Complex Number – JEE Advanced Previous Year Questions with Solutions

Class 9-10, JEE & NEET

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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]

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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]

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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]

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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

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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]

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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]

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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]

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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]

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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]

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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]

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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]

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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]

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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)]

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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)]

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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)]

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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)]

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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)]

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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)]

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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)]

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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]

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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]

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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]

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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]

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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]

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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]

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Sol. (A,C,D)

• January 28, 2021 at 6:39 am

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• January 28, 2021 at 6:41 am

Yes bro

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• November 13, 2020 at 5:39 am

Thank you soo much

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• September 23, 2020 at 9:41 pm

nice

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• August 22, 2020 at 7:13 pm

awerome and very helpful to jee aspirants
THANK YOU

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• May 19, 2020 at 4:01 pm

Thanku

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• 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

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