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Complex Numbers

JEE Mathematics · 150 questions · Page 14 of 15 · Click an option or "Show Solution" to reveal answer

Q131
Let z = x + iy be a non-zero complex number such that z2=i∣z∣2{z^2} = i{\left| z \right|^2}, where i = −1\sqrt { - 1} , then z lies on the :
A line, y = –x
B real axis
C line, y = x
D imaginary axis
Correct Answer
Option C
Solution

Given z = x + iy and

z2=i∣z∣2{z^2} = i{\left| z \right|^2}

⇒\Rightarrow (x + iy)2 = i(x2 + y2) ⇒\Rightarrow x2 - y2 + 2ixy = i(x2 + y2) + 0 Comparing both side we get, x2 - y2 = 0 ⇒\Rightarrow x2 = y2 and 2xy = (x2 + y2) ⇒\Rightarrow (x - y)2 = 0 ⇒\Rightarrow x = y ∴\therefore z lies on line x = y

Q132
Let (−2−13i)3=x+iy27(i=−1),  {\left( { - 2 - {1 \over 3}i} \right)^3} = {{x + iy} \over {27}}\left( {i = \sqrt { - 1} } \right),\,\, where x and y are real numbers, then y −- x equals :
A −- 85
B 85
C −- 91
D 91
Correct Answer
Option D
Solution
(−2−i3)3=−(6+i)327{\left( { - 2 - {i \over 3}} \right)^3} = - {{{{\left( {6 + i} \right)}^3}} \over {27}}
=−198−107i27=x+iy27= {{ - 198 - 107i} \over {27}} = {{x + iy} \over {27}}

Hence,

y−x=198−107=91y - x = 198 - 107 = 91
Q133
If a and b are real numbers such that (2+α)4=a+bα{\left( {2 + \alpha } \right)^4} = a + b\alpha where α=−1+i32\alpha = {{ - 1 + i\sqrt 3 } \over 2} then a + b is equal to :
A 33
B 9
C 24
D 57
Correct Answer
Option B
Solution
α=ω\alpha = \omega

as given

α=−1+i32\alpha = {{ - 1 + i\sqrt 3 } \over 2}
⇒(2+ω)4=a+bω (ω3=1)\Rightarrow {(2 + \omega )^4} = a + b\omega \,({\omega ^3} = 1)
⇒24+4.23ω+6.22ω3+4.2. ω3+ω4=a+bω\Rightarrow {2^4} + {4.2^3}\omega + {6.2^2}{\omega ^3} + 4.2.\,{\omega ^3} + {\omega ^4} = a + b\omega
⇒16+32ω+24ω2+8+ω=a+bω\Rightarrow 16 + 32\omega + 24{\omega ^2} + 8 + \omega = a + b\omega
⇒24+24ω2+33ω=a+bω\Rightarrow 24 + 24{\omega ^2} + 33\omega = a + b\omega
⇒−24ω+33ω=a+bω\Rightarrow - 24\omega + 33\omega = a + b\omega
⇒a=0, b=9\Rightarrow a = 0,\,b = 9
Q134
If z1{z_1} and z2{z_2} are two non-zero complex numbers such that  ∣z1+z2∣=∣z1∣+∣z2∣\,\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|, then arg z1{z_1} - arg z2{z_2} is equal to :
A π2 {\pi \over 2}\,
B −π - \pi
C 0
D −π2{{ - \pi } \over 2}
Correct Answer
Option C
Solution

Given that,

 ∣z1+z2∣=∣z1∣+∣z2∣\,\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|
 ∣z1+z2∣\,\left| {{z_1} + {z_2}} \right|

is the vector sum of

z1{z_1}

and

z2{z_2}

. So

 ∣z1+z2∣\,\left| {{z_1} + {z_2}} \right|

should be

<<
∣z1∣+∣z2∣\left| {{z_1}} \right| + \left| {{z_2}} \right|

but here they are equal so

z1{z_1}

and

z2{z_2}

are collinear. S if

z1{z_1}

makes an angle θ\theta with x axis then

z2{z_2}

will also make θ\theta angle. ∴\therefore arg

z1{z_1}

- arg

z2{z_2}

= θ\theta - θ\theta = 0

Q135
The value of ∑k=110(sin⁡2kπ11+i  cos⁡2kπ11)\sum\limits_{k = 1}^{10} {\left( {\sin {{2k\pi } \over {11}} + i\,\,\cos {{2k\pi } \over {11}}} \right)} is :
A i
B 1
C - 1
D - i
Correct Answer
Option D
Solution
∑k=110(sin⁡2kπ11+icos⁡2kπ11)\sum\limits_{k = 1}^{10} {\left( {\sin {{2k\pi } \over {11}} + i\cos {{2k\pi } \over {11}}} \right)}
=i∑k=110(cos⁡2kπ11−i sin⁡2kπ11)= i\sum\limits_{k = 1}^{10} {\left( {\cos {{2k\pi } \over {11}} - i\,\sin {{2k\pi } \over {11}}} \right)}
=i∑k=110e−2kπ11i=i{∑k=010e−2kπ11−1}= i\sum\limits_{k = 1}^{10} {{e^{ - {{2k\pi } \over {11}}}}} i = i\left\{ {\sum\limits_{k = 0}^{10} {{e^{ - {{2k\pi } \over {11}}}}} - 1} \right\}
=i[1+e−2π11i+e−4π11i+.....11  terms]−i= i\left[ {1 + {e^{ - {{2\pi } \over {11}}i}} + e - {{4\pi } \over {11}}i + .....11\,\,terms} \right] - i
=i[1−(e−2π11)111−e−2π11i]−i= i\left[ {{{1 - {{\left( {{e^{ - {{2\pi } \over {11}}}}} \right)}^{11}}} \over {1 - {e^{ - {{2\pi } \over {11}}i}}}}} \right] - i
=i[1−e−2πi1−e−2π11i]−i= i\left[ {{{1 - {e^{ - 2\pi i}}} \over {1 - {e^{ - {{2\pi } \over {11}}i}}}}} \right] - i
=i×0−i= i \times 0 - i

[as

      \,\,\,\,\,\,
e−2πi=1{e^{ - 2\pi i}} = 1

]

=−i= - i
Q136
If the four complex numbers z,z‾,z‾−2Re(z‾)z,\overline z ,\overline z - 2{\mathop{\rm Re}\nolimits} \left( {\overline z } \right) and z−2Re(z)z-2Re(z) represent the vertices of a square of side 4 units in the Argand plane, then ∣z∣|z| is equal to :
A 42\sqrt 2
B 4
C 2
D 22\sqrt 2
Correct Answer
Option D
Solution

Let

z=x+iyz = x + iy

Length of side = 4

AB=4AB = 4
∣z−z‾∣=4|z - \overline z | = 4
∣2y∣ =4;|2y|\, = 4;

⇒\Rightarrow

∣y∣ =2|y|\, = 2
BC=4BC = 4

⇒\Rightarrow

∣z‾−(z‾−2Re(z‾)∣ =4|\overline z - (\overline z - 2{\mathop{\rm Re}\nolimits} (\overline z )|\, = 4

⇒\Rightarrow

∣2x∣ =4; |2x|\, = 4;\,

⇒\Rightarrow

∣x∣ =2|x|\, = 2

∴\therefore

∣z∣ = x2+y2=4+4=22|z|\, = \,\sqrt {{x^2} + {y^2}} = \sqrt {4 + 4} = 2\sqrt 2
Q137
The conjugate of a complex number is 1i−1{1 \over {i - 1}} then that complex number is :
A −1i−1{{ - 1} \over {i - 1}}
B 1i+1 {1 \over {i + 1}}\,
C −1i+1{{ - 1} \over {i + 1}}
D 1i−1{1 \over {i - 1}}
Correct Answer
Option C
Solution
(1i−1)=1−i−1=−1i+1\left( {{1 \over {i - 1}}} \right) = {1 \over { - i - 1}} = {{ - 1} \over {i + 1}}
Q138
If the cube roots of unity are 1, ω , ω2\omega \,,\,{\omega ^2} then the roots of the equation (x−1)3{(x - 1)^3} + 8 = 0, are :
A −1,−1+2  ω,−1−2  ω2 - 1, - 1 + 2\,\,\omega , - 1 - 2\,\,{\omega ^2}
B −1,−1,−1 - 1, - 1, - 1
C −1,1−2ω,1−2ω2 - 1,1 - 2\omega ,1 - 2{\omega ^2}
D −1,1+2ω,1+2ω2 - 1,1 + 2\omega ,1 + 2{\omega ^2}
Correct Answer
Option C
Solution
(x−1)3+8=0{\left( {x - 1} \right)^3} + 8 = 0
⇒(x−1)=(−2)(1)1/3\Rightarrow \left( {x - 1} \right) = \left( { - 2} \right){\left( 1 \right)^{1/3}}
⇒x−1=−2   \Rightarrow x - 1 = - 2\,\,\,

or

   −2ω    \,\,\, - 2\omega \,\,\,\,

or

    −2ω2\,\,\,\, - 2{\omega ^2}

or

   x=−1   \,\,\,x = - 1\,\,\,

or

   1−2ω   \,\,\,1 - 2\omega \,\,\,

or

   1−2ω2.\,\,\,1 - 2{\omega ^2}.
Q139
If z is a complex number of unit modulus and argument θ\theta , then arg (1+z1+z‾)\left( {{{1 + z} \over {1 + \overline z }}} \right) equals :
A −θ   - \theta \,\,
B π2−θ {\pi \over 2} - \theta \,
C θ \theta \,
D  π−θ  \,\pi - \theta \,\,
Correct Answer
Option C
Solution

Given

    ∣z∣=1,  arg⁡ z=θ\,\,\,\,\left| z \right| = 1,\,\,\arg \,z = \theta

As we know,

    z→=1z\,\,\,\,\overrightarrow z = {1 \over z}

∴\therefore

    arg⁡(1+z1+z→)=arg⁡(1+z1+1z)\,\,\,\,\arg \left( {{{1 + z} \over {1 + \overrightarrow z }}} \right) = \arg \left( {{{1 + z} \over {1 + {1 \over z}}}} \right)
=arg⁡(z)=θ.= \arg \left( z \right) = \theta .
Q140
If z=x−iyz = x - iy and z13=p+iq{z^{{1 \over 3}}} = p + iq, then (xp+yq)(p2+q2){{\left( {{x \over p} + {y \over q}} \right)} \over {\left( {{p^2} + {q^2}} \right)}} is equal to :
A - 2
B - 1
C 2
D 1
Correct Answer
Option A
Solution

Given

z13=p+iq{z^{{1 \over 3}}} = p + iq

⇒\Rightarrow z = (p + iq)3 = p3 + (iq)3 +3p(iq)(p + iq) = p3 - iq3 +3ip2q - 3pq2 = p(p2 - 3q2) - iq(q2 - 3p2) Given that

z=x−iyz = x - iy

∴\therefore

x−iyx - iy

= p(p2 - 3q2) - iq(q2 - 3p2) By comparing both sides we get,

xp=p2−3q2{x \over p} = {p^2} - 3{q^2}

and

yq=q2−3p2{y \over q} = {q^2} - 3{p^2}

∴\therefore

(xp+yq)(p2+q2){{\left( {{x \over p} + {y \over q}} \right)} \over {\left( {{p^2} + {q^2}} \right)}}

=

p2−3q2+q2−3p2p2+q2{{{p^2} - 3{q^2} + {q^2} - 3{p^2}} \over {{p^2} + {q^2}}}

=

−2q2−2p2p2+q2{{ - 2{q^2} - 2{p^2}} \over {{p^2} + {q^2}}}

=

−2(q2+p2)p2+q2{{ - 2\left( {{q^2} + {p^2}} \right)} \over {{p^2} + {q^2}}}

=

−2-2
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