New chat PYQ Library Doubts Rank report

Sign in to see Recents

Sign in to save progress →

Complex Numbers question

2022 · 25 Jun · Shift 1 · Q38

Guest · try one question or challengeGuest · try one question or challenge

PYQ Library
JEE Main
Mathematics
Complex Numbers
2022 · 25 Jun · Shift 1 · Q38

Complex Numbers question

2022 · 25 Jun · Shift 1 · Q38

JEE MainMathematicsComplex NumbersMCQ+4 / −1

Let a circle C in complex plane pass through the points

{z_1} = 3 + 4i

,

{z_2} = 4 + 3i

and

{z_3} = 5i

. If

z( e {z_1})

is a point on C such that the line through z and z1 is perpendicular to the line through z2 and z3, then

arg(z)

is equal to :

A
${\tan ^{ - 1}}\left( {{2 \over {\sqrt 5 }}} \right) - \pi$
B
${\tan ^{ - 1}}\left( {{{24} \over 7}} \right) - \pi$
C
${\tan ^{ - 1}}\left( 3 \right) - \pi$
D
${\tan ^{ - 1}}\left( {{3 \over 4}} \right) - \pi$

More from Complex Numbers

Let z1 and z2 be two complex numbers such that $\overline{z}_{1} = i \overline{z}_{2}$ and $ar g (\frac{z _{1}}{z _{2}}) = π$ . Then :2022 · MCQ
Let O be the origin and A be the point $z_{1} = 1 + 2 i$ . If B is the point $z_{2}$ , $Re o l imi t s (z_{2}) < 0$ , such that OAB is a right angled isosceles triangle with OB as hypotenuse, then which of the following is NOT…2022 · MCQ
If $z = x + i y$ satisfies $∣ z ∣ - 2 = 0$ and $∣ z - i ∣ - ∣ z + 5 i ∣ = 0$ , then :2022 · MCQ
Let $A = {z \in C : \frac{z + 1}{z - 1} < 1}$ and $B = {z \in C : ar g (\frac{z - 1}{z + 1}) = \frac{2 π}{3}}$ . Then A $\cap$ B is :2022 · MCQ
If $z^{2} + z + 1 = 0$ , $z \in C$ , then $n = 1 \sum 15 (z^{n} + (- 1)^{n} \frac{1}{z ^{n}})^{2}$ is equal to .2022 · Numerical
Let the minimum value $v_{0}$ of $v = ∣ z ∣^{2} + ∣ z - 3 ∣^{2} + ∣ z - 6 i ∣^{2}, z \in C$ is attained at $z = z_{0}$ . Then $2 z_{0}^{2} - \overset{z}{ˉ}_{0}^{3} + 3^{2} + v_{0}^{2}$ is equal to :2022 · MCQ
Let $S = {z \in C : z^{2} + \overset{z}{ˉ} = 0}$ . Then $z \in S \sum (Re (z) + Im (z))$ is equal to .2022 · Numerical
Let S be the set of all $(α, β), π < α, β < 2 π$ , for which the complex number $\frac{1 - i s i n α}{1 + 2 i s i n α}$ is purely imaginary and $\frac{1 + i c o s β}{1 - 2 i c o s β}$ is purely real. Let $Z_{α β} = sin 2 α + i cos 2 β, (α, β) \in S$ …2022 · MCQ