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Complex Numbers

JEE Mathematics · 150 questions · Page 8 of 15 · Click an option or "Show Solution" to reveal answer

Q71
Let z1 and z2 be two complex numbers such that z‾1=iz‾2{\overline z _1} = i{\overline z _2} and arg⁡(z1z‾2)=π\arg \left( {{{{z_1}} \over {{{\overline z }_2}}}} \right) = \pi . Then :
A arg⁡z2=π4\arg {z_2} = {\pi \over 4}
B arg⁡z2=−3π4\arg {z_2} = - {{3\pi } \over 4}
C arg⁡z1=π4\arg {z_1} = {\pi \over 4}
D arg⁡z1=−3π4\arg {z_1} = - {{3\pi } \over 4}
Correct Answer
Option C
Solution

∵\because

z1z2=−i⇒z1=−iz2{{{z_1}} \over {{z_2}}} = - i \Rightarrow {z_1} = - i{z_2}
⇒arg⁡(z1)=−π2+arg⁡(z2)\Rightarrow \arg ({z_1}) = - {\pi \over 2} + \arg ({z_2})

..... (i) Also

arg⁡(z1)−arg⁡(z‾2)=π\arg ({z_1}) - \arg ({\overline z _2}) = \pi
⇒arg⁡(z1)+arg⁡(z2)=π\Rightarrow \arg ({z_1}) + \arg ({z_2}) = \pi

..... (ii) From (i) and (ii), we get

arg⁡(z1)=π4\arg ({z_1}) = {\pi \over 4}

and

arg⁡(z2)=3π4\arg ({z_2}) = {{3\pi } \over 4}
Q72
Let z1=2+3i\mathrm{z_1=2+3i} and z2=3+4i\mathrm{z_2=3+4i}. The set S={z∈C:∣z−z1∣2−∣z−z2∣2=∣z1−z2∣2}\mathrm{S = \left\{ {z \in \mathbb{C}:{{\left| {z - {z_1}} \right|}^2} - {{\left| {z - {z_2}} \right|}^2} = {{\left| {{z_1} - {z_2}} \right|}^2}} \right\}} represents a
A hyperbola with the length of the transverse axis 7
B hyperbola with eccentricity 2
C straight line with the sum of its intercepts on the coordinate axes equals −18-18
D straight line with the sum of its intercepts on the coordinate axes equals 1414
Correct Answer
Option D
Solution

∣z−z1∣2−∣z−z2∣2=∣z1−z2∣2\left|z-z_{1}\right|^{2}-\left|z-z_{2}\right|^{2}=\left|z_{1}-z_{2}\right|^{2} ⇒(x−2)2+(y−3)2−(x−3)2−(y−4)2=1+1\Rightarrow(x-2)^{2}+(y-3)^{2}-(x-3)^{2}-(y-4)^{2}=1+1 ⇒−4x+4+9−6y−9+6x−16+8y=2\Rightarrow-4 x+4+9-6 y-9+6 x-16+8 y=2 ⇒2x+2y=14\Rightarrow 2 x+2 y=14 ⇒x+y=7\Rightarrow x+y=7

Q73
The real part of the complex number (1+2i)8 . (1−2i)2(3+2i) . (4−6i)‾{{{{(1 + 2i)}^8}\,.\,{{(1 - 2i)}^2}} \over {(3 + 2i)\,.\,\overline {(4 - 6i)} }} is equal to :
A 50013{{500} \over {13}}
B 11013{{110} \over {13}}
C 556{{55} \over {6}}
D 55013{{550} \over {13}}
Correct Answer
Option D
Solution

Given,

(1+2i)8 . (1−2i)2(3+2i) . (4−6i)‾{{{{(1 + 2i)}^8}\,.\,{{(1 - 2i)}^2}} \over {(3 + 2i)\,.\,\overline {(4 - 6i)} }}
=(1+2i)2(1−2i)2(1+2i)6(3+2i)(4+6i)= {{{{(1 + 2i)}^2}{{(1 - 2i)}^2}{{(1 + 2i)}^6}} \over {(3 + 2i)(4 + 6i)}}
=(1−4i2)2(1+2i)612+18i+8i+12i2= {{{{(1 - 4{i^2})}^2}{{(1 + 2i)}^6}} \over {12 + 18i + 8i + 12{i^2}}}
=(1+5)2[(1+2i)2]312+26i−12= {{{{(1 + 5)}^2}{{\left[ {{{(1 + 2i)}^2}} \right]}^3}} \over {12 + 26i - 12}}
=25(1+4i2+4i)326i= {{25{{(1 + 4{i^2} + 4i)}^3}} \over {26i}}
=25(1−4+4i)326i= {{25{{(1 - 4 + 4i)}^3}} \over {26i}}
=25(−3+4i)326i= {{25{{( - 3 + 4i)}^3}} \over {26i}}
=2526i[(−3)3+(4i)3+3 . (−3)2 . 4i+3(−3) . (4i)2]= {{25} \over {26i}}\left[ {{{( - 3)}^3} + {{(4i)}^3} + 3\,.\,{{( - 3)}^2}\,.\,4i + 3( - 3)\,.\,{{(4i)}^2}} \right]
=2526i(−27−64i+108i+144)= {{25} \over {26i}}( - 27 - 64i + 108i + 144)
=2526i(117+44i)= {{25} \over {26i}}(117 + 44i)
=25i26i2(117+44i)= {{25i} \over {26{i^2}}}(117 + 44i)
=25i−26(117+44i)= {{25i} \over { - 26}}(117 + 44i)
=25×117i−26−25×44i226= {{25 \times 117i} \over { - 26}} - {{25 \times 44{i^2}} \over {26}}
=25×117i−26+22×2513= {{25 \times 117i} \over { - 26}} + {{22 \times 25} \over {13}}
=25×117i−26+55013= {{25 \times 117i} \over { - 26}} + {{550} \over {13}}

∴\therefore Real part

=55013= {{550} \over {13}}
Q74
For z∈Cz \in \mathbb{C} if the minimum value of (∣z−32∣+∣z−p2i∣)(|z-3 \sqrt{2}|+|z-p \sqrt{2} i|) is 525 \sqrt{2}, then a value Question: of pp is _____________.
A 3
B 72\dfrac{7}{2}
C 4
D 92\dfrac{9}{2}
Correct Answer
Option C
Solution

It is sum of distance of z from

(32,0)\left( {3\sqrt 2 ,0} \right)

and

(0,p2)\left( {0,p\sqrt 2 } \right)

For minimising, z should lie on AB and

AB=52AB = 5\sqrt 2
(AB)2=18+2p2{(AB)^2} = 18 + 2{p^2}
p= ± 4p = \, \pm \,4
Q75
Let O be the origin and A be the point z1=1+2i{z_1} = 1 + 2i. If B is the point z2{z_2}, $${\mathop{\rm Re}\nolimits} ({z_2})
A arg⁡z2=π−tan⁡−13\arg {z_2} = \pi - {\tan ^{ - 1}}3
B arg⁡(z1−2z2)=−tan⁡−143\arg ({z_1} - 2{z_2}) = - {\tan ^{ - 1}}{4 \over 3}
C ∣z2∣=10|{z_2}| = \sqrt {10}
D ∣2z1−z2∣=5|2{z_1} - {z_2}| = 5
Correct Answer
Option D
Solution
z2−0(1+2i)−0=∣OB∣∣OA∣eiπ4{{{z_2} - 0} \over {(1 + 2i) - 0}} = {{|OB|} \over {|OA|}}{e^{{{i\pi } \over 4}}}
⇒z21+2i=2eiπ4\Rightarrow {{{z_2}} \over {1 + 2i}} = \sqrt 2 {e^{{{i\pi } \over 4}}}

OR

z2=(1+2i)(1+i){z_2} = (1 + 2i)(1 + i)
=−1+3i= - 1 + 3i
arg⁡z2=π−tan⁡−13\arg {z_2} = \pi - {\tan ^{ - 1}}3
∣z2∣=10|{z_2}| = \sqrt {10}
z1−2z2=(1+2i)+2−6i=3−4i{z_1} - 2{z_2} = (1 + 2i) + 2 - 6i = 3 - 4i
arg⁡(z1−2z2)=−tan⁡−143\arg ({z_1} - 2{z_2}) = - {\tan ^{ - 1}}{4 \over 3}
∣2z1−z2∣=∣2+4i+1−3i∣=∣3+i∣=10|2{z_1} - {z_2}| = |2 + 4i + 1 - 3i| = |3 + i| = \sqrt {10}
Q76
If z=x+iyz=x+i y satisfies ∣z∣−2=0|z|-2=0 and ∣z−i∣−∣z+5i∣=0|z-i|-|z+5 i|=0, then :
A x+2y−4=0x+2 y-4=0
B x2+y−4=0x^{2}+y-4=0
C x+2y+4=0x+2 y+4=0
D x2−y+3=0x^{2}-y+3=0
Correct Answer
Option C
Solution
∣z−i∣=∣z+5i∣|z - i| = |z + 5i|

So,

z\mathrm{z}

lies on

⊥r{ \bot ^r}

bisector of

(0,1)(0,1)

and

(0,−5)(0, - 5)

i.e., line

y=−2y = - 2

as

∣z∣=2|z| = 2
⇒z=−2i\Rightarrow z = - 2i
x=0x = 0

and

y=−2y = - 2

so,

x+2y+4=0x + 2y + 4 = 0
Q77
Let S be the set of all $$(\alpha, \beta), \pi
A 3
B 3 i
C 1
D 2 −- i
Correct Answer
Option C
Solution

∵\because

1−isin⁡α1+2isin⁡α{{1 - i\sin \alpha } \over {1 + 2i\sin \alpha }}

is purely imaginary ∴\therefore

1−isin⁡α1+2isin⁡α+1+isin⁡α1−2isin⁡α=0{{1 - i\sin \alpha } \over {1 + 2i\sin \alpha }} + {{1 + i\sin \alpha } \over {1 - 2i\sin \alpha }} = 0
⇒1−2sin⁡2α=0\Rightarrow 1 - 2{\sin ^2}\alpha = 0

∴\therefore

α=5π4, 7π4\alpha = {{5\pi } \over 4},\,{{7\pi } \over 4}

and

1+icos⁡β1−2icos⁡β{{1 + i\cos \beta } \over {1 - 2i\cos \beta }}

is purely real

1+icos⁡β1−2icos⁡β−1−icos⁡β1+2icos⁡β=0{{1 + i\cos \beta } \over {1 - 2i\cos \beta }} - {{1 - i\cos \beta } \over {1 + 2i\cos \beta }} = 0
⇒cos⁡β=0\Rightarrow \cos \beta = 0

∴\therefore

β=3π2\beta = {{3\pi } \over 2}

∴\therefore

S={(5π2,3π2),(7π4,3π2)}S = \left\{ {\left( {{{5\pi } \over 2},{{3\pi } \over 2}} \right),\left( {{{7\pi } \over 4},{{3\pi } \over 2}} \right)} \right\}
Zαβ=1−i{Z_{\alpha \beta }} = 1 - i

and

Zαβ=−1−i{Z_{\alpha \beta }} = - 1 - i

∴\therefore

∑(α,β)∈S(iZαβ+1iZ‾αβ)=i(−2i)+1i[11+i+1−1+i]\sum\limits_{(\alpha ,\beta ) \in S} {\left( {i{Z_{\alpha \beta }} + {1 \over {i{{\overline Z }_{\alpha \beta }}}}} \right) = i( - 2i) + {1 \over i}\left[ {{1 \over {1 + i}} + {1 \over { - 1 + i}}} \right]}
=2+1i2i−2=1= 2 + {1 \over i}{{2i} \over { - 2}} = 1
Q78
If z≠0z \neq 0 be a complex number such that ∣z−1z∣=2\left|z-\dfrac{1}{z}\right|=2, then the maximum value of ∣z∣|z| is :
A 2\sqrt{2}
B 1
C 2−1\sqrt{2}-1
D 2+1\sqrt{2}+1
Correct Answer
Option D
Solution

We know,

∣∣z1∣−∣z2∣∣≤∣z1+z2∣≤∣z1∣+∣z2∣\left| {|{z_1}| - |{z_2}|} \right| \le \left| {{z_1} + {z_2}} \right| \le |{z_1}| + |{z_2}|

∴\therefore

∣∣z∣−1∣z∣∣≤∣z−1z∣\left| {|z| - {1 \over {|z|}}} \right| \le \left| {z - {1 \over z}} \right|
⇒∣∣z∣−1∣z∣∣≤2\Rightarrow \left| {|z| - {1 \over {|z|}}} \right| \le 2

[Given

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

]

⇒∣∣z∣2−1∣z∣∣≤2\Rightarrow \left| {{{|z{|^2} - 1} \over {|z|}}} \right| \le 2
⇒−2≤∣z∣2−1∣z∣≤2\Rightarrow - 2 \le {{|z{|^2} - 1} \over {|z|}} \le 2

∴\therefore

∣z∣2−1∣z∣≤2{{|z{|^2} - 1} \over {|z|}} \le 2
⇒∣z∣2−1≤2∣z∣\Rightarrow |z{|^2} - 1 \le 2|z|
⇒∣z∣2−2∣z∣−1≤0\Rightarrow |z{|^2} - 2|z| - 1 \le 0
⇒∣z∣2−2∣z∣+1−2≤0\Rightarrow |z{|^2} - 2|z| + 1 - 2 \le 0
⇒(∣z∣−1)2−2≤0\Rightarrow {(|z| - 1)^2} - 2 \le 0
⇒−2≤∣z∣−1≤2\Rightarrow - \sqrt 2 \le |z| - 1 \le \sqrt 2
⇒1−2≤∣z∣≤1+2\Rightarrow 1 - \sqrt 2 \le |z| \le 1 + \sqrt 2

..... (1) or

−2≤∣z∣2−1∣z∣- 2 \le {{|z{|^2} - 1} \over {|z|}}
⇒∣z∣2−1≤−2∣z∣\Rightarrow |z{|^2} - 1 \le - 2|z|
⇒∣z∣2+2∣z∣−1≤0\Rightarrow |z{|^2} + 2|z| - 1 \le 0
⇒∣z∣2+2∣z∣+1−2≤0\Rightarrow |z{|^2} + 2|z| + 1 - 2 \le 0
⇒(∣z∣+1)2−2≤0\Rightarrow {(|z| + 1)^2} - 2 \le 0
⇒−2≤∣z∣+1≤+ 2\Rightarrow - \sqrt 2 \le |z| + 1 \le + \,\sqrt 2
⇒−2−1≤∣z∣≤2−1\Rightarrow - \sqrt 2 - 1 \le |z| \le \sqrt 2 - 1

...... (2) From (1) and (2) we get, Maximum value of

∣z∣=2+1|z| = \sqrt 2 + 1

and minimum value of

∣z∣=−2−1|z| = - \sqrt 2 - 1
Q79
Let $$\mathrm{S}=\{z=x+i y:|z-1+i| \geq|z|,|z|
A (−2,122]\left(-\sqrt{2}, \dfrac{1}{2 \sqrt{2}}\right]
B (−12,14]\left(-\dfrac{1}{\sqrt{2}}, \dfrac{1}{4}\right]
C (−2,12]\left(-\sqrt{2}, \dfrac{1}{2}\right]
D (−12,122]\left(-\dfrac{1}{\sqrt{2}}, \dfrac{1}{2 \sqrt{2}}\right]
Correct Answer
Option B
Solution

S:{z=x+iy:∣z−1+i∣≥∣z∣,∣z∣<2,∣z−i∣=∣z−1∣}S:\{z=x+i y:|z-1+i| \geq|z|,|z|<2,|z-i|=|z-1|\}

∣z−1+i∣≥∣z∣|z-1+i| \geq|z|

∣z∣<2|z|<2

∣z−i∣=∣z−1∣|z-i| = |z-1|

∵w∈S\because \quad w \in S and w=2x+iyw=2 x+i y 2x<12∴x<142 x<\dfrac{1}{2} \quad\quad \therefore x<\dfrac{1}{4} (2x)2+(−2x)2<4(2 x)^{2}+(-2 x)^{2}<4 4x2+4x2<44 x^{2}+4 x^{2}<4 x2<12⇒x∈(−12,12)x^{2}<\dfrac{1}{2} \Rightarrow x \in\left(-\dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}\right) ∴x∈(−12,14]\therefore \quad x \in\left(-\dfrac{1}{2}, \dfrac{1}{4}\right]

Q80
Let a,ba,b be two real numbers such that $$ab
A (1+74)\left(\dfrac{1+\sqrt{7}}{4}\right)
B 12\dfrac{1}{2}
C 0
D −-1
Correct Answer
Option C
Solution

∣1+aib+i∣=1∣1+ia∣=∣b+i∣a2+1=b2+1⇒a=±b⇒b=−a as ab<0(a+ib) lies on ∣z−1∣=∣2z∣∣a+ib−1∣=2∣a+ib∣(a−1)2+b2=4(a2+b2)(a−1)2=a2=4(2a2)1−2a=6a2⇒6a2+2a−1=0a=−2±2812=−1±76a=7−16 and b=1−76\begin{aligned} & \left|\dfrac{1+a i}{b+i}\right|=1 \\\\ & |1+i a|=|b+i| \\\\ & a^2+1=b^2+1 \Rightarrow \mathrm{a}=\pm \mathrm{b} \Rightarrow \mathrm{b}=-\mathrm{a} \quad \text{ as } \mathrm{ab}<0 \\\\ & (\mathrm{a} +\mathrm{ib}) \text{ lies on }|z-1|=|2 z| \\\\ & |a+i b-1|=2|a+i b| \\\\ & (a-1)^2+b^2=4\left(a^2+b^2\right) \\\\ & (a-1)^2=a^2=4\left(2 a^2\right) \\\\ & 1-2 a=6 a^2 \Rightarrow 6 a^2+2 a-1=0 \\\\ & a=\dfrac{-2 \pm \sqrt{28}}{12}=\dfrac{-1 \pm \sqrt{7}}{6} \\\\ & a=\dfrac{\sqrt{7}-1}{6} \text{ and } b=\dfrac{1-\sqrt{7}}{6}\end{aligned}

[a]=0∴1+[a]4b=64(1−7)=−(1+74)\begin{aligned} & {[a]=0} \\\\ & \therefore \frac{1+[a]}{4 b}=\frac{6}{4(1-\sqrt{7})}=-\left(\frac{1+\sqrt{7}}{4}\right) \end{aligned}

Similarly when

a=−1−76  and  b=1+76a = {{ - 1 - \sqrt 7 } \over 6}\,\text{ and }\,b = {{1 + \sqrt 7 } \over 6}

then [aa] = -1 ∴\therefore

1+[a]4b=1−14×1+76{{1 + \left[ a \right]} \over {4b}} = {{1 - 1} \over {4 \times {{1 + \sqrt 7 } \over 6}}}

= 0

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