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Q. If $\left|Z-\frac{4}{Z}\right|=2,$ then the maximum value of $|Z|$ is equal to :-.
(1) 2
(2) $2+\sqrt{2}$
(3) $\sqrt{3}+1$
(4) $\sqrt{5}+1$
[AIEEE -2009]
Ans. (4)
$\left|z-\frac{4}{z}\right| \geq|z|-\left|\frac{4}{z}\right|$
$2 \geq|z|-\frac{4}{|z|}$
$2|z| \geq|z|^{2}-4$
$|z|^{2}-2|z|-4 \leq 0$
$|z| \leq \sqrt{5}+1$
Q. The number of complex numbers z such that $|z-1|=|z+1|=|z-i|$ equals :-
(1) 0 (2)1 (3) 2 (4) $\infty$
[AIEEE -2010]
Ans. (2)
z is the circumcentre (0, 0) of triangle ABC so their exist only one complex number.
Q. Let $\alpha, \beta$ be real and $z$ be a complex number. If $z^{2}+\alpha z+\beta=0$ has two distinct roots on the line $\operatorname{Re} z=1,$ then it is necessary that :-
(1) $|\beta|=1$
(2) $\beta \in(1, \infty)$
(3) $\beta \in(0,1)$
(4) $\beta \in(-1,0)$
[AIEEE -2011]
Ans. (2)
Let $z^{2}+\alpha z+\beta=0$ has $\left(1+i y_{1}\right)$ and $\left(1+i y_{2}\right)$
so $z_{1} z_{2}=\beta$
$\left(1+i y_{1}\right)\left(1+i y_{2}\right)=\beta$
$\beta=1-\mathrm{y}_{1} \mathrm{y}_{2}+\mathrm{i}\left(\mathrm{y}_{1}+\mathrm{y}_{2}\right)(\because \beta \text { is purely real })$
here $\mathrm{y}_{1}+\mathrm{y}_{2}=0$
$\mathrm{y}_{1}=-\mathrm{y}_{2}$
$\beta=1-\mathrm{y}_{1} \mathrm{y}_{2}$
$\beta=1+\mathrm{y}_{1}^{2}$
$\beta>1$
$\Rightarrow \beta \in(1, \infty)$
Q. If $\omega(\neq 1)$ is a cube root of unity, and $(1+\omega)^{7}=\mathrm{A}+\mathrm{B} \omega .$ Then $(\mathrm{A}, \mathrm{B})$ equals :-
(1) (1, 0) (2) (–1, 1) (3) (0, 1) (4) (1, 1)
[AIEEE -2011]
Ans. (4)
$(1+\omega)^{7}=\mathrm{A}+\mathrm{B} \omega$
$\left(-\omega^{2}\right)^{7}=\mathrm{A}+\mathrm{B} \omega$
$-\omega^{2}=\mathrm{A}+\mathrm{B} \omega$
$1+\omega=\mathrm{A}+\mathrm{B} \omega$
$\mathrm{A}=1$
$\mathrm{B}=1$ (1, 1)
Q. If $z \neq 1$ and $\frac{z^{2}}{z-1}$ is real, then the point represented by the complex number $z$ lies :
(1) on the imaginary axis.
(2) either on the real axis or on a circle passing through the origin.
(3) on a circle with centre at the origin.
(4) either on the real axis or on a circle not passing through the origin.
[AIEEE -2012]
Ans. (2)
$\frac{z^{2}}{z-1}$ is purely real where $(Z \neq 1)$
Q. If $z$ is a complex number of unit modulus and argument $\theta,$ then $\arg \left(\frac{1+z}{1+\bar{z}}\right)$ equals
(1) $-\theta$
(2) $\frac{\pi}{2}-\theta$
(3) $\theta$
(4) $\pi-\theta$
[JEE (Main)-2013]
Ans. (3)
$\bar{z}=\frac{1}{z} \Rightarrow \arg \left(\frac{1+z}{1+\frac{1}{z}}\right) \quad \Rightarrow \operatorname{argz} \Rightarrow \theta$
Q. If $\mathrm{z}$ is a complex number such that $|\mathrm{z}| \geq 2,$ then the minimum value of $\left|\mathrm{z}+\frac{1}{2}\right|:$
(1) is equal to $\frac{5}{2}$
(2) lies in the interval (1, 2)
(3) is strictly greater than $\frac{5}{2}$
(4) is strictly greater than $\frac{3}{2}$ but less than
[JEE (Main)-2014]
Ans. (2)
$\left|z+\frac{1}{2}\right| \geq|| z\left|-\frac{1}{2}\right|$
Min. value of $\left|z+\frac{1}{2}\right|$ occurs at $|z|=2$
$\because|z| \geq 2$
$\therefore\left|z+\frac{1}{2}\right|_{\text {min }}=\left|2-\frac{1}{2}\right|=\frac{3}{2}$
Q. A complex number $z$ is said to be unimodular if $|z|=1 .$ Suppose $z_{1}$ and $z_{2}$ are complex numbers such that $\frac{z_{1}-2 z_{2}}{2-z_{1} \bar{z}_{2}}$ is unimodular and $z_{2}$ is not unimodular. Then the point $z_{1}$ lies on a :
(1) circle of radius 2
(2) circle of radius $\sqrt{2}$
(3) straight line parallel to x-axis
(4) straight line parallel to y-axis
[JEE (Main)-2015]
Ans. (1)
$\frac{\left|z_{1}-2 z_{2}\right|}{\left|2-z_{1} \bar{z}_{2}\right|}=1$
$\Rightarrow \quad\left|z_{1}-2 z_{2}\right|^{2}=\left|2-z_{1} \bar{z}_{2}\right|^{2}$
$\Rightarrow\left(z_{1}-2 z_{2}\right)\left(\bar{z}_{1}-2 \bar{z}_{2}\right)=\left(2-z_{1} \bar{z}_{2}\right)\left(2-\bar{z}_{1} z_{2}\right)$
$\Rightarrow\left|z_{1}\right|^{2}+4\left|z_{2}\right|^{2}-4-\left|z_{1}\right|^{2}\left|z_{2}\right|^{2}=0$
$\Rightarrow 4\left(\left|z_{2}\right|^{2}-1\right)-\left|z_{1}\right|^{2}\left(\left|z_{2}\right|^{2}-1\right)=0$
$\Rightarrow\left|z_{1}\right|^{2}-4=0 \Rightarrow\left|z_{1}\right|=2$ is a circle of radius 2 and centre at origin.
Alter
$\frac{\left|z_{1}-2 z_{2}\right|}{\left|2-z_{1} \bar{z}_{2}\right|}=1$
$\left(z_{1}-2 z_{2}\right)\left(\bar{z}_{1}-2 \bar{z}_{2}\right)=\left(2-z_{1} \bar{z}_{2}\right)\left(2-\bar{z}_{1} z_{2}\right)$
$\left|z_{1}\right|^{2}-2 z_{1} \bar{z}_{2}-2 z_{2} \bar{z}_{1}+4\left|z_{2}\right|^{2}$
$=4-2 z_{1} \bar{z}_{2}-2 z_{1} \bar{z}_{2}+\left|z_{1}\right|^{2}\left|z_{2}\right|^{2}$
$\left|z_{2}\right|^{2}\left(1-\left|z_{2}\right|^{2}\right)-4\left(1-\left|z_{2}\right|^{2}\right)=0$
$\left.\Rightarrow\left|z_{1}\right|=2 \quad \text { (as }\left|z_{2}\right| \neq 1\right)$
$\Rightarrow$ which is circle of radius 2
Q. A value of $\theta$ for which $\frac{2+3 \text { isin } \theta}{1-2 i \sin \theta}$ is purely imaginary, is :
(1) $\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
(2) $\frac{\pi}{3}$
(3) $\frac{\pi}{6}$
(4) $\sin ^{-1}\left(\frac{\sqrt{3}}{4}\right)$
[JEE (Main)-2016]
Ans. (1)
$\begin{aligned} \mathrm{Z}=& \frac{2+3 \mathrm{i} \sin \theta}{1-2 \mathrm{i} \sin \theta} \\ \Rightarrow \mathrm{Z} &=\frac{(2+3 \mathrm{i} \sin \theta)(1+2 \mathrm{i} \sin \theta)}{1+4 \sin ^{2} \theta} \\ &=\frac{\left(2-6 \sin ^{2} \theta\right)+7 \mathrm{i} \sin \theta}{1+4 \sin ^{2} \theta} \end{aligned}$
for purely imaginary $Z, \operatorname{Re}(Z)=0$
$\Rightarrow 2-6 \sin ^{2} \theta=0 \Rightarrow \sin \theta=\pm \frac{1}{\sqrt{3}}$
$\Rightarrow \theta=\pm \sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
Q. Let $\omega$ be a complex number such that $2 \omega+1=z$ where $z=\sqrt{-3} .$ If $\left|\begin{array}{ccc}{1} & {1} & {1} \\ {1} & {-\omega^{2}-1} & {\omega^{2}} \\ {1} & {\omega^{2}} & {\omega^{7}}\end{array}\right|=3 \mathrm{k},$ then $\mathrm{k}$ is equal to :
(1) 1 (2) –z (3) z (4) –1
[JEE (Main)-2017]
Ans. (2)
Here $\omega$ is complex cube root of unity
$\quad \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3}$
$=\left|\begin{array}{ccc}{3} & {0} & {0} \\ {1} & {-\omega^{2}-1} & {\omega^{2}} \\ {1} & {\omega^{2}} & {\omega}\end{array}\right|=3(-1-\omega-\omega)=-3 \mathrm{z} \Rightarrow \mathrm{k}=-\mathrm{z}$
Comments
Kritika
Nov. 11, 2020, 8:28 p.m.
Please provide more question for jee mains & jee advanced level