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]

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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]

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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]

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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

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• February 25, 2021 at 10:34 am

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• December 23, 2020 at 12:22 am

Thanks

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2
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• November 11, 2020 at 8:28 pm

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• November 17, 2020 at 10:17 am

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• October 22, 2020 at 9:08 pm

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1
• October 14, 2020 at 7:59 pm

Nice

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• September 14, 2020 at 6:24 pm

Thanks you

7
• September 11, 2020 at 10:17 pm

Goood one

0
• August 28, 2020 at 6:14 pm

Sir send some more questions regarding complex numbers

15
• February 25, 2021 at 10:35 am

yes

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• August 27, 2020 at 8:33 pm

thank u but plz update the questions till2020

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• August 25, 2020 at 9:30 am

Thanks you sir

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• August 6, 2020 at 12:04 pm

Thanku so much

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• August 4, 2020 at 9:48 pm

Excellent approach…

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• July 25, 2020 at 9:24 pm

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2
• July 25, 2020 at 9:23 pm

good

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• June 20, 2020 at 10:47 am

Update all questions of jee till 2020

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• June 20, 2020 at 10:46 am

Good

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• June 11, 2020 at 2:41 pm

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• January 19, 2021 at 7:31 pm

Hi

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• April 19, 2020 at 2:34 pm

Best questions

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• April 10, 2020 at 5:01 pm

Thank you

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