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Complex Numbers

JEE Mathematics · 150 questions · Page 5 of 15 · Click an option or "Show Solution" to reveal answer

Q41
Let z be a complex number such that |z| + z = 3 + i (where i = −1\sqrt { - 1} ). Then |z| is equal to :
A 343{{\sqrt {34} } \over 3}
B 53{5 \over 3}
C 54{5 \over 4}
D 414{{\sqrt {41} } \over 4}
Correct Answer
Option B
Solution
∣z∣+z=3+i\left| z \right| + z = 3 + i
z=3−∣z∣+iz = 3 - \left| z \right| + i

Let

3−∣z∣=a⇒∣z∣=(3−a)3 - \left| z \right| = a \Rightarrow \left| z \right| = \left( {3 - a} \right)
⇒z=a+i⇒∣z∣=a2+1\Rightarrow z = a + i \Rightarrow \left| z \right| = \sqrt {{a^2} + 1}
⇒9+a2−6a=a2+1⇒a=86=43\Rightarrow 9 + {a^2} - 6a = {a^2} + 1 \Rightarrow a = {8 \over 6} = {4 \over 3}
⇒∣z∣=3−43=53\Rightarrow \left| z \right| = 3 - {4 \over 3} = {5 \over 3}
Q42
The value of (−1+i31−i)30{\left( {{{ - 1 + i\sqrt 3 } \over {1 - i}}} \right)^{30}} is :
A –215i
B –215
C 215i
D 65
Correct Answer
Option A
Solution
(−1+i31−i)30{\left( {{{ - 1 + i\sqrt 3 } \over {1 - i}}} \right)^{30}}

=

(2ω1−i)30{\left( {{{2\omega } \over {1 - i}}} \right)^{30}}

=

230.ω30((1−i)2)15{{{2^{30}}.{\omega ^{30}}} \over {{{\left( {{{\left( {1 - i} \right)}^2}} \right)}^{15}}}}

=

230.1(1+i2−2i)15{{{2^{30}}.1} \over {{{\left( {1 + {i^{^2}} - 2i} \right)}^{15}}}}

=

230.1−215.i15{{{2^{30}}.1} \over { - {2^{15}}.{i^{15}}}}

= –215i

Q43
The region represented by {z = x + iy ∈ \in C : |z| – Re(z) ≤ \le 1} is also given by the inequality : {z = x + iy ∈ \in C : |z| – Re(z) ≤ \le 1}
A y2 ≤ \le 2(x+12)2\left( {x + {1 \over 2}} \right)
B y2 ≤ \le x+12{x + {1 \over 2}}
C y2 ≥ \ge 2(x + 1)
D y2 ≥ \ge x + 1
Correct Answer
Option A
Solution

Given z = x + iy |z| – Re(z) ≤\le 1 ⇒\Rightarrow

x2+y2\sqrt {{x^2} + {y^2}}

- x ≤\le 1 ⇒\Rightarrow

x2+y2\sqrt {{x^2} + {y^2}}

≤\le 1 + x ⇒\Rightarrow x2 + y2 ≤\le 1 + 2x + x2 ⇒\Rightarrow y2 ≤\le 2x + 1 ⇒\Rightarrow y2 ≤\le 2

(x+12)\left( {x + {1 \over 2}} \right)
Q44
Let u=2z+iz−kiu = {{2z + i} \over {z - ki}}, z = x + iy and k > 0. If the curve represented by Re(u) + Im(u) = 1 intersects the y-axis at the points P and Q where PQ = 5, then the value of k is :
A 2
B 4
C 1/2
D 3/2
Correct Answer
Option A
Solution

Given, z = x + iy and

u=2z+iz−kiu = {{2z + i} \over {z - ki}}
=2(x+iy)+i(x+iy)−ki= {{2(x + iy) + i} \over {(x + iy) - ki}}
=2x+i(2y+1)x+i(y−k)×x−i(y−k)x−i(y−k)= {{2x + i(2y + 1)} \over {x + i(y - k)}} \times {{x - i(y - k)} \over {x - i(y - k)}}
=2x2+(2y+1)(y−k)+i(2xy+x−2xy+2kx)x2+(y−k)2= {{2{x^2} + (2y + 1)(y - k) + i(2xy + x - 2xy + 2kx)} \over {{x^2} + {{(y - k)}^2}}}

Given,

Re(u)+Im(u)=1{\mathop{\rm Re}\nolimits} (u) + {\mathop{\rm Im}\nolimits} (u) = 1
⇒2x2+(2y+1)(y−k)x2+(y−k)2+x+2kxx2+(y−k)2=1\Rightarrow {{2{x^2} + (2y + 1)(y - k)} \over {{x^2} + {{(y - k)}^2}}} + {{x + 2kx} \over {{x^2} + {{(y - k)}^2}}} = 1
⇒2x2+(2y+1)(y−k)+x+2kx=x2+(y−k)2\Rightarrow 2{x^2} + (2y + 1)(y - k) + x + 2kx = {x^2} + {(y - k)^2}

This curve intersect the y-axis at point P and Q, so at point P and Q x = 0 Putting x = 0 at the above equation, ∴\therefore

(2y+1)(y−k)=(y−k)2(2y + 1)(y - k) = {(y - k)^2}
⇒2y2+y−2yk−k=y2+k2−2ky\Rightarrow 2{y^2} + y - 2yk - k = {y^2} + {k^2} - 2ky
⇒y2+y−(k+k2)=0\Rightarrow {y^2} + y - (k + {k^2}) = 0

Let roots of this quadratic equation y1 and y2 ∴\therefore Point P (0, y1) and Q (0, y2) and

y1+y2=1{y_{_1}} + {y_2} = 1

,

y1y2=−k−k2{y_1}{y_2} = - k - {k^2}

∴\therefore

(y1−y2)2=(y1+y2)2−4y1y2{({y_1} - {y_2})^2} = {({y_1} + {y_2})^2} - 4{y_1}{y_2}
=1+4k+4k2= 1 + 4k + 4{k^2}
⇒∣y1−y2∣=1+4k+4k2\Rightarrow |{y_1} - {y_2}| = \sqrt {1 + 4k + 4{k^2}}

Given, PQ = 5

⇒∣y1−y2∣=5\Rightarrow |{y_1} - {y_2}| = 5
⇒1+4k+4k2=5\Rightarrow \sqrt {1 + 4k + 4{k^2}} = 5
⇒k2+k−6=0\Rightarrow {k^2} + k - 6 = 0
⇒k=−3, 2\Rightarrow k = - 3,\,2

So, k = 2 (Given k > 0)

Q45
If z1 , z2 are complex numbers such that Re(z1) = |z1 – 1|, Re(z2) = |z2 – 1| , and arg(z1 - z2) = π6{\pi \over 6}, then Im(z1 + z2 ) is equal to :
A 32{{\sqrt 3 } \over 2}
B 13{1 \over {\sqrt 3 }}
C 23{2 \over {\sqrt 3 }}
D 23{2\sqrt 3 }
Correct Answer
Option D
Solution

Let

z1=x1+iy1, z2=x2+iy2{z_1} = {x_1} + i{y_1},\,{z_2} = {x_2} + i{y_2}

Given Re(z1) = |z1 – 1| ∴\therefore x1 = |(x1 - 1) + iy1| ⇒\Rightarrow x1 =

(x1−1)2+y12\sqrt {{{\left( {{x_1} - 1} \right)}^2} + y_1^2}

⇒\Rightarrow

x12=(x1−1)2+y12{x_1}^2 = {({x_1} - 1)^2} + {y_1}^2
⇒y12−2x1+1=0\Rightarrow {y_1}^2 - 2{x_1} + 1 = 0

Also given Re(z2) = |z2 – 1| ∴\therefore x2 = |(x2 - 1) + iy2| ⇒\Rightarrow

x22=(x2−1)2+y22{x_2}^2 = {({x^2} - 1)^2} + {y_2}^2

⇒\Rightarrow

y22−2x2−1=0y_2^2 - 2{x_2} - 1 = 0

Performing equation (1) - (2),

(y12−y22)+2(x2−x1)=0({y_1}^2 - {y_2}^2) + 2({x_2} - {x_1}) = 0

⇒\Rightarrow

(y1+y2)(y1−y2)=2(x1−x2)({y_1} + {y_2})({y_1} - {y_2}) = 2({x_1} - {x_2})

⇒\Rightarrow

y1+y2=2(x1−x2y1−y2){y_1} + {y_2} = 2\left( {{{{x_1} - {x_2}} \over {{y_1} - {y_2}}}} \right)

Now given,

arg⁡(z1−z2)=π6\arg ({z_1} - {z_2}) = {\pi \over 6}

⇒\Rightarrow

tan⁡−1(y1−y2x1−x2)=π6{\tan ^{ - 1}}\left( {{{{y_1} - {y_2}} \over {{x_1} - {x_2}}}} \right) = {\pi \over 6}
⇒y1−y2x1−x2=13\Rightarrow {{{y_1} - {y_2}} \over {{x_1} - {x_2}}} = {1 \over {\sqrt 3 }}

∴\therefore

y1+y2=23{y_1} + {y_2} = 2\sqrt 3
Q46
The imaginary part of (3+2−54)12−(3−2−54)12{\left( {3 + 2\sqrt { - 54} } \right)^{{1 \over 2}}} - {\left( {3 - 2\sqrt { - 54} } \right)^{{1 \over 2}}} can be :
A -26\sqrt 6
B 6
C 6\sqrt 6
D -6\sqrt 6
Correct Answer
Option A
Solution
3+2−543 + 2\sqrt { - 54}
=9−6+2−54= 9 - 6 + 2\sqrt { - 54}
=9+(6i)2+2.3.6i= 9 + {\left( {\sqrt 6 i} \right)^2} + 2.3.\sqrt 6 i
=32+(6i)2+2.3.6i= {3^2} + {\left( {\sqrt 6 i} \right)^2} + 2.3.\sqrt 6 i
=(3+6i)2= {\left( {3 + \sqrt 6 i} \right)^2}

Similarly,

(3−2−54)=(3−6i)2\left( {3 - 2\sqrt { - 54} } \right) = {\left( {3 - \sqrt 6 i} \right)^2}
∴(3+2−54)12−(3−2−54)12\therefore {\left( {3 + 2\sqrt { - 54} } \right)^{{1 \over 2}}} - {\left( {3 - 2\sqrt { - 54} } \right)^{{1 \over 2}}}
=±(3+6i)−[±(3−6i)]= \pm \left( {3 + \sqrt 6 i} \right) - \left[ { \pm \left( {3 - \sqrt 6 i} \right)} \right]
=6,−6,26i,−26i= 6, - 6,2\sqrt 6 i, - 2\sqrt 6 i

∴\therefore Possible imaginary parts are

26i,−26i2\sqrt 6 i, - 2\sqrt 6 i
Q47
If 3+isin⁡θ4−icos⁡θ{{3 + i\sin \theta } \over {4 - i\cos \theta }}, θ\theta ∈ \in [0, 2θ\theta ], is a real number, then an argument of sinθ\theta + icosθ\theta is :
A π−tan⁡−1(34)\pi - {\tan ^{ - 1}}\left( {{3 \over 4}} \right)
B −tan⁡−1(34) - {\tan ^{ - 1}}\left( {{3 \over 4}} \right)
C tan⁡−1(43){\tan ^{ - 1}}\left( {{4 \over 3}} \right)
D π−tan⁡−1(43)\pi - {\tan ^{ - 1}}\left( {{4 \over 3}} \right)
Correct Answer
Option D
Solution

Let z =

3+isin⁡θ4−icos⁡θ{{3 + i\sin \theta } \over {4 - i\cos \theta }}

=

3+isin⁡θ4−icos⁡θ×(4+icos⁡θ)(4+icos⁡θ){{3 + i\sin \theta } \over {4 - i\cos \theta }} \times {{\left( {4 + i\cos \theta } \right)} \over {\left( {4 + i\cos \theta } \right)}}

=

(12−sin⁡θcos⁡θ)+i(4sin⁡θ+3cos⁡θ)16+cos⁡2θ{{\left( {12 - \sin \theta \cos \theta } \right) + i\left( {4\sin \theta + 3\cos \theta } \right)} \over {16 + {{\cos }^2}\theta }}

As Z is purely real ∴\therefore 4sinθ\theta + 3cosθ\theta = 0 ⇒\Rightarrow tan θ\theta =

−34- {3 \over 4}

∴\therefore θ\theta lies in the 2nd quadrant then arg(sinθ + icosθ) = π\pi +

tan⁡−1(cos⁡θsin⁡θ){\tan ^{ - 1}}\left( {{{\cos \theta } \over {\sin \theta }}} \right)

=

π−tan⁡−1(43)\pi - {\tan ^{ - 1}}\left( {{4 \over 3}} \right)
Q48
If z be a complex number satisfying |Re(z)| + |Im(z)| = 4, then |z| cannot be :
A 10\sqrt {10}
B 7\sqrt {7}
C 172\sqrt {{{17} \over 2}}
D 8\sqrt {8}
Correct Answer
Option B
Solution

Let z = x + iy given that |Re(z)| + |Im(z)| = 4 ∴\therefore |x| + |y| = 4 Maximum value of |z| = 4 Minimum value of |z| = perpendicular distance of line AB from (0, 0) =

222\sqrt 2

∴\therefore |z|

∈\in
[22,4]\left[ {2\sqrt 2 ,4} \right]

∴\therefore |z| cannot be

7\sqrt {7}

.

Q49
If the equation, x2 + bx + 45 = 0 (b ∈ \in R) has conjugate complex roots and they satisfy |z +1| = 210\sqrt {10} , then :
A b2 – b = 42
B b2 + b = 12
C b2 + b = 72
D b2 – b = 30
Correct Answer
Option D
Solution

x2 + bx = 45 = 0 (b

∈\in

R) has roots α\alpha + iβ\beta, α\alpha – iβ\beta sum of roots = – b = 2α\alpha product of roots = 45 = α\alpha2 + β\beta2 Let z = x + iy ∴\therefore |x + iy +1| = 2

10\sqrt {10}
∣x+iy+1∣2=(210)2{\left| {x + iy + 1} \right|^2} = {\left( {2\sqrt {10} } \right)^2}

⇒\Rightarrow (x + 1)2 + y2 = 40 ⇒\Rightarrow (α\alpha + 1)2 + β\beta2 = 40 [putting real part α\alpha in place of x and imaginary part β\beta in place of y] ⇒\Rightarrow α\alpha2 + 2α\alpha + 1 + β\beta2 = 40 ⇒\Rightarrow 45 + 2α\alpha + 1 = 40 ⇒\Rightarrow α\alpha = -3 ∴\therefore -b = 2α\alpha = 2×\times(-3) = -6 ⇒\Rightarrow b = 6 By checking options we found b2 – b = 30.

Q50
For n∈N\mathrm{n} \in \mathbf{N}, let Sn={z∈C:∣z−3+2i∣=n4}\mathrm{S}_{\mathrm{n}}=\left\{z \in \mathbf{C}:|z-3+2 i|=\frac{\mathrm{n}}{4}\right\} and Tn={z∈C:∣z−2+3i∣=1n}\mathrm{T}_{\mathrm{n}}=\left\{z \in \mathbf{C}:|z-2+3 i|=\frac{1}{\mathrm{n}}\right\}. Then the number of elements in the set {n∈N:Sn∩Tn=ϕ}\left\{n \in \mathbf{N}: S_{n} \cap T_{n}=\phi\right\} is :
A 0
B 2
C 3
D 4
Correct Answer
Option D
Solution

Sn={z∈C:∣z−3+2i∣=n4}S_{n}=\left\{z \in C:|z-3+2 i|=\dfrac{n}{4}\right\} represents a circle with centre C1(3,−2)C_{1}(3,-2) and radius r1=n4r_{1}=\dfrac{n}{4} Similarly TnT_{n} represents circle with centre C2(2,−3)C_{2}(2,-3) and radius r2=1nr_{2}=\dfrac{1}{n}  As Sn∩Tn=ϕ\text{ As } S_{n} \cap T_{n}=\phi

C1C2>r1+r2 OR C1C2n4+1n OR 2<∣n4−1n∣n=1,2,3,4n may take infinite values \begin{array}{llr} C_{1} C_{2}>r_{1}+r_{2}& \text{ OR } & C_{1} C_{2}\frac{n}{4}+\frac{1}{n} & \text{ OR } & \sqrt{2}<\left|\frac{n}{4}-\frac{1}{n}\right| \\\\ n=1,2,3,4&& n \text{ may take infinite values } \end{array}
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