Complex Number – JEE Main Previous Year Question with Solutions
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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]

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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]

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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]

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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]

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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]

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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]

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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]

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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

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• December 15, 2020 at 5:18 pm

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• December 26, 2020 at 4:01 pm

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• November 11, 2020 at 8:28 pm

Please provide more question for jee mains & jee advanced level

• 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