Practice Start Test

Complex Numbers

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

Q141
If α,β∈C\alpha ,\beta \in C are the distinct roots of the equation x2 - x + 1 = 0, then α101+β107{\alpha ^{101}} + {\beta ^{107}} is equal to :
A 2
B -1
C 0
D 1
Correct Answer
Option D
Solution

Given equation, x2 −- x + 1 = 0 Roots of this equation x =

1±3i2{{1 \pm \sqrt 3 i} \over 2}
∴   \therefore\,\,\,
∝ = 1+3 i2\propto \, = \,{{1 + \sqrt 3 \,i} \over 2}

and

β= 1−3 i2\beta = \,{{1 - \sqrt 3 \,i} \over 2}

We know;

ω=−1+3 i2=−(1−3 i2)=−β\omega = {{ - 1 + \sqrt 3 \,i} \over 2} = - \left( {{{1 - \sqrt 3 \,i} \over 2}} \right) = - \beta

and

ω2=−1−3 i2=−(1+3 i2)=−∝{\omega ^2} = {{ - 1 - \sqrt 3 \,i} \over 2} = - \left( {{{1 + \sqrt 3 \,i} \over 2}} \right) = - \propto
∴   \therefore\,\,\,
∝ =−ω2\propto \, = - {\omega ^2}

and

β = −ω\beta \, = \, - \omega
∴   \therefore\,\,\,
∝101+β107{ \propto ^{101}} + {\beta ^{107}}
=(−ω2)101+(−ω)107= {\left( { - {\omega ^2}} \right)^{101}} + {\left( { - \omega } \right)^{107}}
=(−1)101.(ω2)101+(−1)107.(ω)107= {\left( { - 1} \right)^{101}}.{\left( {{\omega ^2}} \right)^{101}} + {\left( { - 1} \right)^{107}}.{\left( \omega \right)^{107}}
=−1.(ω2)101−ω107= - 1.{\left( {{\omega ^2}} \right)^{101}} - {\omega ^{107}}
=−(ω202+ω107)= - \left( {{\omega ^{202}} + {\omega ^{107}}} \right)
=−(ω3.67.ω+ω3.35.ω2)= - \left( {{\omega ^{3.67}}.\omega + {\omega ^{3.35}}.{\omega ^2}} \right)
=−(ω+ω2)   = - \left( {\omega + {\omega ^2}} \right)\,\,\,

[ as

   \,\,\,
ω3n=1{\omega ^{3n}} = 1

]

=−(−1)= - \left( { - 1} \right)
      \,\,\,\,\,\,

[as

   \,\,\,
1+ω+ω2=01 + \omega + {\omega ^2} = 0

]

=1= 1
Q142
The number of complex numbers z such that ∣z−1∣=∣z+1∣=∣z−i∣\left| {z - 1} \right| = \left| {z + 1} \right| = \left| {z - i} \right| equals :
A 1
B 2
C ∞\infty
D 0
Correct Answer
Option A
Solution

Let

z=x+iyz=x+iy
∣z−1∣=∣z+1∣(x−1)2+y2\left| {z - 1} \right| = \left| {z + 1} \right|{\left( {x - 1} \right)^2} + {y^2}
                     =(x+1)2+y2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {x + 1} \right)^2} + {y^2}
⇒Re z=0              ⇒x=0\Rightarrow {\mathop{\rm Re}\nolimits} \,z = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow x = 0
∣z−1∣=∣z−i∣(x−1)2+y2\left| {z - 1} \right| = \left| {z - i} \right|{\left( {x - 1} \right)^2} + {y^2}
                     =x2+(y−1)2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {x^2} + {\left( {y - 1} \right)^2}
⇒x=y\Rightarrow x = y
∣z+1∣=∣z−i∣(x+1)2+y2\left| {z + 1} \right| = \left| {z - i} \right|{\left( {x + 1} \right)^2} + {y^2}
                     =x2+(y−1)2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {x^2} + {\left( {y - 1} \right)^2}

Only

(0,0)(0,0)

will satisfy all conditions. ⇒\Rightarrow Number of complex number

z=1z=1
Q143
If (1+i1−i)x=1{\left( {{{1 + i} \over {1 - i}}} \right)^x} = 1 then :
A x = 2n + 1, where n is any positive integer
B x = 4n , where n is any positive integer
C x = 2n, where n is any positive integer
D x = 4n + 1, where n is any positive integer.
Correct Answer
Option B
Solution
(1+i1−i)x=1{\left( {{{1 + i} \over {1 - i}}} \right)^x} = 1

⇒\Rightarrow

[(1+i)(1+i)(1−i)(1+i)]x=1{\left[ {{{\left( {1 + i} \right)\left( {1 + i} \right)} \over {\left( {1 - i} \right)\left( {1 + i} \right)}}} \right]^x} = 1

⇒\Rightarrow

[(1+i)21−i2]x=1{\left[ {{{{{\left( {1 + i} \right)}^2}} \over {1 - {i^2}}}} \right]^x} = 1

⇒\Rightarrow

[1+2i+i21+1]x=1{\left[ {{{1 + 2i + {i^2}} \over {1 + 1}}} \right]^x} = 1

⇒\Rightarrow

[1+2i−12]x=1{\left[ {{{1 + 2i - 1} \over 2}} \right]^x} = 1
⇒(i)x=1\Rightarrow {\left( i \right)^x} = 1

We know

i=−1i = \sqrt { - 1}

∴\therefore

i2=−1{i^2} = - 1

⇒\Rightarrow

i3=−1×i=−i{i^3} = - 1 \times i = - i

⇒\Rightarrow

i4=−i×i=−i2=−(−1)=1{i^4} = - i \times i = - {i^2} = - \left( { - 1} \right) = 1

So when power of

ii

is 4 or multiple of 4 then it's value is = 1 ∴\therefore

(i)x=1{\left( i \right)^x} = 1
=(i)4n= {\left( i \right)^{4n}}

where n is a positive integer.

Q144
If  ∣z+4∣  ≤  3 \,\left| {z + 4} \right|\,\, \le \,\,3\,, then the maximum value of ∣z+1∣\left| {z + 1} \right| is :
A 6
B 0
C 4
D 10
Correct Answer
Option A
Solution
zz

lies on or inside the circle with center

(−4,0)(-4,0)

and radius

33

units. From the Argand diagram maximum value of

∣z+1∣\left| {z + 1} \right|

is

66
Q145
If z2+z+1=0{z^2} + z + 1 = 0, where z is complex number, then value of (z+1z)2+(z2+1z2)2+(z3+1z3)2+..........+(z6+1z6)2{\left( {z + {1 \over z}} \right)^2} + {\left( {{z^2} + {1 \over {{z^2}}}} \right)^2} + {\left( {{z^3} + {1 \over {{z^3}}}} \right)^2} + .......... + {\left( {{z^6} + {1 \over {{z^6}}}} \right)^2} is :
A 18
B 54
C 6
D 12
Correct Answer
Option D
Solution
z2+z+1=0⇒z=ω   {z^2} + z + 1 = 0 \Rightarrow z = \omega \,\,\,

or

   ω2\,\,\,{\omega ^2}

So,

z+1z=ω+ω2=−1z + {1 \over z} = \omega + {\omega ^2} = - 1
z2+1z2=ω2+ω=−1,{z^2} + {1 \over {{z^2}}} = {\omega ^2} + \omega = - 1,
z3+1z3=ω3+ω3=2{z^3} + {1 \over {{z^3}}} = {\omega ^3} + {\omega ^3} = 2
z4+1z4=−1,{z^4} + {1 \over {{z^4}}} = - 1,
z5+1z5=−1{z^5} + {1 \over {{z^5}}} = - 1

and

    z6+1z6=2\,\,\,\,{z^6} + {1 \over {{z^6}}} = 2

∴\therefore The given sum

=1+1+4+1+1+4=12= 1 + 1 + 4 + 1 + 1 + 4 = 12
Q146
Let z1 and z2 be any two non-zero complex numbers such that 3∣z1∣=4∣z2∣.3\left| {{z_1}} \right| = 4\left| {{z_2}} \right|. If z=3z12z2+2z23z1z = {{3{z_1}} \over {2{z_2}}} + {{2{z_2}} \over {3{z_1}}} then :
A Im(z)=0{\rm I}m\left( z \right) = 0
B ∣z∣=172\left| z \right| = \sqrt {{17 \over 2}}
C ∣z∣=\left| z \right| = 129+16cos⁡2θ{1 \over 2}\sqrt {9 + 16{{\cos }^2}\theta }
D Re(z) == 0
Correct Answer
Option C
Solution

Given,

3∣z1∣=4∣z2∣3\left| {{z_1}} \right| = 4\left| {{z_2}} \right|

⇒\Rightarrow

∣z1∣∣z2∣=43{{\left| {{z_1}} \right|} \over {\left| {{z_2}} \right|}} = {4 \over 3}

⇒\Rightarrow

∣3z1∣∣2z2∣=43×32=2{{\left| {3{z_1}} \right|} \over {\left| {2{z_2}} \right|}} = {4 \over 3} \times {3 \over 2} = 2

As we know, for any compled number

3z12z2=∣3z1∣∣2z2∣{{3{z_1}} \over {2{z_2}}} = {{\left| {3{z_1}} \right|} \over {\left| {2{z_2}} \right|}}

(cosθ\theta + i sinθ\theta) = 2(cosθ\theta + i sinθ\theta) ∴\therefore

2z23z1{{2{z_2}} \over {3{z_1}}}

=

12(cos⁡θ+isin⁡θ){1 \over {2\left( {\cos \theta + i\sin \theta } \right)}}

=

12(cos⁡θ+isin⁡θ)×(cos⁡θ−isin⁡θ)(cos⁡θ−isin⁡θ){1 \over {2\left( {\cos \theta + i\sin \theta } \right)}} \times {{\left( {\cos \theta - i\sin \theta } \right)} \over {\left( {\cos \theta - i\sin \theta } \right)}}

=

(12cos⁡θ−i2sin⁡θ){\left( {{1 \over 2}\cos \theta - {i \over 2}\sin \theta } \right)}

Now, given

z=3z12z2+2z23z1z = {{3{z_1}} \over {2{z_2}}} + {{2{z_2}} \over {3{z_1}}}

= 2(cosθ\theta + i sinθ\theta) +

(12cos⁡θ−i2sin⁡θ){\left( {{1 \over 2}\cos \theta - {i \over 2}\sin \theta } \right)}

=

52cos⁡θ+32isin⁡θ{{5 \over 2}\cos \theta + {3 \over 2}i\sin \theta }

So, |z| =

254cos⁡2θ+94sin⁡2θ\sqrt {{{25} \over 4}{{\cos }^2}\theta + {9 \over 4}{{\sin }^2}\theta }

=

129+16cos⁡2θ{1 \over 2}\sqrt {9 + 16{{\cos }^2}\theta }

z is neither purely real nor purely imaginary and |z| depends on θ\theta.

Q147
Let z0 be a root of the quadratic equation, x2 + x + 1 = 0, If z = 3 + 6iz081_0^{81} −- 3iz093_0^{93}, then arg z is equal to :
A π4{\pi \over 4}
B π6{\pi \over 6}
C π3{\pi \over 3}
D 0
Correct Answer
Option A
Solution

1 + x + x2 = 0 x =

−1±1−42=−1±i32{{ - 1 \pm \sqrt {1 - 4} } \over 2} = {{ - 1 \pm i\sqrt 3 } \over 2}

z0 = w, w2 Now z = 3 + 6iz

081_0^{81}

−- 3iz

093_0^{93}

z = 3 + 6iw81 −- 3iw93 (w93 = w81 = 1) ⇒\Rightarrow z = 3 + 3i then arg(z) = tan−-1

(33)\left( {{3 \over 3}} \right)

= tan−-1 (1) =

π4{\pi \over 4}
Q148
If zz and ω\omega are two non-zero complex numbers such that ∣zω∣=1\left| {z\omega } \right| = 1 and Arg(z)−Arg(ω)=π2,Arg(z) - Arg(\omega ) = {\pi \over 2}, then  z ‾ω\,\overline {z\,} \omega is equal to
A −i- i
B 1
C - 1
D ii
Correct Answer
Option A
Solution

Given that,

∣zω∣=1\left| {z\omega } \right| = 1

⇒\Rightarrow

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

= 1 ⇒\Rightarrow

∣z∣\left| z \right|

=

1∣ω∣{1 \over {\left| \omega \right|}}

and

Arg(z)−Arg(ω)=π2Arg(z) - Arg(\omega ) = {\pi \over 2}

⇒\Rightarrow

Arg(zω)Arg\left( {{z \over \omega }} \right)
=π2= {\pi \over 2}

When argument of a complex number is

π2{\pi \over 2}

, it means it is making an angle of

π2{\pi \over 2}

with the real axis in the counterclockwise, so it is along the imaginary axis and positive side of imaginary axis. So,

zω{{z \over \omega }}

is a purely imaginary number that means there is no real part in this complex number. So we can assume,

zω{{z \over \omega }}

=

kiki

⇒\Rightarrow

∣zω∣{\left| {{z \over \omega }} \right|}

=

∣ki∣\left| {ki} \right|

⇒\Rightarrow

∣zω∣{\left| {{z \over \omega }} \right|}

=

∣k∣∣i∣\left| k \right|\left| i \right|

⇒\Rightarrow

∣zω∣{\left| {{z \over \omega }} \right|}

=

kk

[ as

∣i∣\left| i \right|

= 1 ] ⇒\Rightarrow

∣z∣\left| z \right|

×\times

1∣ω∣{1 \over {\left| \omega \right|}}

=

kk

⇒\Rightarrow

∣z∣\left| z \right|

×\times

∣z∣\left| z \right|

=

kk

[ as

1∣ω∣{1 \over {\left| \omega \right|}}

=

∣z∣\left| z \right|

] ⇒\Rightarrow

∣z∣2{\left| z \right|^2}

=

kk

⇒\Rightarrow

∣z∣\left| z \right|

=

k\sqrt k

∴\therefore

∣ω∣\left| \omega \right|

=

1k{1 \over {\sqrt k }}

As

zω{{z \over \omega }}

is imaginary so we can write,

zω{{z \over \omega }}

=

−z‾ω‾- {{\overline z } \over {\overline \omega }}

[ When

zz

is imaginary then

zz

=

−z‾-\overline z

] ⇒\Rightarrow

z‾ω\overline z \omega

=

−zω‾- z\overline \omega

⇒\Rightarrow

z‾ω\overline z \omega

=

−zω-{{z \over \omega }}

.

ω‾\overline \omega

.ω\omega ⇒\Rightarrow

z‾ω\overline z \omega

=

−zω-{{z \over \omega }}

.

∣ω∣2{\left| \omega \right|^2}

⇒\Rightarrow

z‾ω\overline z \omega

=

−ki-ki

.

(1k)2{\left( {{1 \over {\sqrt k }}} \right)^2}

⇒\Rightarrow

z‾ω\overline z \omega

=

−ki-ki

.

1k{1 \over k}

⇒\Rightarrow

z‾ω\overline z \omega

=

−i-i

Method 2 : Given that,

∣zω∣=1\left| {z\omega } \right| = 1

⇒\Rightarrow

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

= 1 ⇒\Rightarrow

∣z∣\left| z \right|

=

1∣ω∣{1 \over {\left| \omega \right|}}

and

Arg(z)−Arg(ω)=π2Arg(z) - Arg(\omega ) = {\pi \over 2}

⇒\Rightarrow

Arg(zω)Arg\left( {{z \over \omega }} \right)
=π2= {\pi \over 2}

When argument of a complex number is

π2{\pi \over 2}

, it means it is making an angle of

π2{\pi \over 2}

with the real axis in the counterclockwise, so it is along the imaginary axis and positive side of imaginary axis. So,

zω{{z \over \omega }}

is a purely imaginary number that means there is no real part in this complex number. So we can assume,

zω{{z \over \omega }}

=

kiki

⇒\Rightarrow

∣zω∣{\left| {{z \over \omega }} \right|}

=

∣ki∣\left| {ki} \right|

⇒\Rightarrow

∣zω∣{\left| {{z \over \omega }} \right|}

=

∣k∣∣i∣\left| k \right|\left| i \right|

⇒\Rightarrow

∣zω∣{\left| {{z \over \omega }} \right|}

=

kk

[ as

∣i∣\left| i \right|

= 1 ] ⇒\Rightarrow

∣z∣\left| z \right|

×\times

1∣ω∣{1 \over {\left| \omega \right|}}

=

kk

⇒\Rightarrow

∣z∣\left| z \right|

×\times

∣z∣\left| z \right|

=

kk

[ as

1∣ω∣{1 \over {\left| \omega \right|}}

=

∣z∣\left| z \right|

] ⇒\Rightarrow

∣z∣2{\left| z \right|^2}

=

kk

⇒\Rightarrow

∣z∣\left| z \right|

=

k\sqrt k

∴\therefore

∣ω∣\left| \omega \right|

=

1k{1 \over {\sqrt k }}

(1) Magnitude of

z‾ω\overline z \omega

=

∣z‾∣∣ω∣\left| {\overline z } \right|\left| \omega \right|

=

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

[ as

∣z∣\left| z \right|

=

∣z‾∣\left| {\overline z } \right|

] =

k\sqrt k

.

1k{{1 \over {\sqrt k }}}

= 1 ∴\therefore The distance from the origin of

z‾ω{\overline z \omega }

is 1. (2) Argument of

z‾ω{\overline z \omega }

=

Arg(z‾ω)Arg\left( {\overline z \omega } \right)

=

Arg(z‾)+Arg(ω)Arg\left( {\overline z } \right) + Arg\left( \omega \right)

=

−Arg(z)+Arg(ω)-Arg\left( z \right) + Arg\left( \omega \right)

=

−(Arg(z)−Arg(ω))- \left( {Arg\left( z \right) - Arg\left( \omega \right)} \right)

=

−π2- {\pi \over 2}

∴\therefore

z‾ω{\overline z \omega }

is at (0, -1) on the negative side of imaginary axis and making an angle of

π2{\pi \over 2}

clockwise. ∴\therefore

z‾ω{\overline z \omega }

= 0 + (-1)×\times

ii

=

−i-i
Q149
The point represented by 2 + i in the Argand plane moves 1 unit eastwards, then 2 units northwards and finally from there 222\sqrt 2 units in the south-westwardsdirection. Then its new position in the Argand plane is at the point represented by :
A 2 + 2i
B 1 + i
C −-1 −- i
D −-2 −-2i
Correct Answer
Option B
Solution

Here, z −- (3 + 3i) =

222\sqrt 2

(cos(−-135o) + i sin (−- 135o)) =

222\sqrt 2

(−-

12{1 \over {\sqrt 2 }}

−-

i2{i \over {\sqrt 2 }}

) = −- 2 −- 2i ⇒\Rightarrow z = 3 + 3 i −- 2 −- 2 i = 1 + i Note : Polar form of a complex number : z = r (cosθ\theta + i sinθ\theta) Here r = modulus of z and θ\theta argument of z.

Q150
If z−αz+α(α∈R){{z - \alpha } \over {z + \alpha }}\left( {\alpha \in R} \right) is a purely imaginary number and | z | = 2, then a value of α\alpha is :
A 12{1 \over 2}
B 2\sqrt 2
C 2
D 1
Correct Answer
Option C
Solution
z−αz+α+z‾−αz‾+α=0{{z - \alpha } \over {z + \alpha }} + {{\overline z - \alpha } \over {\overline z + \alpha }} = 0
zz‾+zα−αz‾−α2+zz‾−zα+z‾α−α2=0z\overline z + z\alpha - \alpha \overline z - {\alpha ^2} + z\overline z - z\alpha + \overline z \alpha - {\alpha ^2} = 0
∣z∣2=α2,{\left| z \right|^2} = {\alpha ^2},
a=±2a = \pm 2
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