Practice Start Test

Complex Numbers

JEE Mathematics · 150 questions · Page 12 of 15 · Click an option or "Show Solution" to reveal answer

Q111
Let z∈Cz \in C be such that z2+3iz−2+i=2+3i\dfrac{z^2+3 i}{z-2+i}=2+3 i. Then the sum of all possible values of z2z^2 is :
A −19+2i -19+2 i
B −19−2i-19-2 i
C 19−2i19-2 i
D 19+2i19+2 i
Correct Answer
Option B
Solution
z2+3iz−2+i=2+3iz2+3i=(z−2+i)(2+3i)z2+3i=2z−4+2i+3iz−6i−3z2+3i=(2z−7)+i(3z−4)z2−(2+3i)z+(7+7i)=0 This is a quadratic in z. z1+z2=2+3iz1+z2=7+7iz12+z22=(z1+z2)2−2z1z2=(2+3i)2−2(7+7i)=4−9+12i−14−14i=−19−2i\begin{aligned} &\begin{aligned} & \frac{z^2+3 i}{z-2+i}=2+3 i \\ & z^2+3 i=(z-2+i)(2+3 i) \\ & z^2+3 i=2 z-4+2 i+3 i z-6 i-3 \\ & z^2+3 i=(2 z-7)+i(3 z-4) \\ & z^2-(2+3 i) z+(7+7 i)=0 \end{aligned}\\ &\text{ This is a quadratic in } z \text{. }\\ &\begin{aligned} & z_1+z_2=2+3 i \\ & z_1+z_2=7+7 i \\ & z_1^2+z_2^2=\left(z_1+z_2\right)^2-2 z_1 z_2 \\ & =(2+3 i)^2-2(7+7 i) \\ & =4-9+12 i-14-14 i \\ & =-19-2 i \end{aligned} \end{aligned}
Q112
Let S1={z∈C:∣z∣≤5},S2={z∈C:Im⁡(z+1−3i1−3i)≥0}S_1=\{z \in \mathbf{C}:|z| \leq 5\}, S_2=\left\{z \in \mathbf{C}: \operatorname{Im}\left(\frac{z+1-\sqrt{3} i}{1-\sqrt{3} i}\right) \geq 0\right\} and S3={z∈C:Re⁡(z)≥0}S_3=\{z \in \mathbf{C}: \operatorname{Re}(z) \geq 0\}. Then the area of the region S1∩S2∩S3S_1 \cap S_2 \cap S_3 is :
A 125π24\dfrac{125 \pi}{24}
B 125π6\dfrac{125 \pi}{6}
C 125π12\dfrac{125 \pi}{12}
D 125π4\dfrac{125 \pi}{4}
Correct Answer
Option C
Solution
S1={z∈C:∣z∣≤5}S_1=\{z \in C:|z| \leq 5\}
S2=Im⁡(z+1−3i1−3i)≥0S_2=\operatorname{Im}\left(\frac{z+1-\sqrt{3} i}{1-\sqrt{3} i}\right) \geq 0

Take

z=x+iyz=x+i y
=x+iy+1−3i1−3i×1+3i1+3i=\frac{x+i y+1-\sqrt{3} i}{1-\sqrt{3} i} \times \frac{1+\sqrt{3} i}{1+\sqrt{3} i}
=x+iy+1−3i+3ix−3y+3i+31+3=y+3x≥0S3=x≥0\begin{aligned} = & \frac{x+i y+1-\sqrt{3} i+\sqrt{3} i x-\sqrt{3} y+\sqrt{3} i+3}{1+3} \\ & =y+\sqrt{3} x \geq 0 \\ S_3= & x \geq 0 \end{aligned}
y+3x=0y=−3x Slope =−390+θ=120∘θ=30∘90−θ=60∘\begin{aligned} & y+\sqrt{3} x=0 \\ & y=-\sqrt{3} x \\ & \text{ Slope }=-\sqrt{3} \\ & 90+\theta=120^{\circ} \\ & \theta=30^{\circ} \\ & 90-\theta=60^{\circ} \end{aligned}

In first quadrant we have angle

π2\frac{\pi}{2}

so, total angle

90∘+60∘=150∘90^{\circ}+60^{\circ}=150^{\circ}

So, area

πr22π×5π6=125π12\frac{\pi r^2}{2 \pi} \times \frac{5 \pi}{6}=\frac{125 \pi}{12}
Q113
If z1,z2z_1, z_2 are two distinct complex number such that ∣z1−2z212−z1zˉ2∣=2\left|\dfrac{z_1-2 z_2}{\dfrac{1}{2}-z_1 \bar{z}_2}\right|=2, then
A either z1z_1 lies on a circle of radius 12\dfrac{1}{2} or z2z_2 lies on a circle of radius 1.
B z1z_1 lies on a circle of radius 12\dfrac{1}{2} and z2z_2 lies on a circle of radius 1.
C either z1z_1 lies on a circle of radius 1 or z2z_2 lies on a circle of radius 12\dfrac{1}{2}.
D both z1z_1 and z2z_2 lie on the same circle.
Correct Answer
Option C
Solution
∣z1−2z212−z1zˉ2∣=2∣z1−2z2∣=∣1−2z1zˉ2∣⇒(z1−2z2)(zˉ1−2zˉ2)=(1−2z1zˉ2)(1−2zˉ1z2)⇒∣z1∣2+4∣z2∣2−2zˉ1z2−2zˉ2z1=1+4∣z1∣2∣z2∣2−2z1zˉ2−2zˉ1z2⇒∣z1∣2+4∣z2∣2−4∣z1∣2∣z2∣2−1=0⇒(∣z1∣2−1)(1−4∣z2∣2)=0⇒∣z1∣=1 and ∣z2∣=12\begin{aligned} & \left|\frac{z_1-2 z_2}{\frac{1}{2}-z_1 \bar{z}_2}\right|=2 \\ & \left|z_1-2 z_2\right|=\left|1-2 z_1 \bar{z}_2\right| \\ & \Rightarrow\left(z_1-2 z_2\right)\left(\bar{z}_1-2 \bar{z}_2\right)=\left(1-2 z_1 \bar{z}_2\right)\left(1-2 \bar{z}_1 z_2\right) \\ & \Rightarrow\left|z_1\right|^2+4\left|z_2\right|^2-2 \bar{z}_1 z_2-2 \bar{z}_2 z_1 \\ & \quad=1+4\left|z_1\right|^2\left|z_2\right|^2-2 z_1 \bar{z}_2-2 \bar{z}_1 z_2 \\ & \Rightarrow\left|z_1\right|^2+4\left|z_2\right|^2-4\left|z_1\right|^2\left|z_2\right|^2-1=0 \\ & \Rightarrow\left(\left|z_1\right|^2-1\right)\left(1-4\left|z_2\right|^2\right)=0 \\ & \Rightarrow\left|z_1\right|=1 \text{ and }\left|z_2\right|=\frac{1}{2} \end{aligned}
Q114
Let ∣z1−8−2i∣≤1 |z_1 − 8−2i| \leq 1 and ∣z2−2+6i∣≤2 |z_2−2+6i| \leq 2 , z1,z2∈C z_1, z_2 \in \mathbb{C} . Then the minimum value of ∣z1−z2∣ |z_1 − z_2| is :
A 3
B 10
C 7
D 13
Correct Answer
Option C
Solution
∵AB=100=10∴∣Z1−Z2∣min⁡=10−2−1=7\begin{aligned} & \because \mathrm{AB}=\sqrt{100}=10 \\ & \therefore\left|\mathrm{Z}_1-\mathrm{Z}_2\right|_{\min }=10-2-1=7 \end{aligned}
Q115
Let z1,z2z_1, z_2 and z3z_3 be three complex numbers on the circle ∣z∣=1|z|=1 with arg⁡(z1)=−π4,arg⁡(z2)=0\arg \left(z_1\right)=\dfrac{-\pi}{4}, \arg \left(z_2\right)=0 and arg⁡(z3)=π4\arg \left(z_3\right)=\dfrac{\pi}{4}. If ∣z1zˉ2+z2zˉ3+z3zˉ1∣2=α+β2,α,β∈Z\left|z_1 \bar{z}_2+z_2 \bar{z}_3+z_3 \bar{z}_1\right|^2=\alpha+\beta \sqrt{2}, \alpha, \beta \in Z, then the value of α2+β2\alpha^2+\beta^2 is :
A 41
B 29
C 24
D 31
Correct Answer
Option B
Solution

To solve the problem, we start with the given information about the complex numbers z1,z2, z_1, z_2, and z3 z_3 , which lie on the unit circle ∣z∣=1 |z| = 1 .

Their arguments are as follows: arg⁡(z1)=−π4 \arg(z_1) = -\dfrac{\pi}{4} arg⁡(z2)=0 \arg(z_2) = 0 arg⁡(z3)=π4 \arg(z_3) = \dfrac{\pi}{4} Thus, the complex numbers can be represented as: z1=e−iπ4=12−i2 z_1 = e^{-i\dfrac{\pi}{4}} = \dfrac{1}{\sqrt{2}} - \dfrac{i}{\sqrt{2}} z2=ei⋅0=1 z_2 = e^{i\cdot 0} = 1 z3=eiπ4=12+i2 z_3 = e^{i\dfrac{\pi}{4}} = \dfrac{1}{\sqrt{2}} + \dfrac{i}{\sqrt{2}} Next, calculate the conjugates needed: zˉ2=1 \bar{z}_2 = 1 zˉ3=12−i2 \bar{z}_3 = \dfrac{1}{\sqrt{2}} - \dfrac{i}{\sqrt{2}} zˉ1=12+i2 \bar{z}_1 = \dfrac{1}{\sqrt{2}} + \dfrac{i}{\sqrt{2}} We need to evaluate: z1zˉ2+z2zˉ3+z3zˉ1 z_1 \bar{z}_2 + z_2 \bar{z}_3 + z_3 \bar{z}_1 Calculate each term separately: z1zˉ2=(12−i2)⋅1=12−i2 z_1 \bar{z}_2 = \left(\dfrac{1}{\sqrt{2}} - \dfrac{i}{\sqrt{2}}\right) \cdot 1 = \dfrac{1}{\sqrt{2}} - \dfrac{i}{\sqrt{2}} z2zˉ3=1⋅(12−i2)=12−i2 z_2 \bar{z}_3 = 1 \cdot \left(\dfrac{1}{\sqrt{2}} - \dfrac{i}{\sqrt{2}}\right) = \dfrac{1}{\sqrt{2}} - \dfrac{i}{\sqrt{2}} z3zˉ1=(12+i2)⋅(12+i2)=(1+i)22=1+2i−12=i z_3 \bar{z}_1 = \left(\dfrac{1}{\sqrt{2}} + \dfrac{i}{\sqrt{2}}\right) \cdot \left(\dfrac{1}{\sqrt{2}} + \dfrac{i}{\sqrt{2}}\right) = \dfrac{(1 + i)^2}{2} = \dfrac{1 + 2i - 1}{2} = i Sum the evaluated terms: z1zˉ2+z2zˉ3+z3zˉ1=(12−i2)+(12−i2)+i z_1 \bar{z}_2 + z_2 \bar{z}_3 + z_3 \bar{z}_1 = \left(\dfrac{1}{\sqrt{2}} - \dfrac{i}{\sqrt{2}}\right) + \left(\dfrac{1}{\sqrt{2}} - \dfrac{i}{\sqrt{2}}\right) + i Simplify: =2+i−2i=2+i(1−2) = \sqrt{2} + i - \sqrt{2}i = \sqrt{2} + i(1-\sqrt{2}) Calculate the modulus squared: ∣2+i(1−2)∣2=(2)2+(1−2)2 \left| \sqrt{2} + i (1 - \sqrt{2}) \right|^2 = (\sqrt{2})^2 + (1-\sqrt{2})^2 =2+(1−22+2)=5−22 = 2 + (1 - 2\sqrt{2} + 2) = 5 - 2\sqrt{2} Thus, the expression ∣z1zˉ2+z2zˉ3+z3zˉ1∣2 |z_1 \bar{z}_2 + z_2 \bar{z}_3 + z_3 \bar{z}_1|^2 simplifies as follows: α=5 \alpha = 5 β=−2 \beta = -2 Finally, compute α2+β2 \alpha^2 + \beta^2 : α2+β2=52+(−2)2=25+4=29 \alpha^2 + \beta^2 = 5^2 + (-2)^2 = 25 + 4 = 29 Therefore, the value of α2+β2 \alpha^2 + \beta^2 is 29.

Q116
Let the curve z(1+i)+zˉ(1−i)=4,z∈Cz(1+i)+\bar{z}(1-i)=4, z \in C, divide the region ∣z−3∣≤1|z-3| \leq 1 into two parts of areas α\alpha and β\beta. Then ∣α−β∣|\alpha-\beta| equals :
A 1+π31+\dfrac{\pi}{3}
B 1+π61+\dfrac{\pi}{6}
C 1+π21+\dfrac{\pi}{2}
D 1+π41+\dfrac{\pi}{4}
Correct Answer
Option C
Solution
 Let z=x+iy(x+iy)(1+i)+(x−iy)(1−i)=4x+ix+iy−y+x−ix−iy−y=42x−2y=4x−y=2∣z−3∣≤1(x−3)2+y2≤1\begin{aligned} & \text{ Let } z=x+i y \\ & (x+i y)(1+i)+(x-i y)(1-i)=4 \\ & x+i x+i y-y+x-i x-i y-y=4 \\ & 2 x-2 y=4 \\ & x-y=2 \\ & |z-3| \leq 1 \\ & (x-3)^2+y^2 \leq 1 \end{aligned}
 Area of shaded region =π⋅124−12⋅1⋅1=π4−12 Area of unshaded region inside the circle =34π⋅12+12⋅1⋅1=3π4+12∴ difference of area =(3π4+12)−(π4−12)=π2+1\begin{aligned} &\text{ Area of shaded region }=\frac{\pi \cdot 1^2}{4}-\frac{1}{2} \cdot 1 \cdot 1=\frac{\pi}{4}-\frac{1}{2}\\ &\text{ Area of unshaded region inside the circle }\\ &\begin{aligned} & =\frac{3}{4} \pi \cdot 1^2+\frac{1}{2} \cdot 1 \cdot 1=\frac{3 \pi}{4}+\frac{1}{2} \\ & \therefore \text{ difference of area }=\left(\frac{3 \pi}{4}+\frac{1}{2}\right)-\left(\frac{\pi}{4}-\frac{1}{2}\right) \end{aligned}\\ &=\frac{\pi}{2}+1 \end{aligned}
Q117
If α+iβ\alpha + i\beta and γ+iδ\gamma + i\delta are the roots of x2−(3−2i)x−(2i−2)=0x^2 - (3 - 2i)x - (2i - 2) = 0, i=−1i = \sqrt{-1}, then αγ+βδ\alpha \gamma + \beta \delta is equal to:
A 2
B -6
C 6
D -2
Correct Answer
Option A
Solution
x2−(3−2i)x−(2i−2)=0x=(3−2i)±(3−2i)2−4(1)(−(2i−2))2(1)==(3−2i)±9−4−12i+8i−82==3−2i±−3−4i2=3−2i±(1)2+(2i)2−2(1)(2i)2=3−2i±(1−2i)2⇒3−2i+1−2i2 or 3−2i−1+2i2⇒2−2i or 1+0i So αγ+βδ=2(1)+(−2)(0)=2\begin{aligned} & x^2-(3-2 i) x-(2 i-2)=0 \\ & x=\frac{(3-2 i) \pm \sqrt{(3-2 i)^2-4(1)(-(2 i-2))}}{2(1)} \\ & ==\frac{(3-2 i) \pm \sqrt{9-4-12 i+8 i-8}}{2} \\ & ==\frac{3-2 i \pm \sqrt{-3-4 i}}{2} \\ & =\frac{3-2 i \pm \sqrt{(1)^2+(2 i)^2-2(1)(2 i)}}{2} \\ & =\frac{3-2 \mathrm{i} \pm(1-2 \mathrm{i})}{2} \\ & \Rightarrow \frac{3-2 \mathrm{i}+1-2 \mathrm{i}}{2} \text{ or } \frac{3-2 \mathrm{i}-1+2 \mathrm{i}}{2} \\ & \Rightarrow 2-2 \mathrm{i} \text{ or } 1+0 \mathrm{i} \\ & \text{ So } \alpha \gamma+\beta \delta=2(1)+(-2)(0)=2 \end{aligned}
Q118
Let ∣zˉ−i2zˉ+i∣=13,z∈C\left|\dfrac{\bar{z}-i}{2 \bar{z}+i}\right|=\dfrac{1}{3}, z \in C, be the equation of a circle with center at CC. If the area of the triangle, whose vertices are at the points (0,0),C(0,0), C and (α,0)(\alpha, 0) is 11 square units, then α2\alpha^2 equals:
A 12125\dfrac{121}{25}
B 100
C 8125\dfrac{81}{25}
D 50
Correct Answer
Option B
Solution
∣zˉ−i2zˉ+i∣=13∣zˉ−izˉ+i2∣=233∣x−iy−i∣=2∣x−iy+i2∣9(x2+(y+1)2)=4(x2+(y−1/3)2)9x2+9y2+18y+9=4x2+4y2−4y+15x2+5y2+22y+8=0x2+y2+225y+85=0 centre ⇒(0,−115)\begin{aligned} & \left|\frac{\bar{z}-i}{2 \bar{z}+i}\right|=\frac{1}{3} \\ & \left|\frac{\bar{z}-i}{\bar{z}+\frac{i}{2}}\right|=\frac{2}{3} \\ & 3|x-i y-i|=2\left|x-i y+\frac{i}{2}\right| \\ & 9\left(x^2+(y+1)^2\right)=4\left(x^2+(y-1 / 3)^2\right) \\ & 9 x^2+9 y^2+18 y+9=4 x^2+4 y^2-4 y+1 \\ & 5 x^2+5 y^2+22 y+8=0 \\ & x^2+y^2+\frac{22}{5} y+\frac{8}{5}=0 \\ & \text{ centre } \Rightarrow\left(0,-\frac{11}{5}\right) \end{aligned}
∣12∣0010−11/51α01∣∣=11\left| {{1 \over 2}\left| \begin{array}{lll}0 & 0 & 1 \\ 0 & { - 11/5} & 1 \\ \alpha & 0 & 1 \end{array} \right|} \right| = 11
⇒(−115α)2=(11×2)2⇒α2=100\begin{aligned} & \Rightarrow\left(-\frac{11}{5} \alpha\right)^2=(11 \times 2)^2 \\ & \Rightarrow \alpha^2=100 \end{aligned}
Q119
The number of complex numbers zz, satisfying ∣z∣=1|z|=1 and ∣zzˉ+zˉz∣=1\left|\dfrac{z}{\bar{z}}+\dfrac{\bar{z}}{z}\right|=1, is :
A 8
B 10
C 4
D 6
Correct Answer
Option A
Solution

∣z∣=1∣zzˉ+zˉz∣=1⇒∣z2+(zˉ)2∣=1 Let z=x+iy⇒∣(x+iy)2+(x−iy)2∣=1⇒∣2x2−2y2∣=1⇒∣x2−y2∣=12⇒x2−y2=±12 and x2+y2=1\begin{aligned} & |z|=1 \\ & \left|\dfrac{z}{\bar{z}}+\dfrac{\bar{z}}{z}\right|=1 \\ & \Rightarrow\left|z^2+(\bar{z})^2\right|=1 \\ & \text{ Let } z=x+i y \\ & \Rightarrow\left|(x+i y)^2+(x-i y)^2\right|=1 \\ & \Rightarrow\left|2 x^2-2 y^2\right|=1 \\ & \Rightarrow\left|x^2-y^2\right|=\dfrac{1}{2} \\ & \Rightarrow x^2-y^2=\dfrac{ \pm 1}{2} \\ & \text{ and } x^2+y^2=1\end{aligned} Case I: x2−y2=12x^2-y^2=\dfrac{1}{2} Case II: x2−y2=−12x^2-y^2=-\dfrac{1}{2} Hence, we get 8 complex numbers.

Q120
If α\alpha and β\beta are the roots of the equation 2z2−3z−2i=02 z^2-3 z-2 i=0, where i=−1i=\sqrt{-1}, then 16⋅Re⁡(α19+β19+α11+β11α15+β15)⋅lm⁡(α19+β19+α11+β11α15+β15)16 \cdot \operatorname{Re}\left(\dfrac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right) \cdot \operatorname{lm}\left(\dfrac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right) is equal to
A 441
B 312
C 409
D 398
Correct Answer
Option A
Solution
 Sol. 2z2−32−2i=02(z−iz)=3α−iα=32⇒α2−1α2−2i=94⇒α2−1α2−2i=94⇒94+2i=α2−1α2⇒8116−4+9i=α4+1α4−2⇒4916+9i=α4+1α4 Similarly ⇒4916+9i=β4+1β4\begin{aligned} & \text{ Sol. } 2 z^2-32-2 i=0 \\ & 2\left(z-\frac{i}{z}\right)=3 \\ & \alpha-\frac{i}{\alpha}=\frac{3}{2} \\ & \Rightarrow \alpha^2-\frac{1}{\alpha^2}-2 i=\frac{9}{4} \\ & \Rightarrow \alpha^2-\frac{1}{\alpha^2}-2 i=\frac{9}{4} \\ & \Rightarrow \frac{9}{4}+2 i=\alpha^2-\frac{1}{\alpha^2} \\ & \Rightarrow \frac{81}{16}-4+9 i=\alpha^4+\frac{1}{\alpha^4}-2 \\ & \Rightarrow \frac{49}{16}+9 i=\alpha^4+\frac{1}{\alpha^4} \\ & \text{ Similarly } \\ & \Rightarrow \frac{49}{16}+9 i=\beta^4+\frac{1}{\beta^4} \end{aligned}
⇒α19+β19+α11+β11α15+β15=α15(α4+1α4)+β15(β4+1β4)α15+β15=(α15+β15)(4916+9i)(α15+β15) Real =4916 Im =9 Ans. 441\begin{aligned} & \Rightarrow \frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}=\frac{\alpha^{15}\left(\alpha^4+\frac{1}{\alpha^4}\right)+\beta^{15}\left(\beta^4+\frac{1}{\beta^4}\right)}{\alpha^{15}+\beta^{15}} \\ & =\frac{\left(\alpha^{15}+\beta^{15}\right)\left(\frac{49}{16}+9 \mathrm{i}\right)}{\left(\alpha^{15}+\beta^{15}\right)} \\ & \text{ Real }=\frac{49}{16} \\ & \text{ Im }=9 \\ & \text{ Ans. } 441 \end{aligned}
Ready for a full JEE mock test? Timed · full syllabus · instant results
Take a Mock Test →