Complex Number – JEE Main Previous Year Question with Solutions
JEE Main Previous Year Question of Math with Solutions are available at eSaral. Practicing JEE Main Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.Get detailed Class 11th &12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.Download eSaral app for free study material and video tutorials.
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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (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$$\Rightarrow \beta \in(1, \infty) Q. If \omega(\neq 1) is a cube root of unity, and (1+\omega)^{7}=\mathrm{A}+\mathrm{B} \omega . Then (\mathrm{A}, \mathrm{B}) equals :-(1) (1, 0) (2) (–1, 1) (3) (0, 1) (4) (1, 1) [AIEEE -2011] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (4)(1+\omega)^{7}=\mathrm{A}+\mathrm{B} \omega$$\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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (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] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (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}}$$\Rightarrow \theta=\pm \sin ^{-1}\left(\frac{1}{\sqrt{3}}\right) Q. Let \omega be a complex number such that 2 \omega+1=z where z=\sqrt{-3} . If \left|\begin{array}{ccc}{1} & {1} & {1} \\ {1} & {-\omega^{2}-1} & {\omega^{2}} \\ {1} & {\omega^{2}} & {\omega^{7}}\end{array}\right|=3 \mathrm{k}, then \mathrm{k} is equal to :(1) 1 (2) –z (3) z (4) –1 [JEE (Main)-2017] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (2)Here \omega is complex cube root of unity\quad \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3}$$=\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}

• March 6, 2021 at 1:12 pm

.

• February 25, 2021 at 10:34 am

sir send some more question from previous year jee main

• December 23, 2020 at 12:22 am

Thanks

• December 15, 2020 at 5:17 pm

sdfsdfsdf

• December 15, 2020 at 5:18 pm

wfdsdfsd

• December 26, 2020 at 4:01 pm

Skskskksks and I oop

• November 11, 2020 at 8:28 pm

• October 22, 2020 at 9:08 pm

Hahahaha

• October 14, 2020 at 7:59 pm

Nice

• September 14, 2020 at 6:24 pm

Thanks you

• September 11, 2020 at 10:17 pm

Goood one

• August 28, 2020 at 6:14 pm

Sir send some more questions regarding complex numbers

• February 25, 2021 at 10:35 am

yes

• August 27, 2020 at 8:33 pm

thank u but plz update the questions till2020

• August 25, 2020 at 9:30 am

Thanks you sir

• August 6, 2020 at 12:04 pm

Thanku so much

• August 4, 2020 at 9:48 pm

Excellent approach…

• July 25, 2020 at 9:24 pm

OK

• July 25, 2020 at 9:23 pm

good

• June 20, 2020 at 10:47 am

Update all questions of jee till 2020

• June 20, 2020 at 10:46 am

Good

• June 11, 2020 at 2:41 pm