Practice Start Test

Complex Numbers

JEE Mathematics · 150 questions · Page 2 of 15 · Click an option or "Show Solution" to reveal answer

Q11
Let C be the set of all complex numbers. Let S1={z∈C∣∣z−3−2i∣2=8}{S_1} = \{ z \in C||z - 3 - 2i{|^2} = 8\} S2={z∈C∣Re(z)≥5}{S_2} = \{ z \in C|{\mathop{\rm Re}\nolimits} (z) \ge 5\} and S3={z∈C∣∣z−z‾∣≥8}{S_3} = \{ z \in C||z - \overline z | \ge 8\} . Then the number of elements in S1∩S2∩S3{S_1} \cap {S_2} \cap {S_3} is equal to :
A 1
B 0
C 2
D Infinite
Correct Answer
Option A
Solution
S1:∣z−3−2i∣2=8{S_1}:|z - 3 - 2i{|^2} = 8
∣z−3−2i∣=22|z - 3 - 2i| = 2\sqrt 2
(x−3)2+(y−2)2=(22)2{(x - 3)^2} + {(y - 2)^2} = {(2\sqrt 2 )^2}
S2:x≥5{S_2}:x \ge 5
S3:∣z−z‾∣≥8{S_3}:|z - \overline z | \ge 8
∣2iy∣≥8|2iy| \ge 8
2∣y∣≥82|y| \ge 8

∴\therefore

y≥4y \ge 4

,

y≤−4y \le - 4
n(S1∩S2∩S3)=1n\left( {{S_1} \cap {S_2} \cap {S_3}} \right) = 1
Q12
If z≠1z \ne 1 and  z2z−1 \,{{{z^2}} \over {z - 1}}\, is real, then the point represented by the complex number z lies :
A either on the real axis or a circle passing through the origin.
B on a circle with centre at the origin
C either on real axis or on a circle not passing through the origin.
D on the imaginary axis.
Correct Answer
Option A
Solution

Let

z=x+iyz = x + iy

∴\therefore

    z2=x2−y2+2ixy\,\,\,\,{z^2} = {x^2} - {y^2} + 2ixy

Now

z2z−1{{{z^2}} \over {z - 1}}

is real

⇒Im(z2z−1)=0\Rightarrow {\mathop{\rm Im}\nolimits} \left( {{{{z^2}} \over {z - 1}}} \right) = 0
⇒Im(x2−y2+2ixy(x−1)+iy)=0\Rightarrow {\mathop{\rm Im}\nolimits} \left( {{{{x^2} - {y^2} + 2ixy} \over {\left( {x - 1} \right) + iy}}} \right) = 0
⇒Im[(x2−y2+2ixy)(x−1)−iy)]=0\Rightarrow {\mathop{\rm Im}\nolimits} \left[ {\left( {{x^2} - {y^2} + 2ixy} \right)\left. {\left( {x - 1} \right) - iy} \right)} \right] = 0
⇒2xy(x−1)−y(x2−y2)=0\Rightarrow 2xy\left( {x - 1} \right) - y\left( {{x^2} - {y^2}} \right) = 0
⇒y(x2+y2−2x)=0\Rightarrow y\left( {{x^2} + {y^2} - 2x} \right) = 0
⇒y=0; x2+y2−2x=0\Rightarrow y = 0;\,{x^2} + {y^2} - 2x = 0

∴\therefore

    \,\,\,\,
zz

lies either on real axis or on a circle through origin.

Q13
If a > 0 and z = (1+i)2a−i{{{{\left( {1 + i} \right)}^2}} \over {a - i}}, has magnitude 25\sqrt {{2 \over 5}} , then z‾\overline z is equal to :
A −15+35i - {1 \over 5} + {3 \over 5}i
B −15−35i - {1 \over 5} - {3 \over 5}i
C 15−35i{1 \over 5} - {3 \over 5}i
D −35−15i - {3 \over 5} - {1 \over 5}i
Correct Answer
Option B
Solution
z=(1+i)2a−i×a+ia+iz = {{{{\left( {1 + i} \right)}^2}} \over {a - i}} \times {{a + i} \over {a + i}}
⇒z=(1−1+2i)(a+i)a2+1=2ai−2a2+1\Rightarrow z = {{\left( {1 - 1 + 2i} \right)\left( {a + i} \right)} \over {{a^2} + 1}} = {{2ai - 2} \over {{a^2} + 1}}
⇒∣z∣=(−2a2+1)2+(2aa2+1)2=4+4a2(a2+1)2\Rightarrow \left| z \right| = \sqrt {{{\left( {{{ - 2} \over {{a^2} + 1}}} \right)}^2} + {{\left( {{{2a} \over {{a^2} + 1}}} \right)}^2}} = \sqrt {{{4 + 4{a^2}} \over {{{({a^2} + 1)}^2}}}}
⇒4(1+a2)(a2+1)2=2a2+1\Rightarrow \sqrt {{{4(1 + {a^2})} \over {{{({a^2} + 1)}^2}}}} = {2 \over {\sqrt {{a^2} + 1} }}

.... (i) Now given

∣z∣=25\left| z \right| = \sqrt {{2 \over 5}}

so

25=21+a2\sqrt {{2 \over 5}} = {2 \over {\sqrt {1 + {a^2}} }}

from equation (i) By squaring both sides

⇒25=41+a2\Rightarrow {2 \over 5} = {4 \over {1 + {a^2}}}
⇒1+a2=10\Rightarrow 1 + {a^2} = 10

a2 = 9 ⇒\Rightarrow a = ±\pm 3 ∵\because (a > 0) then a = 3 Now

z‾=(1−i)23+i\overline z = {{{{\left( {1 - i} \right)}^2}} \over {3 + i}}

=

(1+i2−2i)3+i{{\left( {1 + {i^2} - 2i} \right)} \over {3 + i}}

=

(1+i2−2i)(3−i)(3+i)(3−i){{\left( {1 + {i^2} - 2i} \right)\left( {3 - i} \right)} \over {\left( {3 + i} \right)\left( {3 - i} \right)}}

=

(−2i)(3−i)(9−i2){{\left( { - 2i} \right)\left( {3 - i} \right)} \over {\left( {9 - {i^2}} \right)}}

=

(−2i)(3−i)10{{\left( { - 2i} \right)\left( {3 - i} \right)} \over {10}}

=

−6i+2i210{{ - 6i + 2{i^2}} \over {10}}

=

−6i−210{{ - 6i - 2} \over {10}}

=

−15−35i-{1 \over 5} - {3 \over 5}i
Q14
The equation arg⁡(z−1z+1)=π4\arg \left( {{{z - 1} \over {z + 1}}} \right) = {\pi \over 4} represents a circle with :
A centre at (0, −-1) and radius 2\sqrt 2
B centre at (0, 1) and radius 2\sqrt 2
C centre (0, 0) and radius 2\sqrt 2
D centre at (0, 1) and radius 2
Correct Answer
Option B
Solution

In

Δ\Delta

OAC

sin⁡(π4)=1AC\sin \left( {{\pi \over 4}} \right) = {1 \over {AC}}
⇒AC=2\Rightarrow AC = \sqrt 2

Also,

tan⁡π4=OAOC=1OC\tan {\pi \over 4} = {{OA} \over {OC}} = {1 \over {OC}}

⇒\Rightarrow OC = 1 ∴\therefore centre (0, 1); Radius =

2\sqrt 2
Q15
Let z be complex number such that ∣z−iz+2i∣=1\left| {{{z - i} \over {z + 2i}}} \right| = 1 and |z| = 52{5 \over 2}. Then the value of |z + 3i| is :
A 232\sqrt 3
B 10\sqrt {10}
C 154{{15} \over 4}
D 72{7 \over 2}
Correct Answer
Option D
Solution

Given

∣z−iz+2i∣=1\left| {{{z - i} \over {z + 2i}}} \right| = 1

|z – i| = |z + 2i| (let z = x + iy) ⇒\Rightarrow x2 + (y – 1)2 = x2 + (y + 2)2 ⇒\Rightarrow y =

−12- {1 \over 2}

Also given |z| =

52{5 \over 2}

⇒\Rightarrow x2 + y2 =

254{{25} \over 4}

⇒\Rightarrow x2 = 6 ∴\therefore z =

±6\pm \sqrt 6

-

−12i- {1 \over 2}i

|z + 3i| =

6+254\sqrt {6 + {{25} \over 4}}

=

72{7 \over 2}
Q16
If (3+i)100=299(p+iq){\left( {\sqrt 3 + i} \right)^{100}} = {2^{99}}(p + iq), then p and q are roots of the equation :
A x2−(3−1)x−3=0{x^2} - \left( {\sqrt 3 - 1} \right)x - \sqrt 3 = 0
B x2+(3+1)x+3=0{x^2} + \left( {\sqrt 3 + 1} \right)x + \sqrt 3 = 0
C x2+(3−1)x−3=0{x^2} + \left( {\sqrt 3 - 1} \right)x - \sqrt 3 = 0
D x2−(3+1)x+3=0{x^2} - \left( {\sqrt 3 + 1} \right)x + \sqrt 3 = 0
Correct Answer
Option A
Solution
(2eiπ/6)100=299(p+iq){\left( {2{e^{i\pi /6}}} \right)^{100}} = {2^{99}}(p + iq)
2100(cos⁡50π3+isin⁡50π3)=299(p+iq){2^{100}}\left( {\cos {{50\pi } \over 3} + i\sin {{50\pi } \over 3}} \right) = {2^{99}}(p + iq)
p+iq=2(cos⁡2π3+isin⁡2π3)p + iq = 2\left( {\cos {{2\pi } \over 3} + i\sin {{2\pi } \over 3}} \right)

p = −-1, q =

3\sqrt 3
x2−(3−1)x−3=0{x^2} - (\sqrt 3 - 1)x - \sqrt 3 = 0
Q17
The equation |z – i| = |z – 1|, i = −1\sqrt { - 1} , represents :
A a circle of radius 1
B the line through the origin with slope – 1
C a circle of radius 12{1 \over 2}
D the line through the origin with slope 1
Correct Answer
Option D
Solution

Let the complex number z = x + iy Now given | (x + iy) - 1 | = | (x+iy) - i | ⇒\Rightarrow (x - 1)2 + y2 = x2 + (y - 1)2 ⇒\Rightarrow x = y ∴\therefore the correct answer is option (D)

Q18
Let z1 and z2 be two complex numbers satisfying | z1 | = 9 and | z2 – 3 – 4i | = 4. Then the minimum value of | z1 – z2 | is :
A 0
B 1
C 2
D 2\sqrt 2
Correct Answer
Option A
Solution
∣z1∣=9,  ∣z2−(3+4i)∣=4\left| {{z_1}} \right| = 9,\,\,\left| {{z_2} - \left( {3 + 4i} \right)} \right| = 4
C1,(0,0){C_1},(0,0)

radius r1 = 9 C2 (3, 4), radius r2 = 4 C1C2 =

∣r1−r2∣=5\left| {{r_1} - {r_2}} \right| = 5

∴\therefore Circle touches internally ∴\therefore

∣z1−z2∣min⁡=0{\left| {{z_1} - {z_2}} \right|_{\min }} = 0
Q19
z and w are two nonzero complex numbers such that  ∣z∣=∣w∣\,\left| z \right| = \left| w \right| and Arg z + Arg w =π\pi then z equals
A ω‾\overline \omega
B −ω‾ - \overline \omega
C ω\omega
D −ω - \omega
Correct Answer
Option B
Solution

Let

∣z∣=∣ω∣=r\left| z \right| = \left| \omega \right| = r

∴\therefore

z=reiθ,ω=reiϕz = r{e^{i\theta }},\omega = r{e^{i\phi }}

where

  θ+ϕ=π.\,\,\theta + \phi = \pi .

∴\therefore

z=rei(π−ϕ)=reiπ.z = r{e^{i\left( {\pi - \phi } \right)}} = r{e^{i\pi }}.
e−iϕ=−re−iϕ=−ω‾.{e^{ - i\phi }} = - r{e^{ - i\phi }} = - \overline \omega .

[as

    ω‾=re−iϕ\,\,\,\,\overline \omega = r{e^{ - i\phi }}

]

Q20
The locus of the centre of a circle which touches the circle ∣z−z1∣=a\left| {z - {z_1}} \right| = a and∣z−z2∣=b \left| {z - {z_2}} \right| = b\, externally (z, z1 & z2 z,\,{z_1}\,\& \,{z_2}\, are complex numbers) will be :
A an ellipse
B a hyperbola
C a circle
D none of these
Correct Answer
Option B
Solution

Let the circle be

∣z−z3∣=r.\left| {z - {z_3}} \right| = r.

Then according to given conditions

∣z3−z1∣=r+a\left| {{z_3} - {z_1}} \right| = r + a

(Shown in the image) and

∣z3−z2∣=r+b.\left| {{z_3} - {z_2}} \right| = r + b.

(Shown in the image) Eliminating

r,r,

we get

∣z3−z1∣−∣z3−z2∣=a−b.\left| {{z_3} - {z_1}} \right| - \left| {{z_3} - {z_2}} \right| = a - b.

∴\therefore Locus of center

z3{z_3}

is

∣z−z1∣−∣z−z2∣=a−b\left| {z - {z_1}} \right| - \left| {z - {z_2}} \right| = a - b

= constant.

Definition of hyperbola says, when difference of distance between two points is constant from a particular point then that particular point will lie on a hyperbola.

Here distance of z1 from z3 is =

r+ar + a

and distance of z2 from z3 is =

r+br + b

Now their difference = (

r+ar + a

) - (

r+br + b

) =

a−ba - b

= a constant ∴\therefore Locus of z3 is a hyperbola.

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