Practice Start Test

Complex Numbers

JEE Mathematics · 150 questions · Page 1 of 15 · Click an option or "Show Solution" to reveal answer

Q1
If ∣z−4∣<∣z−2∣\left| {z - 4} \right| < \left| {z - 2} \right|, its solution is given by :
A Re(z)>0{\mathop{\rm Re}\nolimits} (z) > 0
B Re(z)<0{\mathop{\rm Re}\nolimits} (z) < 0
C Re(z)>3{\mathop{\rm Re}\nolimits} (z) > 3
D Re(z)>2{\mathop{\rm Re}\nolimits} (z) > 2
Correct Answer
Option C
Solution

Given

∣z−4∣<∣z−2∣\left| {z - 4} \right| < \left| {z - 2} \right|

Let

   z=x+iy\,\,\,z = x + iy
⇒∣(x−4)+iy)∣<∣(x−2)+iy∣\Rightarrow \left| {\left. {\left( {x - 4} \right) + iy} \right)} \right| < \left| {\left( {x - 2} \right) + iy} \right|
⇒(x−4)2+y2<(x−2)2+y2\Rightarrow {\left( {x - 4} \right)^2} + {y^2} < {\left( {x - 2} \right)^2} + {y^2}
⇒x2−8x+16<x2−4x+4\Rightarrow {x^2} - 8x + 16 < {x^2} - 4x + 4
⇒12<4x\Rightarrow 12 < 4x
⇒x>3\Rightarrow x > 3
⇒Re(z)>3\Rightarrow {\mathop{\rm Re}\nolimits} \left( z \right) > 3
Q2
Let z ∈ \in C with Im(z) = 10 and it satisfies 2z−n2z+n{{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
Correct Answer
Option C
Solution

Let Re (z) = x, then

2(x+10i)−n2(x+10i)+n=2i−1{{2(x + 10i) - n} \over {2(x + 10i) + n}} = 2i - 1
⇒(2x−n)+20i=−(2x+n)−40−20i+2ni\Rightarrow \left( {2x - n} \right) + 20i = - \left( {2x + n} \right) - 40 - 20i + 2ni
⇒2x−n=2x−n−40\Rightarrow 2x - n = 2x - n - 40

& 20 = -20 + 2n ⇒\Rightarrow x = -10 & n = 20

Q3
Let α ,β\alpha \,,\beta be real and z be a complex number. If z2+αz+β=0{z^2} + \alpha z + \beta = 0 has two distinct roots on the line Re z = 1, then it is necessary that :
A β ∈(−1,0)\beta \, \in ( - 1,0)
B ∣β ∣=1\left| {\beta \,} \right| = 1
C β ∈(1,∞)\beta \, \in (1,\infty )
D β ∈(0,1)\beta \, \in (0,1)
Correct Answer
Option C
Solution

As real part of roots is

11

Let roots are

1+π,1+q1 + \pi,1 + q

∴\therefore sum of roots

=1+π+1+qi=−α= 1 + \pi + 1 + qi = - \alpha

which is real

⇒q=−p  \Rightarrow q = - p\,\,

or root are

1+π1+\pi

and

1−π1-\pi

product of roots

=1+p2=β∈(1,∞)= 1 + {p^2} = \beta \in \left( {1,\infty } \right)
p≠0p \ne 0

as roots are distinct.

Q4
If ω(≠1)\omega ( \ne 1) is a cube root of unity, and (1+ω)7=A+Bω {(1 + \omega )^7} = A + B\omega \,. Then (A,B)(A,B) equals :
A (1 ,1)
B (1, 0)
C (- 1 ,1)
D (0 ,1)
Correct Answer
Option A
Solution
(1+ω)7=A+Bω;    (−ω2)7=A+Bω{\left( {1 + \omega } \right)^7} = A + B\omega ;\,\,\,\,{\left( { - {\omega ^2}} \right)^7} = A + B\omega
−ω2=A+Bω;          1+ω=A+Bω- {\omega ^2} = A + B\omega ;\,\,\,\,\,\,\,\,\,\,1 + \omega = A + B\omega
⇒A=1,B=1.\Rightarrow A = 1,B = 1.
Q5
The least positive integer n for which (1+i31−i3)n=1,{\left( {{{1 + i\sqrt 3 } \over {1 - i\sqrt 3 }}} \right)^n} = 1, is :
A 2
B 3
C 5
D 6
Correct Answer
Option B
Solution
(1+i31−i3)n=1{\left( {{{1 + i\sqrt 3 } \over {1 - i\sqrt 3 }}} \right)^n} = 1
⇒     (1+i321−i32)n=1\Rightarrow \,\,\,\,\,{\left( {{{{{1 + i\sqrt 3 } \over 2}} \over {{{1 - i\sqrt 3 } \over 2}}}} \right)^n} = 1

We know, ω\omega = −-

1−i32{{1 - i\sqrt 3 } \over 2}

and ω\omega2 = −-

(1+i3)2{{\left( {1 + i\sqrt 3 } \right)} \over 2}

⇒\Rightarrow

    \,\,\,\,
(−ω2−ω)n=1{\left( {{{ - {\omega ^2}} \over { - \omega }}} \right)^n} = 1

⇒\Rightarrow (ω\omega)n = 1 = ω\omega3 ⇒\Rightarrow

    \,\,\,\,

n = 3

Q6
Let A={z∈C:1≤∣z−(1+i)∣≤2}A = \{ z \in C:1 \le |z - (1 + i)| \le 2\} and B={z∈A:∣z−(1−i)∣=1}B = \{ z \in A:|z - (1 - i)| = 1\} . Then, B :
A is an empty set
B contains exactly two elements
C contains exactly three elements
D is an infinite set
Correct Answer
Option D
Solution

Let,

z=x+iyz = x + iy

Given,

1≤∣z−(1+i)∣≤21 \le \left| {z - (1 + i)} \right| \le 2
⇒1≤∣x+iy−1−i∣≤2\Rightarrow 1 \le \left| {x + iy - 1 - i} \right| \le 2
⇒1≤∣(x−1)+i(y−1)∣≤2\Rightarrow 1 \le \left| {(x - 1) + i(y - 1)} \right| \le 2
⇒1≤(x−1)2+(y−1)2≤2\Rightarrow 1 \le \sqrt {{{(x - 1)}^2} + {{(y - 1)}^2}} \le 2

It represent two concentric circle both have center at (1, 1) and radius 1 and 2. Also given,

∣z−(1−i)∣=1\left| {z - (1 - i)} \right| = 1
⇒∣x+iy−1+i∣=1\Rightarrow \left| {x + iy - 1 + i} \right| = 1
⇒∣(x−1)+i(y+1)∣=1\Rightarrow \left| {(x - 1) + i(y + 1)} \right| = 1
⇒(x−1)2+(y+1)2=1\Rightarrow \sqrt {{{(x - 1)}^2} + {{(y + 1)}^2}} = 1

This represent a circle with center at (1, −-1) and radius = 1. In the common region infinite values of B possible.

Q7
Let the minimum value v0v_{0} of v=∣z∣2+∣z−3∣2+∣z−6i∣2,z∈Cv=|z|^{2}+|z-3|^{2}+|z-6 i|^{2}, z \in \mathbb{C} is attained at z=z0{ }{z}=z_{0}. Then ∣2z02−zˉ03+3∣2+v02\left|2 z_{0}^{2}-\bar{z}_{0}^{3}+3\right|^{2}+v_{0}^{2} is equal to :
A 1000
B 1024
C 1105
D 1196
Correct Answer
Option A
Solution

Let

z=x+iyz = x + iy
v=x2+y2+(x−3)2+y2+x2+(y−6)2v = {x^2} + {y^2} + {(x - 3)^2} + {y^2} + {x^2} + {(y - 6)^2}
=(3x2−6x+9)+(3y2−12y+36)= (3{x^2} - 6x + 9) + (3{y^2} - 12y + 36)
=3(x2+y2−2x−4y+15)= 3({x^2} + {y^2} - 2x - 4y + 15)
=3[(x−1)2+(y−2)2+10]= 3[{(x - 1)^2} + {(y - 2)^2} + 10]
vmin⁡{v_{\min }}

at

z=1+2i=z0z = 1 + 2i = {z_0}

and

v0=30{v_0} = 30

so

∣2(1+2i)2−(1−2i)3+3∣2+900|2{(1 + 2i)^2} - {(1 - 2i)^3} + 3{|^2} + 900
=∣2(−3+4i)−(1−8i3−6i(1−2i)+3∣2+900= |2( - 3 + 4i) - (1 - 8{i^3} - 6i(1 - 2i) + 3{|^2} + 900
=∣−6+8i−(1+8i−6i−12)+3∣2+900= | - 6 + 8i - (1 + 8i - 6i - 12) + 3{|^2} + 900
=∣8+6i∣2+900= |8 + 6i{|^2} + 900
=1000= 1000
Q8
Let a circle C in complex plane pass through the points z1=3+4i{z_1} = 3 + 4i, z2=4+3i{z_2} = 4 + 3i and z3=5i{z_3} = 5i. If z(≠z1)z( \ne {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)arg(z) is equal to :
A tan⁡−1(25)−π{\tan ^{ - 1}}\left( {{2 \over {\sqrt 5 }}} \right) - \pi
B tan⁡−1(247)−π{\tan ^{ - 1}}\left( {{{24} \over 7}} \right) - \pi
C tan⁡−1(3)−π{\tan ^{ - 1}}\left( 3 \right) - \pi
D tan⁡−1(34)−π{\tan ^{ - 1}}\left( {{3 \over 4}} \right) - \pi
Correct Answer
Option B
Solution
z1=3+4i{z_1} = 3 + 4i

,

z2=4+3i{z_2} = 4 + 3i

and

z3=5i{z_3} = 5i

Clearly,

C≡x2+y2=25C \equiv {x^2} + {y^2} = 25

Let

z(x,y)z(x,y)
⇒(y−4x−3)(2−4)=−1\Rightarrow \left( {{{y - 4} \over {x - 3}}} \right)\left( {{2 \over { - 4}}} \right) = - 1
⇒y=2x−2≡L\Rightarrow y = 2x - 2 \equiv L

∴\therefore z is intersection of C & L

⇒z≡(−75,−245)\Rightarrow z \equiv \left( {{{ - 7} \over 5},{{ - 24} \over 5}} \right)

∴\therefore

Arg(z)=−π+tan⁡−1(247)Arg(z) = - \pi + {\tan ^{ - 1}}\left( {{{24} \over 7}} \right)
Q9
If z=2+3iz=2+3 i, then z5+(zˉ)5z^{5}+(\bar{z})^{5} is equal to :
A 244
B 224
C 245
D 265
Correct Answer
Option A
Solution
z=(2+3i)z = (2 + 3i)
⇒z5=(2+3i)((2+3i)2)2\Rightarrow {z^5} = (2 + 3i){\left( {{{(2 + 3i)}^2}} \right)^2}
=(2+3i)(−5+12i)2= (2 + 3i){( - 5 + 12i)^2}
=(2+3i)(−119−120i)= (2 + 3i)( - 119 - 120i)
=−238−240i−357i+360= - 238 - 240i - 357i + 360
=122−597i= 122 - 597i
z‾5=122+597i{\overline z ^5} = 122 + 597i
z5+z‾5=244{z^5} + {\overline z ^5} = 244
Q10
If α\alpha, β\beta ∈\in R are such that 1 −- 2i (here i2 = −-1) is a root of z2 + α\alphaz + β\beta = 0, then (α\alpha −- β\beta) is equal to :
A −-7
B 7
C 3
D −-3
Correct Answer
Option A
Solution

1 −- 2i is the root of the equation.

So other root is 1 ++ 2i ∴\therefore Sum of roots = 1 −- 2i + 1 ++ 2i = 2 = -α\alpha Product of roots = (1 −- 2i)(1 ++ 2i) = 1 - 4i2 = 5 = β\beta ∴\therefore α\alpha - β\beta = -2 - 5 = -7

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