Complex Number – JEE Advanced Previous Year Questions with Solutions

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Q. Let $\mathrm{z}=\cos \theta+\mathrm{i} \sin \theta .$ Then the value of $\sum_{\mathrm{m}=1}^{15} \operatorname{Im}\left(\mathrm{z}^{2 \mathrm{m}-1}\right)$ at $\theta=2^{\circ}$ is

(A) $\frac{1}{\sin 2^{\circ}}$

(B) $\frac{1}{3 \sin 2^{\circ}}$

(C) $\frac{1}{2 \sin 2^{\circ}}$

(D) $\frac{1}{4 \sin 2^{\circ}}$

[JEE 2009, 3M,-1M]

Sol. (D)

$\mathrm{z}=\cos \theta+\mathrm{i} \sin \theta=\mathrm{e}^{\mathrm{i} \theta}$

$\therefore \sum_{m=1}^{15} \operatorname{Im}\left(z^{2 m-1}\right)=\sum_{m=1}^{15} \operatorname{Im}\left(e^{i \theta}\right)^{2 m-1}=\sum_{m=1}^{15} \operatorname{Im} e^{i(2 m-1) \theta}$

$=\sin \theta+\sin 3 \theta+\ldots \ldots+\sin 29 \theta$

$=\frac{\sin \left(\frac{\theta+29 \theta}{2}\right) \cdot \sin \left(\frac{15 \times 2 \theta}{2}\right)}{\sin \frac{2 \theta}{2}}$c

$=\frac{\sin (15 \theta) \cdot \sin (15 \theta)}{\sin \theta}, \theta=2^{\circ}$b

$=\frac{1}{4 \sin 2^{\circ}}$


Q. Let $z=x+$ iy be a complex number where $x$ and $y$ are integers. Then the area of the rectangle whose vertices are the roots of the equation $z \bar{z}^{3}+\bar{z} z^{3}=350$ is

(A) 48           (B) 32             (C) 40            (D) 80

[JEE 2009, 3 + 3]

Sol. (A)

$z \bar{z}\left(z^{2}+z^{-2}\right)=350$

$2\left(x^{2}+y^{2}\right)\left(x^{2}-y^{2}\right)=350$

$\left(x^{2}+y^{2}\right)\left(x^{2}-y^{2}\right)=175$

$\because \mathrm{x}, \mathrm{y} \in \mathrm{I}$ the only possible case which gives integral solutions

$\mathrm{x}^{2}+\mathrm{y}^{2}=25$

$\mathrm{x}^{2}-\mathrm{y}^{2}=7$

$\mathrm{x}^{2}=16, \mathrm{y}^{2}=9$

$\mathrm{x}=\pm 4, \mathrm{y}=\pm 3$

Area of rectangle $=8 \times 6=48$


Q. Let $z_{1}$ and $z_{2}$ be two distinct complex numbers and let $z=(1-t) z_{1}+t z_{2}$ for some real number $t$ with $0<t<1 .$ If $\operatorname{Arg}(w)$ denotes the principal argument of a nonzero complex number $w,$ then

(A) $\left|z-z_{1}\right|+\left|z-z_{2}\right|=\left|z_{1}-z_{2}\right|$

(B) $\operatorname{Arg}\left(\mathrm{z}-\mathrm{z}_{1}\right)=\operatorname{Arg}\left(\mathrm{z}-\mathrm{z}_{2}\right)$

(C) $\left|\begin{array}{cc}{\mathrm{z}-\mathrm{z}_{1}} & {\overline{\mathrm{z}}-\overline{\mathrm{z}}_{1}} \\ {\mathrm{z}_{2}-\mathrm{z}_{1}} & {\overline{\mathrm{z}}_{2}-\overline{\mathrm{z}}_{1}}\end{array}\right|=0$

(D) $\operatorname{Arg}\left(\mathrm{z}-\mathrm{z}_{1}\right)=\operatorname{Arg}\left(\mathrm{z}_{2}-\mathrm{z}_{1}\right)$

[JEE 2009, 3 + 3]

Sol. (A,C,D)

$\mathrm{z}=\mathrm{z}_{1}+\mathrm{t}\left(\mathrm{z}_{2}-\mathrm{z}_{1}\right)$

$\frac{\mathrm{z}-\mathrm{z}_{1}}{\mathrm{z}_{2}-\mathrm{z}_{1}}=\mathrm{t}, \mathrm{t} \in(0,1)$

$\Rightarrow \mathrm{z}=\frac{\mathrm{z}_{1}(1-\mathrm{t})+\mathrm{t} z_{2}}{(1+\mathrm{t})+\mathrm{t}}$

point $\mathrm{P}(\mathrm{z})$ divides point $\mathrm{A}\left(\mathrm{z}_{1}\right) \& \mathrm{B}\left(\mathrm{z}_{2}\right)$ internally in ratio $(1-\mathrm{t}): \mathrm{t}$

Hence locus is a line segment such that $\mathrm{P}(\mathrm{z})$ lies between $\mathrm{A}\left(\mathrm{z}_{1}\right) \& \mathrm{B}\left(\mathrm{z}_{2}\right)$ as shown in figure.

Hence options A,C & D are correct.


Q. Let $\omega$ be the complex number $\cos \frac{2 \pi}{3}+$ isin $\frac{2 \pi}{3} .$ Then the number of distinct complex numbers $z$ satisfying $\left|\begin{array}{ccc}{\mathrm{z}+1} & {\omega} & {\omega^{2}} \\ {\omega} & {\mathrm{z}+\omega^{2}} & {1} \\ {\omega^{2}} & {1} & {\mathrm{z}+\omega}\end{array}\right|=0$ is equal to

Sol. 1


Q. Match the statements in Column-I with those in Column II.

[Note : Here z takes values in the complex plane and Im z and Re z denote, respectively, the imaginary part and the real part of z.]

[JEE 10, 3+3+8]

Sol. ((A) Q,R (B) P (C) P,S,T (D) Q,R,S,T )

(A) $\left.|z-i| z\right|^{2}=|z+i| z||^{2}$

$\quad \Rightarrow(z-i|z|)(\bar{z}+i|z|)=(z+i|z|)(\bar{z}-i|z|)$

$\quad \Rightarrow 2 i|z| z=2 i|z| \bar{z}$

$\quad \Rightarrow z=\bar{z} \quad \therefore z$ is purely real.

$\quad \therefore z$ lies on real axis.

(B) Locus is ellipse having focii $(-4,0) \&(4,0)$

2ae $=8$ \& $2 \mathrm{a}=10$

$\Rightarrow \mathrm{a}=5 \quad \& \quad \mathrm{e}=4 / 5$

It is ellipse having eccentricity $4 / 5$.

(C) $\mathrm{w}=2(\cos \theta+\mathrm{i} \sin \theta)$

$\quad \mathrm{z}=2(\cos \theta+\mathrm{i} \sin \theta)-\frac{1}{2(\cos \theta+\mathrm{i} \sin \theta)}$

$\quad \mathrm{x}+\mathrm{iy}=\frac{3}{2} \cos \theta+\frac{\mathrm{i} 5}{2} \sin \theta$

$\quad \Rightarrow \quad \mathrm{x}=\frac{3}{2} \cos \theta \quad \& \quad \mathrm{y}=\frac{5}{2} \sin \theta$

It is a locus $\frac{x^{2}}{9 / 4}+\frac{y^{2}}{25 / 4}=1$

$\frac{9}{4}=\frac{25}{4}\left(1-e^{2}\right) \Rightarrow e=\frac{4}{5}$

since $\mathrm{x}=\frac{3}{2} \cos \theta \quad \Rightarrow \quad|\operatorname{Re}(z)| \leq \frac{3}{2}$

$|\operatorname{Re}(z)| \leq \frac{3}{2} \quad \Rightarrow \quad|\operatorname{Re}(z)| \leq 2$

Consider the circle $\mathrm{x}^{2}+\mathrm{y}^{2}-9=0$

By putting $\mathrm{x}=\frac{3}{2} \cos \theta$

$\& y=\frac{5}{2} \sin \theta$ into $x^{2}+y^{2}-9$

$\frac{9 \cos ^{2} \theta}{4}+\frac{25}{4} \sin ^{2} \theta-9<0$

(D) $z=(\cos \theta+i \sin \theta)+\frac{1}{(\cos \theta+i \sin \theta)}$

$z=2 \cos \theta$

where $z$ is real value $\& z \in[-2,2]$z


Q. Comprehension (3 questions together)

Let a,b and c be three real numbers satisfying

(i) If the point P(a,b,c), with reference to (E), lies on the plane 2x + y + z = 1, then the

value of 7a+b+c is

(A) 0           (B) 12             (C) 7             (D) 6

(ii) Let  be a solution of $x^{3}-1=0$ with Im() > 0. If a = 2 with b and c satisfying (E), then the value of $\frac{3}{\omega^{a}}+\frac{1}{\omega^{b}}+\frac{3}{\omega^{c}}$ is equal to –

(A) –2 (B) 2 (C) 3 (D) –3

(iii) Let b = 6, with a and c satisfying (E). If $\alpha$ and $\beta$ are the roots of the quadratic equation $\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}=0,$ then $\sum_{\mathrm{n}=0}^{\infty}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)^{\mathrm{n}}$ is –

(A) 6       (B) 7        (C) $\frac{6}{7}$           $(D) \infty$

[JEE 2011, 3+3+3]

Sol. ((i) $\mathrm{D} \quad$ (ii) $\mathrm{A} \quad$ (iii) $\mathrm{B}$)

Comprehension $(3$ questions together)

$\mathrm{a}+8 \mathrm{b}+7 \mathrm{c}=0$

$9 \mathrm{a}+2 \mathrm{b}+3 \mathrm{c}=0$

$7 \mathrm{a}+7 \mathrm{b}+7 \mathrm{c}=0$

$\Rightarrow \quad \mathrm{a}=\mathrm{K}, \mathrm{b}=6 \mathrm{K}, \mathrm{c}=-7 \mathrm{K}$

$\quad(\mathrm{K}, 6 \mathrm{K},-7 \mathrm{K})$

$2 \mathrm{x}+\mathrm{y}+\mathrm{z}=1$

$2 \mathrm{K}+6 \mathrm{K}-7 \mathrm{K}=1$

$\begin{aligned} &(\because \text { point lies on the plane }) \\ \Rightarrow & \mathrm{K}=1 \\ \Rightarrow & 7 \mathrm{a}+\mathrm{b}+\mathrm{c}=7 \mathrm{K}+6 \mathrm{K}-7 \mathrm{K}=6 \end{aligned}$

(ii) $x^{3}-1=0$

$\quad \Rightarrow x=1, \omega, \omega^{2}$

$\quad \omega=-\frac{1}{2}+\frac{i \sqrt{3}}{2}$

If $\quad a=2=K$

$\Rightarrow \quad b=12 \quad \& c=-14$

Hence $\frac{3}{\omega^{a}}+\frac{1}{\omega^{b}}+\frac{3}{\omega^{c}}=\frac{3}{\omega^{2}}+\frac{1}{\omega^{12}}+\frac{3}{\omega^{-14}}$

$=3 \omega+1+3 \omega^{2}=-3+1=-2$

$\begin{aligned}( \text { iii) } & \because \mathrm{b}=6 \quad \Rightarrow \quad 6 \mathrm{K}=6 \\ \Rightarrow & \mathrm{K}=1 \\ \Rightarrow & \mathrm{a}=1, \quad \mathrm{b}=6 \quad \& \quad \mathrm{c}=-7 \\ & \mathrm{x}^{2}+6 \mathrm{x}-7=0 \\ & \Rightarrow \alpha+\beta=-6, \alpha \beta=-7 \\ & \Rightarrow \quad \sum_{n=0}^{\infty}\left(\frac{\alpha+\beta}{\alpha \beta}\right)^{n}=\sum_{n=0}^{\infty}\left(\frac{6}{7}\right)^{n}=\frac{1}{1-\frac{6}{7}}=7 \end{aligned}$


Q. If $z$ is any complex number satisfying $|z-3-2 i| \leq 2,$ then the minimum value of $|2 z-6+5 i|$ is

[JEE 2011, 4M]

Sol. 5

We have to find minimum value of

$$

2\left|z-\left(3-\frac{5}{2} i\right)\right|

$$

$=2 \times$ (minimum distance between $\mathrm{z}$ and point

$\left.\qquad\left(3,-\frac{5}{2}\right)\right)$

$=2 \times\left(\text { distance between }(3,0) \text { and }\left(3,-\frac{5}{2}\right)\right)$

$=2 \times \frac{5}{2}=5$ units.


Q. Let $\omega=\mathrm{e}^{\mathrm{i} \pi / 3},$ and $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{x}, \mathrm{y}, \mathrm{z}$ be non-zero complex numbers such that

$a+b+c=x$

$a+b \omega+c \omega^{2}=y$

$a+b \omega^{2}+c \omega=z$

Then the value of $\frac{|\mathrm{x}|^{2}+|\mathrm{y}|^{2}+|\mathrm{z}|^{2}}{|\mathrm{a}|^{2}+|\mathrm{b}|^{2}+|\mathrm{c}|^{2}}$ is

[JEE 2011, 4M]

Sol. (Bonus)

Ans. 3 (Bonus)

(Comment : If $\omega=\mathrm{e}^{\mathrm{i} \pi / 3}$

then $\frac{|x|^{2}+|y|^{2}+|z|^{2}}{|a|^{2}+|b|^{2}+|c|^{2}}$ is not always an integer.

For example if $a=b=c=1$ then the value of $\frac{|x|^{2}+|y|^{2}+|z|^{2}}{|a|^{2}+|b|^{2}+|c|^{2}}$. Now if we consider

$\omega=\mathrm{e}^{\mathrm{i} 2 \pi / 3}$ then the solution is

$|\mathrm{x}|^{2}=(\mathrm{a}+\mathrm{b}+\mathrm{c})(\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}})$

$=|\mathrm{a}|^{2}+\left.|\mathrm{b} \hat{\mathrm{p}}+| \mathrm{c}\right|^{2}+\mathrm{a} \overline{\mathrm{b}}+\overline{\mathrm{a} \overline{\mathrm{c}}}+\overline{\mathrm{b} \overline{\mathrm{a}}}+\mathrm{b} \overline{\mathrm{c}}+\mathrm{c} \overline{\mathrm{a}}+\mathrm{c} \overline{\mathrm{b}}$

$|\mathrm{y}|^{2}=\left(\mathrm{a}+\mathrm{b} \omega+\mathrm{c} \omega^{2}\right)\left(\overline{\mathrm{a}}+\overline{\mathrm{b}} \omega^{2}+\overline{\mathrm{c}} \omega\right)$

$=|a|^{2}+|b|^{2}+|c|^{2}+a \bar{b} \omega^{2}+a \bar{c} \omega+b \bar{a} \omega+b \bar{c} \omega^{2}+c \bar{a} \omega^{2}+c \bar{b} \omega$

$|z|^{2}=\left(a+b \omega^{2}+c \omega\right)\left(\bar{a}+\bar{b} \omega+\bar{c} \omega^{2}\right)$

$=|a|^{2}+|b|^{2}+\left|c^{2}\right|$

$+a \bar{b} \omega+a \bar{c} \omega^{2}+b \bar{a} \omega^{2}+b \bar{c} \omega+c \bar{a} \omega+c \bar{b} \omega^{2}$

$\therefore|x|^{2}+|y|^{2}+|z|^{2}=3\left(|a|^{2}+|b|^{2}+|c|^{2}\right)$

$\Rightarrow \frac{|x|^{2}+|y|^{2}+|z|^{2}}{|a|^{2}+|b|^{2}+|c|^{2}}=3$


Q. Match the statements given in Column I with the values given in Column II

[JEE 2011, 2+2+2+2M]

Sol. ( (A) q (B) p or p,q,r,s,t (C) s (D) t )


Q. Match the statements given in Column I with the intervals/union of intervals given in Column II

[JEE 2011, 2+2+2+2M]

Sol. ($\begin{array}{llll}{\text { (A) } s} & {\text { (B) } t} & {\text { (C) } r} & {\text { (D) } r}\end{array}$)

(A) Let $z=\cos \theta+i \sin \theta$

$\quad \operatorname{Re}\left(\frac{2 i(\cos \theta+i \sin \theta)}{1-(\cos \theta+i \sin \theta)^{2}}\right)=\operatorname{Re}\left(\frac{i \cos \theta-\sin \theta}{\sin ^{2} \theta-i \cos \theta \sin \theta}\right)$

$=\operatorname{Re}\left(-\frac{1}{\sin \theta}\right)=\frac{-1}{\sin \theta}$

$\begin{aligned} \therefore & \text { Set will be }(-\infty,-1] \cup[1, \infty) \\-1 \leq & \frac{8.3^{(x-2)}}{1-3^{2(x-1)}} \leq 1 \quad \mathrm{x} \neq 1 \\ \Rightarrow &-1 \leq \frac{8.3^{x}}{\left(3-3^{x}\right)\left(3+3^{x}\right)} \leq 1 \end{aligned}$

$3^{\mathrm{x}}=\mathrm{t} \quad \therefore \quad \mathrm{t}>0$

$\frac{8 \mathrm{t}}{(3-\mathrm{t})(\mathrm{t}+3)} \geq-1$

$\Rightarrow \quad \mathrm{t} \in(0,3) \cup[9, \infty)$

$\Rightarrow \quad x \in(-\infty, 1) \cup[2, \infty)$

$\quad \frac{8 t}{(3-t)(t+3)} \leq 1$

$\quad \Rightarrow \quad t \in(0,1] \cup(3, \infty)$

$\quad \Rightarrow \quad x \in(-\infty, 0] \cup(1, \infty)$

Taking intersection

$x \in(-\infty, 0] \cup[2, \infty)$

(D) $f(x)=3 x^{5 / 2}-10 x^{3 / 2}$

$f^{\prime}(x)=\frac{15}{2} x^{3 / 2}-\frac{30}{2} x^{1 / 2}>0$

$\Rightarrow \frac{15}{2} \sqrt{x}(x-2) \geq 0 \quad \Rightarrow \quad x \geq 2$


Q. Let z be a complex number such that the imaginary part of $z$ is nonzero and a $=z^{2}+z+1$ is real. Then a cannot take the value $-$

(A) –1

(B) $\frac{1}{3}$

(C) $\frac{1}{2}$

(D) $\frac{3}{4}$

[JEE 2012, 3M, –1M]

Sol. (D)

$\mathrm{z}^{2}+\mathrm{z}+1-\mathrm{a}=0$

$\because \mathrm{z}$ is imaginary $\Rightarrow \mathrm{D}<0$

$1-4(1-\mathrm{a})<0$

$4 \mathrm{a}<3$

$\mathrm{a}<\frac{3}{4}$

Aliter $: a=z^{2}+z+1$

$\because \quad a=\bar{a}(\text { given a is real })$

$\therefore \quad z^{2}+z=\bar{z}^{2}+$

$\Rightarrow \quad z^{2}-\bar{z}^{2}=\bar{z}-z$

$\Rightarrow \quad z+\bar{z}=-1 \quad(\because \operatorname{Im}(z) \text { is non zero })$

$\Rightarrow \quad \operatorname{Re}(\mathrm{z})=-\frac{1}{2}$

$\therefore \quad \mathrm{z}$ can be taken as $-\frac{1}{2}+\mathrm{iy}$

where $\mathrm{y} \in \mathrm{R}$

$\therefore \quad \mathrm{a}=\left(-\frac{1}{2}+\mathrm{i} y\right)^{2}+\left(\frac{-1}{2}+\mathrm{iy}\right)+1$

$\Rightarrow \quad \mathrm{a}=\frac{1}{4}-\frac{1}{2}+1-\mathrm{i} y+\mathrm{i} y-\mathrm{y}^{2}$

$\Rightarrow \quad a=\frac{3}{4}-y^{2} \Rightarrow a<\frac{3}{4}$

$\therefore \quad a \neq \frac{3}{4}$


Q. Let complex numbers $\alpha$ and $\frac{1}{\bar{\alpha}}$ lie on circles $\left(\mathrm{x}-\mathrm{x}_{0}\right)^{2}+\left(\mathrm{y}-\mathrm{y}_{0}\right)^{2}=\mathrm{r}^{2}$ and $\left(\mathrm{x}-\mathrm{x}_{0}\right)^{2}+\left(\mathrm{y}-\mathrm{y}_{0}\right)^{2}=$

$4 \mathrm{r}^{2}$ respectively. If $\mathrm{z}_{0}=\mathrm{x}_{0}+\mathrm{iy}_{0}$ satisfies the equation $2\left|\mathrm{z}_{0}\right|^{2}=\mathrm{r}^{2}+2,$ then $|\alpha|=$

(A) $\frac{1}{\sqrt{2}}$

(B) $\frac{1}{2}$

(C) $\frac{1}{\sqrt{7}}$

(D) $\frac{1}{3}$

[JEE(Advanced) 2013, 2M]

Sol. (C)


Q. Let $\omega$ be a complex cube root of unity with $\omega \neq 1$ and $\mathrm{P}=\left[\mathrm{p}_{\mathrm{ij}}\right]$ be a $\mathrm{n} \times \mathrm{n}$ matrix with $\mathrm{p}_{\mathrm{ij}}=$ $\omega^{\mathrm{i}+\mathrm{j}} .$ Then $\mathrm{P}^{2} \neq 0,$ when $\mathrm{n}=$

(A) 57           (B) 55             (C) 58              (D) 56

[JEE(Advanced) 2013, 3, (–1)]

Sol. (B,C,D)

$\mathrm{P}^{2}=\left[\alpha_{\mathrm{ij}}\right]_{\mathrm{n} \times \mathrm{n}}$

$\alpha_{\mathrm{ij}}=\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{p}_{\mathrm{ik}} \cdot \mathrm{P}_{\mathrm{kj}}$

$=\sum_{\mathrm{k}=1}^{\mathrm{n}} \omega^{\mathrm{i}+\mathrm{k}} \cdot \omega^{\mathrm{k}+\mathrm{j}}=\omega^{\mathrm{i}+\mathrm{j}} \sum_{\mathrm{k}=1}^{\mathrm{n}} \omega^{2 \mathrm{k}}$

$=\omega^{\mathrm{i}+\mathrm{j}}\left(\omega^{2}+\omega^{4}+\omega^{6}+\ldots \ldots+\omega^{2 \mathrm{n}}\right)$

If $\mathrm{n}$ is a multiple of 3 then $\mathrm{P}^{2}=0$

$\Rightarrow \mathrm{n}$ is not a multiple of 3

$\Rightarrow \mathrm{n}$ can be $55,58,56$


Q. Let $\mathrm{w}=\frac{\sqrt{3}+\mathrm{i}}{2}$ and $\mathrm{P}=\left\{\mathrm{w}^{\mathrm{n}}: \mathrm{n}=1,2,3, \ldots . .\right\} .$ Further $\mathrm{H}_{1}=\left\{\mathrm{z} \in \mathrm{C}: \operatorname{Re} \mathrm{z}>\frac{1}{2}\right\}$ and $\mathrm{H}_{2}=\left\{\mathrm{z} \in \mathrm{C}: \operatorname{Re} \mathrm{z}<\frac{-1}{2}\right\},$ where $\mathrm{C}$ is the set of all complex numbers. If $\mathrm{z}_{1} \in \mathrm{P} \cap \mathrm{H}_{1}, \mathrm{z}_{2} \in \mathrm{P}$ $\cap \mathrm{H}_{2}$ and $\mathrm{O}$ represents the origin, then $\angle \mathrm{z}_{1} \mathrm{O} \mathrm{z}_{2}=$

(A) $\frac{\pi}{2}$

(B) $\frac{\pi}{6}$

(C) $\frac{2 \pi}{3}$

(D) $\frac{5 \pi}{6}$

[JEE-Advanced 2013, 4, (–1)]

Sol. (C,D)

$\mathrm{z}_{1}=\left\{\mathrm{w}_{1}, \mathrm{w}_{11}, \mathrm{w}_{12}\right\}$

$\mathrm{z}_{2}=\left\{\mathrm{w}_{5}, \mathrm{w}_{6}, \mathrm{w}_{7}\right\}$

$\angle \mathrm{w}_{1} \mathrm{O} \mathrm{w}_{5}=\frac{2 \pi}{3} \& \angle \mathrm{w}_{1} \mathrm{O} \mathrm{w}_{6}=\frac{5 \pi}{6}$


Let $\mathrm{S}=\mathrm{S}_{1} \cap \mathrm{S}_{2} \cap \mathrm{S}_{3},$ where $\mathrm{S}_{1}=\{\mathrm{z} \in \mathrm{C}:|\mathrm{z}|<4\}, \mathrm{S}_{2}=\left\{\mathrm{z} \in \mathrm{C}: \operatorname{Im}\left[\frac{\mathrm{z}-1+\sqrt{3} \mathrm{i}}{1-\sqrt{3} \mathrm{i}}\right]>0\right\}$ and $\mathrm{S}_{3}=\{\mathrm{z} \in \mathrm{C}: \operatorname{Re} \mathrm{z}>0\}$

Q. $\min _{\mathbf{z} \in \mathrm{S}}|1-3 \mathbf{i}-\mathbf{z}|=$

(A) $\frac{2-\sqrt{3}}{2}$

(B) $\frac{2+\sqrt{3}}{2}$

(C) $\frac{3-\sqrt{3}}{2}$

(D) $\frac{3+\sqrt{3}}{2}$

[JEE(Advanced) 2013, 3, (–1)]

Sol. (C)

$\mathrm{S}_{1}$ is interior of circle centred at $(0,0) \&$ radius $=4$

$\operatorname{Re}(\mathrm{z})>0$ is in $\mathrm{I}^{\mathrm{st}} \& \mathrm{IV}^{\mathrm{th}} \mathrm{quadrant.}$

$\frac{(z-(1-i \sqrt{3}))}{(1-i \sqrt{3})}=\frac{((x-1)+i(y-\sqrt{3}))}{(1-i \sqrt{3})}$

$=\frac{((x-1)+i(y-\sqrt{3}))(1+i \sqrt{3})}{2}$

$\mathrm{I}_{\mathrm{m}}\left(\mathrm{S}_{2}\right)=\sqrt{3} \mathrm{x}+\mathrm{y}>0$

erpendicular distance from $(1,-3)$ to the line is $\mathrm{P}=\frac{\sqrt{3}-3}{2} |=\left(\frac{3-\sqrt{3}}{2}\right)$


Q. Area of S =

(A) $\frac{10 \pi}{3}$

(B) $\frac{20 \pi}{3}$

(C) $\frac{16 \pi}{3}$

(D) $\frac{32 \pi}{3}$

[JEE(Advanced) 2013, 3, (–1)]

Sol. (B)

Area of $\mathrm{S}=\frac{\pi(4)^{2}}{4}+\frac{1}{2}(4)^{2} \frac{\pi}{3}$

$\frac{8 \pi}{3}+4 \pi=\frac{20 \pi}{3}$


Q. The quadratic equation p(x) = 0 with real coefficients has purely imaginary roots. Then the equation p(p(x)) = 0 has

(A) only purely imaginary roots

(B) all real roots

(C) two real and two purely imaginary roots

(D) neither real nor purely imaginary roots.

[JEE(Advanced) 2014, 3(–1)]

Sol. (D)

Let $\mathrm{p}(\mathrm{x})=\mathrm{ax}^{2}+\mathrm{b}+\mathrm{c}$

$\mathrm{p}(\mathrm{x})=0 \Rightarrow \mathrm{x}=\frac{-\mathrm{b}+\sqrt{\mathrm{b}^{2}-4 \mathrm{ac}}}{2 \mathrm{a}}$

so $\mathrm{b}=0$ as roots are purely imaginary

so equation will be $\mathrm{ax}^{2}+\mathrm{c}=0\{\text { where ‘a’ and ‘c’ have same sign }\}$

Now $\mathrm{p}(\mathrm{p}(\mathrm{x}))=0$

$\Rightarrow \quad a p^{2}(x)+c=0 \quad \Rightarrow \quad p(x)=\pm \sqrt{-\frac{c}{a}}$

$a x^{2}+c=\pm \sqrt{-\frac{c}{a}} \quad \Rightarrow \quad x \notin R$

$\begin{aligned} \text { if } \mathrm{x} &=\mathrm{i} \beta \text { then } \\ &-\mathrm{a} \beta^{2}+\mathrm{c}=\pm \sqrt{-\frac{\mathrm{c}}{\mathrm{a}}} \quad \text { not possible } \end{aligned}$

(real) (imaginary)

So neither real nor purely imaginary roots.


Q. Let $\mathrm{z}_{\mathrm{k}}=\cos \left(\frac{2 \mathrm{k} \pi}{10}\right)+\mathrm{i} \sin \left(\frac{2 \mathrm{k} \pi}{10}\right) ; \mathrm{k}=1,2, \ldots \ldots 9$

[JEE(Advanced) 2014, 3(–1)]

Sol. (C)

(P) $\quad e^{\frac{i 2 k \pi}{10}} \cdot e^{\frac{i 2 \pi}{10}}=1 \quad \Rightarrow e^{i \frac{\pi}{10} x 2(k+1)}=1$

$\frac{\pi}{10}(2(\mathrm{k}+\mathrm{j}))=2 \mathrm{n} \pi$b

$(\mathrm{k}+\mathrm{j})=10$

Possible

(Q) $\quad e^{\frac{\mathrm{i} 2 \pi}{10}} \cdot z=\mathrm{e}^{\frac{\mathrm{i} 2 \pi \mathrm{k}}{10}}$ $$

\mathrm{z}=\frac{\mathrm{e}^{\frac{\mathrm{i} 2 \pi \mathrm{k}}{10}}}{\frac{\mathrm{i} 2 \pi}{10}} \text { is possible } $$

(R) $\quad \mathrm{z}^{10}-1=(\mathrm{z}-1)\left(\mathrm{z}-\mathrm{z}_{1}\right)\left(\mathrm{z}-\mathrm{z}_{2}\right) \ldots \ldots .\left(\mathrm{z}-\mathrm{z}_{9}\right)$

put z = 1

$\lim _{z \rightarrow 1} \frac{z^{10}-1}{(z-1)}=\left(1-z_{1}\right)\left(1-z_{2}\right) \ldots\left(1-z_{9}\right)$

$\lim _{z \rightarrow 1} \frac{10 z^{9}}{1}=\left(1-z_{1}\right)\left(1-z_{2}\right) \ldots\left(1-z_{9}\right)$

$=\left|\left(1-z_{1}\right)\left(1-z_{2}\right) \ldots .\left(1-z_{9}\right)\right|=10$

(S) $1+\cos \frac{2 \pi}{10}+\cos \frac{4 \pi}{10}+\ldots \ldots+\cos \frac{18 \pi}{10}=0$

since they are sum of ten, tenth roots of unity

$\sum_{k=1}^{9} \cos \left(\frac{2 k \pi}{10}\right)=-1$

$1+1=2$


Q.

[JEE 2015, 8(Each 2M, –1M)]

Sol. ((A) p,q (B) p,q (C) p,q,s,t (D) q,t)

(A) $\left|\frac{\alpha \sqrt{3}+\beta}{2}\right|=\sqrt{3} \Rightarrow \alpha \sqrt{3}+\frac{\alpha-2}{\sqrt{3}}=\pm 2 \sqrt{3}$

$\Rightarrow \alpha\left(\frac{4}{\sqrt{3}}\right)=\frac{2}{\sqrt{3}} \pm 2 \sqrt{3}$

$\alpha=2,-1 \Rightarrow|\alpha|=1,2$

(B) By continuity $\quad-3 a-2=b+a^{2}$

By differentiability $-6 a=b$

$\quad a^{2}-3 a+2=0 \Rightarrow a=1,2$

(C) $\quad\left(\left(-3+2 \omega+3 \omega^{2}\right) \omega\right)^{4 \mathrm{n}+3}+\left(\left(-3+2 \omega+3 \omega^{2}\right) \omega^{2}\right)^{4 n+3}+\left(\left(-3+2 \omega+3 \omega^{2}\right)^{4 n+3}\right)=0$

$\quad \Rightarrow\left(-3+2 \omega+3 \omega^{2}\right)^{4 n+3}\left[\omega^{4 n+3}+\omega^{8 n+6}+1\right]=0$

$\quad \Rightarrow \omega^{\mathrm{n}}+\omega^{2 \mathrm{n}}+1=0 \Rightarrow n$ is not a multiple of 3

$\begin{aligned} \text { (D) } \frac{2 a b}{a+b}=4,2(5-a) &=b-5 \\ b &=15-2 a \\ 2 a(15-2 a)=4(15-a) & \Rightarrow 15 a-2 a^{2}=30-2 a \\ 2 a^{2}-17 a+30=0 & \Rightarrow 2 a^{2}-12 a-5 a+30=0 \\ 2 a(a-6)-5(a-6) &=0 \end{aligned}$

$\mathrm{a}=\frac{5}{2}, 6$

$\Rightarrow|\mathrm{q}-\mathrm{a}|=|10-2 \mathrm{a}|=5 \mathrm{or} 2$


Q. For any integer $\mathrm{k},$ let $\alpha_{\mathrm{k}}=\cos \left(\frac{\mathrm{k} \pi}{7}\right)+\mathrm{i} \sin \left(\frac{\mathrm{k} \pi}{7}\right),$ where $\mathrm{i}=\sqrt{-1} .$ The value of the expression

$\frac{\sum_{\mathrm{k}=1}^{12}\left|\alpha_{\mathrm{k}+1}-\alpha_{\mathrm{k}}\right|}{\sum_{\mathrm{k}=1}^{3}\left|\alpha_{4 \mathrm{k}-1}-\alpha_{4 \mathrm{k}-2}\right|}$ is

[JEE 2015, 4M, –0M]

Sol. 4

$\alpha_{\mathrm{k}}$ are vertices of 14 sided regular polygon.

$\left|\alpha_{\mathrm{k}+1}-\alpha_{\mathrm{k}}\right|$ length of a side of the regular polygon

$\left|\alpha_{4 \mathrm{k}-1}-\alpha_{4 \mathrm{k}-2}\right|$ length of a side of the regular polygon

$\Rightarrow \quad \frac{12(\mathrm{S})}{3(\mathrm{S})}=4$

Alter

$\alpha_{\mathrm{k}}=\cos \left(\frac{\mathrm{k} \pi}{7}\right)+\mathrm{i} \sin \left(\frac{\mathrm{k} \pi}{7}\right)=e^{\mathrm{i} \frac{\mathrm{k} \pi}{7}}$

$\alpha_{\mathrm{k}+1}=e^{\frac{i(k+1) \pi}{7}}$


Q. Let $z=\frac{-1+\sqrt{3} i}{2},$ where $i=\sqrt{-1},$ and $r, s \in\{1,2,3\} .$ Let $P=\left[\begin{array}{cc}{(-z)^{\mathrm{r}}} & {z^{2 s}} \\ {z^{2 s}} & {z^{\mathrm{r}}}\end{array}\right]$ and I be the identity matrix of order $2 .$ Then the total number of ordered pairs (r,s) for which $P^{2}=-I$ is

[JEE(Advanced 2016]

Sol. 1


Q. Let $a, b \in \square$ and $a^{2}+b^{2} \neq 0 .$ Suppose $S=\left\{z \in \square: z=\frac{1}{a+i b t}, t \in \square, t \neq 0\right\},$ where $i=\sqrt{-1}$ If $z=x+$ iy and $z \in S,$ then $(x, y)$ lies on

(A) the circle with radius $\frac{1}{2 \mathrm{a}}$ and centre $\left(\frac{1}{2 \mathrm{a}}, 0\right)$ for $\mathrm{a}>0, \mathrm{b} \neq 0$

(B) the circle with radius $-\frac{1}{2 \mathrm{a}}$ and centre $\left(-\frac{1}{2 \mathrm{a}}, 0\right)$ for a $<0, \mathrm{b} \neq 0$

(C) the $\mathrm{x}$ -axis for $\mathrm{a} \neq 0, \mathrm{b}=0$

(D) the y-axis for a $=0, \mathrm{b} \neq 0$

[JEE(Advanced 2016]

Sol. (A,C,D)


Q. Let $a, b, x$ and $y$ be real numbers such that $a-b=1$ and $y \neq 0 .$ If the complex number $z=x+i y$ satisfies $\operatorname{Im}\left(\frac{a z+b}{z+1}\right)=y,$ then which of the following is (are) possible value(s) of $x ?$

(A) $-1-\sqrt{1-\mathrm{y}^{2}}$

(B) $1+\sqrt{1+y^{2}}$

(C) $1-\sqrt{1+y^{2}}$

(D) $-1+\sqrt{1-y^{2}}$

[JEE(Advanced 2017]

Sol. (A,D)

$\operatorname{Im}\left(\frac{\mathrm{az}+\mathrm{b}}{\mathrm{z}+1}\right)=\mathrm{y}$ and $\mathrm{z}=\mathrm{x}+\mathrm{iy}$

$\therefore \quad \operatorname{Im}\left(\frac{\mathrm{a}(\mathrm{x}+\mathrm{iy})+\mathrm{b}}{\mathrm{x}+\mathrm{iy}+1}\right)=\mathrm{y}$

$\Rightarrow \quad \operatorname{Im}\left(\frac{(a x+b+i a y)(x+1-i y)}{(x+1)^{2}+y^{2}}\right)=y$

$\begin{aligned} \Rightarrow &-\mathrm{y}(\mathrm{ax}+\mathrm{b})+\mathrm{ay}(\mathrm{x}+1)=\mathrm{y}\left((\mathrm{x}+1)^{2}+\mathrm{y}^{2}\right) \\ \Rightarrow &(\mathrm{a}-\mathrm{b}) \mathrm{y}=\mathrm{y}\left((\mathrm{x}+1)^{2}+\mathrm{y}^{2}\right) \\ & \because \mathrm{y} \neq 0 \text { and } \mathrm{a}-\mathrm{b}=1 \\ \Rightarrow &(\mathrm{x}+1)^{2}+\mathrm{y}^{2}=1 \end{aligned}$

$\Rightarrow \quad x=-1 \pm \sqrt{1-y^{2}}$


Q. For a non-zero complex number z, let arg(z) denotes the principal argument with $-\pi<\arg (\mathrm{z}) \leq \pi$. Then, which of the following statement(s) is (are) FALSE ?

(A) $\arg (-1-i)=\frac{\pi}{4},$ where $i=\sqrt{-1}$

(B) The function $f: \square \rightarrow(-\pi, \pi],$ defined by $f(\mathrm{t})=\arg (-1+i t)$ for all $t \in \square,$ is continuous at all points of $\square,$ where $i=\sqrt{-1}$

(C) For any two non-zero complex numbers $z_{1}$ and $z_{2}, \arg \left(\frac{z_{1}}{z_{2}}\right)-\arg \left(z_{1}\right)+\arg \left(z_{2}\right)$ is an integer multiple of $2 \pi$

(D) For any three given distinct complex numbers $z_{1}, z_{2}$ and $z_{3}$, the locus of the point z satisfying the condition $\arg \left(\frac{\left(z-z_{1}\right)\left(z_{2}-z_{3}\right)}{\left(z-z_{3}\right)\left(z_{2}-z_{1}\right)}\right)=\pi,$ lies on a straight line(D) For any three given distinct complex numbers $z_{1}, z_{2}$ and $z_{3}$, the locus of the point z satisfying the condition $\arg \left(\frac{\left(z-z_{1}\right)\left(z_{2}-z_{3}\right)}{\left(z-z_{3}\right)\left(z_{2}-z_{1}\right)}\right)=\pi,$ lies on a straight line

[JEE(Advanced 2018]

Sol. (A,B,D)

(A) $\arg (-1-i)=-\frac{3 \pi}{4}$

(B) $f(\mathrm{t})=\arg (-1+\mathrm{i} \mathrm{t})=\left\{\begin{array}{ll}{\pi-\tan ^{-1}(\mathrm{t}),} & {\mathrm{t} \geq 0} \\ {-\pi+\tan ^{-1}(\mathrm{t}),} & {\mathrm{t} \geq 0}\end{array}\right.$

Discontinuous at $\mathrm{t}=0$

(C) $\arg \left(\frac{\mathrm{z}_{1}}{\mathrm{z}_{2}}\right)-\arg \left(\mathrm{z}_{1}\right)+\arg \left(\mathrm{z}_{2}\right)$

$=\arg \mathrm{g}_{1}-\arg \left(\mathrm{z}_{2}\right)+2 \mathrm{n} \pi-\arg \left(\mathrm{z}_{1}\right)+\arg \left(\mathrm{z}_{2}\right)=2 \mathrm{n} \pi$

(D) $\arg \left(\frac{\left(\mathrm{z}-\mathrm{z}_{1}\right)\left(\mathrm{z}_{2}-\mathrm{z}_{3}\right)}{\left(\mathrm{z}-\mathrm{z}_{3}\right)\left(\mathrm{z}_{2}-\mathrm{z}_{1}\right)}\right)=\pi$

$\Rightarrow \frac{\left(\mathrm{z}-\mathrm{z}_{1}\right)\left(\mathrm{z}_{2}-\mathrm{z}_{3}\right)}{\left(\mathrm{z}-\mathrm{z}_{3}\right)\left(\mathrm{z}_{2}-\mathrm{z}_{1}\right)}$ is real.

$\Rightarrow \mathrm{z}, \mathrm{z}_{1}, \mathrm{z}_{2}, \mathrm{z}_{3}$ are concyclic.


Q. Let $s, t, r$ be the non-zero complex numbers and $L$ be the set of solutions $z=x+i y$ $(x, y \in \square, i=\sqrt{-1})$ of the equation $s z+t \bar{z}+r=0,$ where $\bar{z}=x-i y .$ Then, which of the following statement(s) is (are) TRUE?

(A) If L has exactly one element, then $|s| \neq|t|$

(B) If $|s|=|t|,$ then $L$ has infinitely many elements

(C) The number of elements in L\cap $\{z:|z-1+i|=5\}$ is at most 2

(D) If L has more than one element, then L has infinitely many elements

[JEE(Advanced 2018]

Sol. (A,C,D)


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Comments
  • February 6, 2020 at 1:30 pm

    In the solution of last question it has been stated that the locus of z is a line but nothing is mentioned about the method through which we concluded that ,so please give some briefing on that