 cos45$^\circ$, 0 $-$ 4sin45$^\circ$) = ($-$5$-$2$\sqrt 2 $, $-$2$\sqrt 2 $)
$ \therefore $ (PQ)2|max = 32 + 16$\sqrt 2 $
$ \Rightarrow $ $\alpha$ = 32
$ \Rightarrow $ $\beta$ = 16
Thus, $\alpha$ + $\beta$ = 48
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 24th February Evening Shift
Let $i = \sqrt { - 1} $. If ${{{{\left( { - 1 + i\sqrt 3 } \right)}^{21}}} \over {{{(1 - i)}^{24}}}} + {{{{\left( {1 + i\sqrt 3 } \right)}^{21}}} \over {{{(1 + i)}^{24}}}} = k$, and $n = [|k|]$ be the greatest integral part of | k |. Then $\sum\limits_{j = 0}^{n + 5} {{{(j + 5)}^2} - \sum\limits_{j = 0}^{n + 5} {(j + 5)} } $ is equal to _________.
Correct Answer: 310
Explanation:
${(1 + i)^2} = 1 + {i^2} + 2i = 1 - 1 + 2i = 2i$
${(1 - i)^2} = 1 + {i^2} - 2i = 1 - 1 - 2i = - 2i$
We know,
$ - {1 \over 2} + {{i\sqrt 3 } \over 2} = \omega $
$ \Rightarrow - 1 + i\sqrt 3 = 2\omega $
and $ - {1 \over 2} - {{i\sqrt 3 } \over 2} = {\omega ^2}$
$ \Rightarrow - 1 - i\sqrt 3 = 2{\omega ^2}$
$ \Rightarrow 1 + i\sqrt 3 = - 2{\omega ^2}$
Now, $K = {{{{\left( { - 1 + i\sqrt 3 } \right)}^{21}}} \over {{{\left( {1 - i} \right)}^{24}}}} + {{{{\left( {1 + i\sqrt 3 } \right)}^{21}}} \over {{{\left( {1 + i} \right)}^{24}}}}$
$ = {{{{(2\omega )}^{21}}} \over {{{\left( {{{(1 - i)}^2}} \right)}^{12}}}} + {{{{( - 2\omega )}^{21}}} \over {{{\left( {{{(1 + i)}^2}} \right)}^{12}}}}$
$ = {{{2^{21}}.{\omega ^{21}}} \over {{{( - 2i)}^{12}}}} + {{{{( - 2)}^{21}}{{({\omega ^2})}^{21}}} \over {{{(2i)}^{12}}}}$ [as ${\omega ^3} = 1$, ${i^4} = 1$]
$ = {{{2^{21}}} \over {{2^{12}}}} - {{{2^{21}}} \over {{2^{12}}}} = 0$
$ \therefore $ $n = \left[ {|K|} \right] = \left[ {|0|} \right] = 0$
Now $\sum\limits_{j = 0}^5 {{{(j + 5)}^2}} - \sum\limits_{j = 0}^5 {(j + 5)} $
= $\sum\limits_{j = 0}^5 {({j^2} + 25 + 10j - j - 5)} $
= $\sum\limits_{j = 0}^5 {({j^2} + 9j + 20)} $
= $\sum\limits_{j = 0}^5 {{j^2}} + 9\sum\limits_{j = 0}^5 {j + 20\sum\limits_{j = 0}^5 1 } $
= ${{5 \times 6 \times 11} \over 6} + 9\left( {{{5 \times 6} \over 2}} \right) + 20 \times 6$
= 55 + 135 + 120
= 310
${(1 - i)^2} = 1 + {i^2} - 2i = 1 - 1 - 2i = - 2i$
We know,
$ - {1 \over 2} + {{i\sqrt 3 } \over 2} = \omega $
$ \Rightarrow - 1 + i\sqrt 3 = 2\omega $
and $ - {1 \over 2} - {{i\sqrt 3 } \over 2} = {\omega ^2}$
$ \Rightarrow - 1 - i\sqrt 3 = 2{\omega ^2}$
$ \Rightarrow 1 + i\sqrt 3 = - 2{\omega ^2}$
Now, $K = {{{{\left( { - 1 + i\sqrt 3 } \right)}^{21}}} \over {{{\left( {1 - i} \right)}^{24}}}} + {{{{\left( {1 + i\sqrt 3 } \right)}^{21}}} \over {{{\left( {1 + i} \right)}^{24}}}}$
$ = {{{{(2\omega )}^{21}}} \over {{{\left( {{{(1 - i)}^2}} \right)}^{12}}}} + {{{{( - 2\omega )}^{21}}} \over {{{\left( {{{(1 + i)}^2}} \right)}^{12}}}}$
$ = {{{2^{21}}.{\omega ^{21}}} \over {{{( - 2i)}^{12}}}} + {{{{( - 2)}^{21}}{{({\omega ^2})}^{21}}} \over {{{(2i)}^{12}}}}$ [as ${\omega ^3} = 1$, ${i^4} = 1$]
$ = {{{2^{21}}} \over {{2^{12}}}} - {{{2^{21}}} \over {{2^{12}}}} = 0$
$ \therefore $ $n = \left[ {|K|} \right] = \left[ {|0|} \right] = 0$
Now $\sum\limits_{j = 0}^5 {{{(j + 5)}^2}} - \sum\limits_{j = 0}^5 {(j + 5)} $
= $\sum\limits_{j = 0}^5 {({j^2} + 25 + 10j - j - 5)} $
= $\sum\limits_{j = 0}^5 {({j^2} + 9j + 20)} $
= $\sum\limits_{j = 0}^5 {{j^2}} + 9\sum\limits_{j = 0}^5 {j + 20\sum\limits_{j = 0}^5 1 } $
= ${{5 \times 6 \times 11} \over 6} + 9\left( {{{5 \times 6} \over 2}} \right) + 20 \times 6$
= 55 + 135 + 120
= 310
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 24th February Morning Shift
If the least and the largest real values of a, for which the
equation z + $\alpha $|z – 1| + 2i = 0 (z $ \in $ C and i = $\sqrt { - 1} $) has a solution, are p and q respectively; then 4(p2 + q2) is equal to __________.
equation z + $\alpha $|z – 1| + 2i = 0 (z $ \in $ C and i = $\sqrt { - 1} $) has a solution, are p and q respectively; then 4(p2 + q2) is equal to __________.
Correct Answer: 10
Explanation:
$x + iy + \alpha \sqrt {{{(x - 1)}^2} + {y^2}} + 2i = 0$
$ \therefore $ y + 2 = 0 and $x + \alpha \sqrt {{{(x - 1)}^2} + {y^2}} = 0$
y = $-$2 & ${x^2} = {\alpha ^2}({x^2} - 2x + 1 + 4)$
${\alpha ^2} = {{{x^2}} \over {{x^2} - 2x + 5}} \Rightarrow {x^2}({\alpha ^2} - 1) - 2x{\alpha ^2} + 5{\alpha ^2} = 0$
$x \in R \Rightarrow D \ge 0$
$4{\alpha ^4} - 4({\alpha ^2} - 1)5{\alpha ^2} \ge 0$
${\alpha ^2}[4{\alpha ^2} - 2{\alpha ^2} + 20] \ge 0$
${\alpha ^2}[ - 16{\alpha ^2} + 20] \ge 0$
${\alpha ^2}\left[ {{\alpha ^2} - {5 \over 4}} \right] \le 0$
$0 \le {\alpha ^2} \le {5 \over 4}$
$ \therefore $ ${\alpha ^2} \in \left[ {0,{5 \over 4}} \right]$
$ \therefore $ $\alpha \in \left[ { - {{\sqrt 5 } \over 2},{{\sqrt 5 } \over 2}} \right]$
then $4[{(q)^2} + {(p)^2}] = 4\left[ {{5 \over 4} + {5 \over 4}} \right] = 10$
$ \therefore $ y + 2 = 0 and $x + \alpha \sqrt {{{(x - 1)}^2} + {y^2}} = 0$
y = $-$2 & ${x^2} = {\alpha ^2}({x^2} - 2x + 1 + 4)$
${\alpha ^2} = {{{x^2}} \over {{x^2} - 2x + 5}} \Rightarrow {x^2}({\alpha ^2} - 1) - 2x{\alpha ^2} + 5{\alpha ^2} = 0$
$x \in R \Rightarrow D \ge 0$
$4{\alpha ^4} - 4({\alpha ^2} - 1)5{\alpha ^2} \ge 0$
${\alpha ^2}[4{\alpha ^2} - 2{\alpha ^2} + 20] \ge 0$
${\alpha ^2}[ - 16{\alpha ^2} + 20] \ge 0$
${\alpha ^2}\left[ {{\alpha ^2} - {5 \over 4}} \right] \le 0$
$0 \le {\alpha ^2} \le {5 \over 4}$
$ \therefore $ ${\alpha ^2} \in \left[ {0,{5 \over 4}} \right]$
$ \therefore $ $\alpha \in \left[ { - {{\sqrt 5 } \over 2},{{\sqrt 5 } \over 2}} \right]$
then $4[{(q)^2} + {(p)^2}] = 4\left[ {{5 \over 4} + {5 \over 4}} \right] = 10$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Evening Slot
Let z = x + iy be a non-zero complex number
such that ${z^2} = i{\left| z \right|^2}$, where i = $\sqrt { - 1} $ , then z lies
on the :
A.
line, y = –x
B.
real axis
C.
line, y = x
D.
imaginary axis
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Morning Slot
The region represented by
{z = x + iy $ \in $ C : |z| – Re(z) $ \le $ 1} is also given by the
inequality : {z = x + iy $ \in $ C : |z| – Re(z) $ \le $ 1}
{z = x + iy $ \in $ C : |z| – Re(z) $ \le $ 1} is also given by the
inequality : {z = x + iy $ \in $ C : |z| – Re(z) $ \le $ 1}
A.
y2 $ \le $ $2\left( {x + {1 \over 2}} \right)$
B.
y2 $ \le $ ${x + {1 \over 2}}$
C.
y2 $ \ge $ 2(x + 1)
D.
y2 $ \ge $ x + 1
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Evening Slot
The value of ${\left( {{{ - 1 + i\sqrt 3 } \over {1 - i}}} \right)^{30}}$ is :
A.
–215i
B.
–215
C.
215i
D.
65
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Morning Slot
If the four complex numbers $z,\overline z ,\overline z - 2{\mathop{\rm Re}\nolimits} \left( {\overline z } \right)$ and $z-2Re(z)$ represent the vertices of a square of
side 4 units in the Argand plane, then $|z|$ is equal to :
A.
4$\sqrt 2 $
B.
4
C.
2
D.
2$\sqrt 2 $
Let $z = x + iy$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Evening Slot
If a and b are real numbers such that
${\left( {2 + \alpha } \right)^4} = a + b\alpha $
where $\alpha = {{ - 1 + i\sqrt 3 } \over 2}$ then a + b is equal to :
${\left( {2 + \alpha } \right)^4} = a + b\alpha $
where $\alpha = {{ - 1 + i\sqrt 3 } \over 2}$ then a + b is equal to :
A.
33
B.
9
C.
24
D.
57
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Morning Slot
Let $u = {{2z + i} \over {z - ki}}$, z = x + iy and k > 0. If the curve represented
by Re(u) + Im(u) = 1 intersects the y-axis at the points P and Q where PQ = 5, then the value of k is :
by Re(u) + Im(u) = 1 intersects the y-axis at the points P and Q where PQ = 5, then the value of k is :
A.
2
B.
4
C.
1/2
D.
3/2
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Evening Slot
If z1
, z2
are complex numbers such that
Re(z1) = |z1 – 1|, Re(z2) = |z2 – 1| , and
arg(z1 - z2) = ${\pi \over 6}$, then Im(z1 + z2 ) is equal to :
Re(z1) = |z1 – 1|, Re(z2) = |z2 – 1| , and
arg(z1 - z2) = ${\pi \over 6}$, then Im(z1 + z2 ) is equal to :
A.
${{\sqrt 3 } \over 2}$
B.
${1 \over {\sqrt 3 }}$
C.
${2 \over {\sqrt 3 }}$
D.
${2\sqrt 3 }$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Evening Slot
The imaginary part of
${\left( {3 + 2\sqrt { - 54} } \right)^{{1 \over 2}}} - {\left( {3 - 2\sqrt { - 54} } \right)^{{1 \over 2}}}$ can be :
${\left( {3 + 2\sqrt { - 54} } \right)^{{1 \over 2}}} - {\left( {3 - 2\sqrt { - 54} } \right)^{{1 \over 2}}}$ can be :
A.
-2$\sqrt 6 $
B.
6
C.
$\sqrt 6 $
D.
-$\sqrt 6 $
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Morning Slot
The value of
${\left( {{{1 + \sin {{2\pi } \over 9} + i\cos {{2\pi } \over 9}} \over {1 + \sin {{2\pi } \over 9} - i\cos {{2\pi } \over 9}}}} \right)^3}$ is :
${\left( {{{1 + \sin {{2\pi } \over 9} + i\cos {{2\pi } \over 9}} \over {1 + \sin {{2\pi } \over 9} - i\cos {{2\pi } \over 9}}}} \right)^3}$ is :
A.
${1 \over 2}\left( {\sqrt 3 - i} \right)$
B.
-${1 \over 2}\left( {\sqrt 3 - i} \right)$
C.
$ - {1 \over 2}\left( {1 - i\sqrt 3 } \right)$
D.
${1 \over 2}\left( {1 - i\sqrt 3 } \right)$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Evening Slot
If z be a complex number satisfying
|Re(z)| + |Im(z)| = 4, then |z| cannot be :
A.
$\sqrt {10} $
B.
$\sqrt {7} $
C.
$\sqrt {{{17} \over 2}} $
D.
$\sqrt {8} $
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Morning Slot
Let z be complex number such that
$\left| {{{z - i} \over {z + 2i}}} \right| = 1$ and |z| = ${5 \over 2}$.
Then the value of |z + 3i| is :
$\left| {{{z - i} \over {z + 2i}}} \right| = 1$ and |z| = ${5 \over 2}$.
Then the value of |z + 3i| is :
A.
$2\sqrt 3 $
B.
$\sqrt {10} $
C.
${{15} \over 4}$
D.
${7 \over 2}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Morning Slot
If the equation, x2 + bx + 45 = 0 (b $ \in $ R) has
conjugate complex roots and they satisfy
|z +1| = 2$\sqrt {10} $ , then :
A.
b2 – b = 42
B.
b2 + b = 12
C.
b2 + b = 72
D.
b2 – b = 30
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Evening Slot
If ${{3 + i\sin \theta } \over {4 - i\cos \theta }}$, $\theta $ $ \in $ [0, 2$\theta $], is a real number, then an argument of
sin$\theta $ + icos$\theta $ is :
sin$\theta $ + icos$\theta $ is :
A.
$\pi - {\tan ^{ - 1}}\left( {{3 \over 4}} \right)$
B.
$ - {\tan ^{ - 1}}\left( {{3 \over 4}} \right)$
C.
${\tan ^{ - 1}}\left( {{4 \over 3}} \right)$
D.
$\pi - {\tan ^{ - 1}}\left( {{4 \over 3}} \right)$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Morning Slot
If ${\mathop{\rm Re}\nolimits} \left( {{{z - 1} \over {2z + i}}} \right) = 1$, where z = x + iy, then the point (x, y) lies on a :
A.
straight line whose slope is ${3 \over 2}$
B.
straight line whose slope is $-{2 \over 3}$
C.
circle whose diameter is ${{\sqrt 5 } \over 2}$
D.
circle whose centre is at $\left( { - {1 \over 2}, - {3 \over 2}} \right)$
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 3rd September Morning Slot
If ${\left( {{{1 + i} \over {1 - i}}} \right)^{{m \over 2}}} = {\left( {{{1 + i} \over {1 - i}}} \right)^{{n \over 3}}} = 1$, (m, n
$ \in $ N) then the
greatest common divisor of the least values of
m and n is _______ .
Correct Answer: 4
Explanation:
${\left( {{{1 + i} \over {1 - i}}} \right)^{m/2}} = {\left( {{{1 + i} \over {1 - i}}} \right)^{n/3}} = 1$
$ \Rightarrow {\left( {{{{{\left( {1 + i} \right)}^2}} \over 2}} \right)^{m/2}} = {\left( {{{{{\left( {1 + i} \right)}^2}} \over { - 2}}} \right)^{n/3}} = 1$
$ \Rightarrow {(i)^{m/2}} = {( - i)^{n/3}} = 1$ = i4 [ As i4 = 1]
$ \Rightarrow {m \over 2} = 4{k_1}\,and\,{n \over 3} = 4{k_2}$
$ \Rightarrow $ m = 8 k1 and n = 12 k2
Least value of m = 8 and n = 12.
$ \therefore $ GCD (8, 12) = 4
$ \Rightarrow {\left( {{{{{\left( {1 + i} \right)}^2}} \over 2}} \right)^{m/2}} = {\left( {{{{{\left( {1 + i} \right)}^2}} \over { - 2}}} \right)^{n/3}} = 1$
$ \Rightarrow {(i)^{m/2}} = {( - i)^{n/3}} = 1$ = i4 [ As i4 = 1]
$ \Rightarrow {m \over 2} = 4{k_1}\,and\,{n \over 3} = 4{k_2}$
$ \Rightarrow $ m = 8 k1 and n = 12 k2
Least value of m = 8 and n = 12.
$ \therefore $ GCD (8, 12) = 4
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Evening Slot
Let z $ \in $ C with Im(z) = 10 and it satisfies ${{2z - n} \over {2z + n}}$ = 2i - 1 for some natural number n. Then :
A.
n = 20 and Re(z) = –10
B.
n = 40 and Re(z) = 10
C.
n = 40 and Re(z) = –10
D.
n = 20 and Re(z) = 10
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Morning Slot
The equation |z – i| = |z – 1|, i = $\sqrt { - 1} $, represents :
A.
a circle of radius 1
B.
the line through the origin with slope – 1
C.
a circle of radius ${1 \over 2}$
D.
the line through the origin with slope 1
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Evening Slot
If z and w are two complex numbers such that |zw| = 1 and arg(z) – arg(w) = ${\pi \over 2}$
, then :
A.
$z\overline w = {{1 - i} \over {\sqrt 2 }}$
B.
$\overline z w = i$
C.
$z\overline w = {{ - 1 + i} \over {\sqrt 2 }}$
D.
$\overline z w = -i$