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Complex Numbers

JEE Mathematics · 150 questions · Page 11 of 15 · Click an option or "Show Solution" to reveal answer

Q101
Let zz be a complex number such that ∣z+2∣=1|z+2|=1 and lm⁡(z+1z+2)=15\operatorname{lm}\left(\dfrac{z+1}{z+2}\right)=\dfrac{1}{5}. Then the value of ∣Re⁡(z+2‾)∣|\operatorname{Re}(\overline{z+2})| is
A 265\dfrac{2 \sqrt{6}}{5}
B 245\dfrac{24}{5}
C 65\dfrac{\sqrt{6}}{5}
D 1+65\dfrac{1+\sqrt{6}}{5}
Correct Answer
Option A
Solution
∣z+2∣=1Im⁡m(z+1z+2)=15∣Re⁡(z+2‾)∣=?\begin{aligned} & |z+2|=1 \\ & \operatorname{Im}_m\left(\frac{z+1}{z+2}\right)=\frac{1}{5} \\ & |\operatorname{Re}(\overline{z+2})|=? \end{aligned}

Let

z=x+iyz=x+i y
∵∣z+2∣=1⇒(x+2)2+y2=1…(1)Im(z+1z+2)=15⇒Im(x+iy+1x+iy+2)=15⇒Im∣[(x+1)+iy][(x+)−iy](x+2)2+y2∣=15\begin{aligned} & \because|z+2|=1 \Rightarrow(x+2)^2+y^2=1 \quad \ldots(1) \\ & I_m\left(\frac{z+1}{z+2}\right)=\frac{1}{5} \Rightarrow I_m\left(\frac{x+i y+1}{x+i y+2}\right)=\frac{1}{5} \\ & \Rightarrow I_m\left|\frac{[(x+1)+i y][(x+)-i y]}{(x+2)^2+y^2}\right|=\frac{1}{5} \end{aligned}
y(x+2)−y(x+)(x+2)2+y2=15.... (2)\frac{y(x+2)-y(x+)}{(x+2)^2+y^2}=\frac{1}{5}\quad \text{.... (2)}
⇒y=15\Rightarrow y=\frac{1}{5}

Substituting in equation (1)

(x+2)2+125=1(x+2)2=2425⇒x=−2±245∣Re⁡(x+iy+2‾)∣=x+2=±245=265\begin{aligned} & (x+2)^2+\frac{1}{25}=1 \\ & (x+2)^2=\frac{24}{25} \\ & \Rightarrow x=-2 \pm \frac{\sqrt{24}}{5} \\ & |\operatorname{Re}(\overline{x+i y+2})| \\ & =x+2= \pm \frac{\sqrt{24}}{5}=\frac{2 \sqrt{6}}{5} \end{aligned}
Q102
Let S=∣z∈C:∣z−1∣=1\mathrm{S}=|\mathrm{z} \in \mathrm{C}:| z-1 \mid=1 and (2−1)(z+zˉ)−i(z−zˉ)=22∣(\sqrt{2}-1)(z+\bar{z})-i(z-\bar{z})=2 \sqrt{2} \mid. Let z1,z2∈Sz_1, z_2 \in \mathrm{S} be such that ∣z1∣=max⁡z∈s∣z∣\left|z_1\right|=\max\limits_{z \in s}|z| and ∣z2∣=min⁡z∈S∣z∣\left|z_2\right|=\min\limits _{z \in S}|z|. Then ∣2z1−z2∣2\left|\sqrt{2} z_1-z_2\right|^2 equals :
A 1
B 4
C 3
D 2
Correct Answer
Option D
Solution

Let z=x+iyz=x+i y

∣z−1∣=1⇒∣x+iy−1∣=1(x−1)2+y2=1.......(1)\begin{aligned} & |z-1|=1 \Rightarrow|x+i y-1|=1 \\\\ & (x-1)^2+y^2=1 .......(1) \end{aligned}

(2−1)(z+zˉ)−i(z−zˉ)=22 (Given) (2−1)(2x)−i(2iy)=22(2−1)x+y=2.........(2)\begin{aligned} & (\sqrt{2}-1)(z+\bar{z})-i(z-\bar{z})=2 \sqrt{2} \text{ (Given) } \\\\ & (\sqrt{2}-1)(2 x)-i(2 i y)=2 \sqrt{2} \\\\ & (\sqrt{2}-1) x+y=\sqrt{2} .........(2)\end{aligned} Solving (1) and (2), we get

(x−1)2+(2−(2−1)x)2=1(x2−2x+1)+2+(2−1)2x2−22(2−1)x=0x2(1+(2−1)2)+x(−2−22(2−1))+2=0x2(4−22)+x(22−6)+2=0x2(2−2)+x(2−3)+1=0\begin{aligned} & (x-1)^2+(\sqrt{2}-(\sqrt{2}-1) x)^2=1 \\\\ & \left(x^2-2 x+1\right)+2+(\sqrt{2}-1)^2 x^2-2 \sqrt{2}(\sqrt{2}-1) x=0 \\\\ & x^2\left(1+(\sqrt{2}-1)^2\right)+x(-2-2 \sqrt{2}(\sqrt{2}-1))+2=0 \\\\ & x^2(4-2 \sqrt{2})+x(2 \sqrt{2}-6)+2=0 \\\\ & x^2(2-\sqrt{2})+x(\sqrt{2}-3)+1=0 \end{aligned}

⇒x=1 and x=12−2..........(3) When x=1,y=1⇒z2=1+i When x=12−2,y=2−12⇒z1=(1+12)+i2\begin{aligned} & \Rightarrow x=1 \text{ and } x=\dfrac{1}{2-\sqrt{2}} ..........(3) \\\\ & \text{ When } x=1, y=1 \Rightarrow z_2=1+i \\\\ & \text{ When } x=\dfrac{1}{2-\sqrt{2}}, y=\sqrt{2}-\dfrac{1}{\sqrt{2}} \\\\ & \Rightarrow z_1=\left(1+\dfrac{1}{\sqrt{2}}\right)+\dfrac{i}{\sqrt{2}}\end{aligned} Now,

∣2z1−z2∣2=∣(12+1)2+i−(1+i)∣2=(2)2=2\begin{aligned} & \left|\sqrt{2} z_1-z_2\right|^2 \\\\ & =\left|\left(\frac{1}{\sqrt{2}}+1\right) \sqrt{2}+i-(1+i)\right|^2 \\\\ & =(\sqrt{2})^2 \\\\ & =2 \end{aligned}
Q103
If S={z∈C:∣z−i∣=∣z+i∣=∣z−1∣}S=\{z \in C:|z-i|=|z+i|=|z-1|\}, then, n(S)n(S) is :
A 1
B 2
C 3
D 0
Correct Answer
Option A
Solution
∣z−i∣=∣z+i∣=∣z−1∣|z-i|=|z+i|=|z-1|
ABC\mathrm{ABC}

is a triangle. Hence its circum-centre will be the only point whose distance from A, B, C will be same. So

n(S)=1n(S)=1
Q104
Let z1z_1 and z2z_2 be two complex numbers such that z1+z2=5z_1+z_2=5 and z13+z23=20+15iz_1^3+z_2^3=20+15 i Then, ∣z14+z24∣\left|z_1^4+z_2^4\right| equals -
A 151515 \sqrt{15}
B 30330 \sqrt{3}
C 25325 \sqrt{3}
D 75
Correct Answer
Option D
Solution
z1+z2=5z13+z23=20+15iz13+z23=(z1+z2)3−3z1z2(z1+z2)z13+z23=125−3z1⋅z2(5)⇒20+15i=125−15z1z2⇒3z1z2=25−4−3i⇒3z1z2=21−3i⇒z1⋅z2=7−i⇒(z1+z2)2=25⇒z12+z22=25−2(7−i)⇒11+2i(z12+z22)2=121−4+44i⇒z14+z24+2(7−i)2=117+44i⇒z14+z24=117+44i−2(49−1−14i)⇒∣z14+z24∣=75\begin{aligned} & z_1+z_2=5 \\ & z_1^3+z_2^3=20+15 i \\ & z_1^3+z_2^3=\left(z_1+z_2\right)^3-3 z_1 z_2\left(z_1+z_2\right) \\ & z_1^3+z_2^3=125-3 z_1 \cdot z_2(5) \\ & \Rightarrow 20+15 i=125-15 z_1 z_2 \\ & \Rightarrow 3 z_1 z_2=25-4-3 i \\ & \Rightarrow 3 z_1 z_2=21-3 i \\ & \Rightarrow z_1 \cdot z_2=7-i \\ & \Rightarrow\left(z_1+z_2\right)^2=25 \\ & \Rightarrow z_1^2+z_2^2=25-2(7-i) \\ & \Rightarrow 11+2 i \\ & \left(z_1^2+z_2^2\right)^2=121-4+44 i \\ & \Rightarrow z_1^4+z_2^4+2(7-i)^2=117+44 i \\ & \Rightarrow z_1^4+z_2^4=117+44 i-2(49-1-14 i) \\ & \Rightarrow\left|z_1^4+z_2^4\right|=75 \end{aligned}
Q105
If z=12−2iz=\dfrac{1}{2}-2 i is such that ∣z+1∣=αz+β(1+i),i=−1|z+1|=\alpha z+\beta(1+i), i=\sqrt{-1} and α,β∈R\alpha, \beta \in \mathbb{R}, then α+β\alpha+\beta is equal to
A 2
B −-4
C 3
D −-1
Correct Answer
Option C
Solution

To begin with, let's analyze the given equation:

∣z+1∣=αz+β(1+i)|z+1|=\alpha z+\beta(1+i)

First, we compute the modulus of the left side: Let's take the given value of z,

z=12−2iz = \frac{1}{2} - 2i

Now, we find

z+1z + 1

:

z+1=(12−2i)+1=32−2iz + 1 = \left(\frac{1}{2} - 2i\right) + 1 = \frac{3}{2} - 2i

Then, the modulus of

z+1z + 1

is calculated as follows:

∣z+1∣=∣32−2i∣=(32)2+(−2)2|z + 1| = \left|\frac{3}{2} - 2i\right| = \sqrt{\left(\frac{3}{2}\right)^2 + (-2)^2}
∣z+1∣=94+4|z + 1| = \sqrt{\frac{9}{4} + 4}
∣z+1∣=94+164|z + 1| = \sqrt{\frac{9}{4} + \frac{16}{4}}
∣z+1∣=254|z + 1| = \sqrt{\frac{25}{4}}
∣z+1∣=52|z + 1| = \frac{5}{2}

Now we need to equate the modulus to the right-hand side of the equation and solve for α\alpha and β\beta.

Let's rewrite the equation:

52=αz+β(1+i)\frac{5}{2} = \alpha z + \beta(1 + i)

Substitute

zz

with its value:

52=α(12−2i)+β(1+i)\frac{5}{2} = \alpha \left(\frac{1}{2} - 2i\right) + \beta(1+i)

Rewrite the equation separating real and imaginary parts:

52=α(12)−2αi+β+βi\frac{5}{2} = \alpha \left(\frac{1}{2}\right) - 2\alpha i + \beta + \beta i
52=(α12+β)+(−2α+β)i\frac{5}{2} = \left(\alpha \frac{1}{2} + \beta\right) + \left(-2\alpha + \beta\right)i

For the above equality to hold, both real and imaginary parts must be equal. Equating real parts:

α12+β=52\alpha \frac{1}{2} + \beta = \frac{5}{2}

Equating imaginary parts:

−2α+β=0-2\alpha + \beta = 0

We now have a system of two linear equations: 1)

α2+β=52\frac{\alpha}{2} + \beta = \frac{5}{2}

2)

−2α+β=0-2\alpha + \beta = 0

Let's solve the system by isolating β\beta from the second equation and then substituting it into the first one:

β=2α\beta = 2\alpha

Now substitute β\beta in the first equation:

α2+2α=52\frac{\alpha}{2} + 2\alpha = \frac{5}{2}
α(12+2)=52\alpha \left(\frac{1}{2} + 2\right) = \frac{5}{2}
α(12+42)=52\alpha \left(\frac{1}{2} + \frac{4}{2}\right) = \frac{5}{2}
α(52)=52\alpha \left(\frac{5}{2}\right) = \frac{5}{2}

To find the value of α\alpha, we divide both sides by

52\frac{5}{2}

:

α=1\alpha = 1

Now, we use the value of α\alpha to find β\beta:

β=2α\beta = 2\alpha
β=2⋅1\beta = 2 \cdot 1
β=2\beta = 2

Finally, we add both α\alpha and β\beta to find

α+β\alpha + \beta

:

α+β=1+2=3\alpha + \beta = 1 + 2 = 3

The value of

α+β\alpha + \beta

is 3. So, the correct answer is Option C) 3.

Q106
Let zz be a complex number such that the real part of z−2iz+2i\dfrac{z-2 i}{z+2 i} is zero. Then, the maximum value of ∣z−(6+8i)∣|z-(6+8 i)| is equal to
A 8
B 12
C 10
D ∞\infty
Correct Answer
Option B
Solution
n=z−2iz+2i Let z=x+iyn=x+(y−2)ix+(y+2)i×(x−(y+2)ix−(y+2)i)Re⁡(n)=x2+(y−2)(y+2)x2+(y+2)2=0⇒x2+(y−2)(y+2)=0\begin{aligned} & n=\frac{z-2 i}{z+2 i} \\ & \text{ Let } z=x+i y \\ & n=\frac{x+(y-2) i}{x+(y+2) i} \times\left(\frac{x-(y+2) i}{x-(y+2) i}\right) \\ & \operatorname{Re}(n)=\frac{x^2+(y-2)(y+2)}{x^2+(y+2)^2}=0 \\ & \Rightarrow x^2+(y-2)(y+2)=0 \end{aligned}
⇒x2+y2−4=0⇒x2+y2=4 also, ∣z−(6+8i)∣≤∣z∣+∣−6−8i∣∣z−(6+8i)∣≤2+10=12\begin{aligned} & \Rightarrow x^2+y^2-4=0 \\ & \Rightarrow x^2+y^2=4 \\ & \text{ also, }|z-(6+8 i)| \leq|z|+|-6-8 i| \\ & |z-(6+8 i)| \leq 2+10=12 \end{aligned}

Hence, Maximum value of

∣z−(6+8i)∣|z-(6+8 i)|

is 12.

Q107
Let α\alpha and β\beta be the sum and the product of all the non-zero solutions of the equation (zˉ)2+∣z∣=0,z∈C(\bar{z})^2+|z|=0, z \in C. Then 4(α2+β2)4(\alpha^2+\beta^2) is equal to :
A 4
B 2
C 6
D 8
Correct Answer
Option A
Solution
(zˉ)2+∣z∣=0... (1)z2+∣zˉ∣=0... (2)\begin{aligned} & (\bar{z})^2+|z|=0 \quad \text{... (1)}\\ & z^2+|\bar{z}|=0 \quad \text{... (2)} \end{aligned}

From equation (1) and (2)

 as ∣z∣=∣zˉ∣⇒(zˉ)2=z2⇒z=zˉ or z=−zˉ⇒Im⁡(z)=0 or Re⁡(z)=0\begin{aligned} & \text{ as }|z|=|\bar{z}| \\ & \Rightarrow \quad(\bar{z})^2=z^2 \\ & \Rightarrow \quad z=\bar{z} \text{ or } z=-\bar{z} \\ & \Rightarrow \operatorname{Im}(z)=0 \text{ or } \operatorname{Re}(z)=0 \end{aligned}

Case I : If

Im⁡(z)=0\operatorname{Im}(z)=0
⇒z=x\Rightarrow z=x

Putting value of

zz

in equation (1)

x2+∣x∣=0⇒x=0[Rejected] \begin{aligned} & x^2+|x|=0 \\ & \Rightarrow x=0 \quad \text{[Rejected] } \end{aligned}

Case II : If

Re⁡(z)=0\operatorname{Re}(z)=0
⇒z=iy\Rightarrow z=i y

Putting value of

zz

in equation (1)

−y2+∣y∣=0y=±1 as y≠0\begin{aligned} & -y^2+|y|=0 \\ & y= \pm 1 \text{ as } y \neq 0 \end{aligned}

Hence,

z=±iz= \pm i

are the solution of the given equation

⇒α=i−i=0 and β=i(−i)=1⇒4(α2+β2)=4(0+1)=4\begin{aligned} & \Rightarrow \alpha=i-i=0 \\ & \text{ and } \beta=i(-i)=1 \\ & \Rightarrow \quad 4\left(\alpha^2+\beta^2\right)=4(0+1) \\ & \quad=4 \end{aligned}

∴\therefore Option (3) is correct

Q108
The area (in sq. units) of the region S={z∈C:∣z−1∣≤2;(z+zˉ)+i(z−zˉ)≤2,lm⁡(z)≥0}S=\{z \in \mathbb{C}:|z-1| \leq 2 ;(z+\bar{z})+i(z-\bar{z}) \leq 2, \operatorname{lm}(z) \geq 0\} is
A 7π4\dfrac{7 \pi}{4}
B 3π2\dfrac{3 \pi}{2}
C 7π3\dfrac{7 \pi}{3}
D 17π8\dfrac{17 \pi}{8}
Correct Answer
Option B
Solution
∣z−1∣≤2⇒(x−1)2+y2=4|z-1| \leq 2 \quad \Rightarrow \quad(x-1)^2+y^2=4
z+zˉ+i(z−zˉ)≤2⇒x−y≤1Im⁡(z)≥0⇒y≥0\begin{aligned} & z+\bar{z}+i(z-\bar{z}) \leq 2 \\ \Rightarrow \quad & x-y \leq 1 \\ & \operatorname{Im}(z) \geq 0 \\ \Rightarrow \quad & y \geq 0 \end{aligned}
 Required area =(3π42π)(π)(2)2=32π\begin{aligned} & \text{ Required area }=\left(\frac{\frac{3 \pi}{4}}{2 \pi}\right)(\pi)(2)^2 \\ & =\frac{3}{2} \pi \end{aligned}
Q109
The sum of all possible values of θ∈[−π,2π]\theta \in[-\pi, 2 \pi], for which 1+icos⁡θ1−2icos⁡θ\dfrac{1+i \cos \theta}{1-2 i \cos \theta} is purely imaginary, is equal to :
A 4π4 \pi
B 3π3 \pi
C 2π2 \pi
D 5π5 \pi
Correct Answer
Option B
Solution
1+icos⁡θ1−2icos⁡θ\frac{1+i \cos \theta}{1-2 i \cos \theta}

is purely imaginary

n=1+icos⁡θ1−2icos⁡θ×1+2icos⁡θ1+2icos⁡θ=1+3icos⁡θ−2cos⁡2θ1+4cos⁡2θn=\frac{1+i \cos \theta}{1-2 i \cos \theta} \times \frac{1+2 i \cos \theta}{1+2 i \cos \theta}=\frac{1+3 i \cos \theta-2 \cos ^2 \theta}{1+4 \cos ^2 \theta}
n=1−2cos⁡2θ1+4cos⁡2θ+i(3cos⁡θ1+4cos⁡2θ)n=\frac{1-2 \cos ^2 \theta}{1+4 \cos ^2 \theta}+i\left(\frac{3 \cos \theta}{1+4 \cos ^2 \theta}\right)
nn

is purely imaginary

⇒1−2cos⁡2θ1+4cos⁡2θ=0\Rightarrow \frac{1-2 \cos ^2 \theta}{1+4 \cos ^2 \theta}=0
⇒cos⁡2θ=12\Rightarrow \cos ^2 \theta=\frac{1}{2}
⇒cos⁡θ=±12\Rightarrow \cos \theta= \pm \frac{1}{\sqrt{2}}

θ\theta can be

π4,−π4,3π4,−3π4,5π4,7π4\frac{\pi}{4}, \frac{-\pi}{4}, \frac{3 \pi}{4}, \frac{-3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}

Sum of all possible values of

θ=3π\theta=3 \pi
Q110
Consider the following two statements : Statement I: For any two non-zero complex numbers z1,z2,(∣z1∣+∣z2∣)∣z1∣z1∣+z2∣z2∣∣≤2(∣z1∣+∣z2∣), and z_1, z_2,(|z_1|+|z_2|)\left|\dfrac{z_1}{\left|z_1\right|}+\dfrac{z_2}{\left|z_2\right|}\right| \leq 2\left(\left|z_1\right|+\left|z_2\right|\right) \text{, and } Statement II : If x,y,zx, y, z are three distinct complex numbers and a,b,c\mathrm{a}, \mathrm{b}, \mathrm{c} are three positive real numbers such that a∣y−z∣=b∣z−x∣=c∣x−y∣\dfrac{\mathrm{a}}{|y-z|}=\dfrac{\mathrm{b}}{|z-x|}=\dfrac{\mathrm{c}}{|x-y|}, then a2y−z+b2z−x+c2x−y=1\dfrac{\mathrm{a}^2}{y-z}+\dfrac{\mathrm{b}^2}{z-x}+\dfrac{\mathrm{c}^2}{x-y}=1. Between the above two statements,
A both Statement I and Statement II are incorrect.
B Statement I is correct but Statement II is incorrect.
C Statement I is incorrect but Statement II is correct.
D both Statement I and Statement II are correct.
Correct Answer
Option B
Solution
a∣y−z∣=b∣z−x∣=c∣x−y∣=λ⇒a2=λ2∣(y−z)∣2b2=λ2∣(z−x)∣2c2=λ2∣(x−y)∣2a2(y−z‾)(y−z)(y−z)=a2(yˉ−zˉ)∣y−z∣2=a2(yˉ−zˉ)a2λ2=λ2(yˉ−zˉ)⇒∑(a2y−z)=λ2(yˉ−zˉ+zˉ−xˉ+xˉ−yˉ)=0≠1\begin{aligned} & \frac{a}{|y-z|}=\frac{b}{|z-x|}=\frac{c}{|x-y|}=\lambda \\ & \Rightarrow a^2=\lambda^2|(y-z)|^2 \\ & b^2=\lambda^2|(z-x)|^2 \\ & c^2=\lambda^2|(x-y)|^2 \\ & \frac{a^2(\overline{y-z})}{(y-z)(y-z)}=\frac{a^2(\bar{y}-\bar{z})}{|y-z|^2}=\frac{a^2(\bar{y}-\bar{z})}{\frac{a^2}{\lambda^2}}=\lambda^2(\bar{y}-\bar{z}) \\ & \Rightarrow \sum\left(\frac{a^2}{y-z}\right)=\lambda^2(\bar{y}-\bar{z}+\bar{z}-\bar{x}+\bar{x}-\bar{y})=0 \neq 1 \\ \end{aligned}

Statement I

(∣z1∣+∣z2∣)∣z1∣z1∣+z2∣z2∣∣≤2(∣z1∣+∣z2∣)⇒z1=∣z1∣eiθ1z2=∣z2∣eiθ2⇒z1∣z1∣=eiθ1⇒z2∣z2∣=eiθ2⇒∣eiθ1+eiθ2∣=∣2+2cos⁡(θ1−θ2)∣=∣2cos⁡(θ1−θ22)∣≤2\begin{aligned} & \left(\left|z_1\right|+\left|z_2\right|\right)\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq 2\left(\left|z_1\right|+\left|z_2\right|\right) \\ & \Rightarrow \quad z_1=\left|z_1\right| e^{i \theta_1} \\ & \quad z_2=\left|z_2\right| e^{i \theta_2} \\ & \Rightarrow \frac{z_1}{\left|z_1\right|}=e^{i \theta_1} \\ & \Rightarrow \frac{z_2}{\left|z_2\right|}=e^{i \theta_2} \\ & \Rightarrow\left|e^{i \theta_1}+e^{i \theta_2}\right| \\ & \quad=\left|\sqrt{2+2 \cos \left(\theta_1-\theta_2\right)}\right| \\ & \quad=\left|2 \cos \left(\frac{\theta_1-\theta_2}{2}\right)\right| \leq 2 \end{aligned}
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