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]

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]

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]

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]

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]

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]

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]

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 :

(2) circle of radius $\sqrt{2}$

(3) straight line parallel to x-axis

(4) straight line parallel to y-axis

[JEE (Main)-2015]

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]

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]

Sol. (2)

Here $\omega$ is complex cube root of unity

$\quad \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3}$

$=\left|\begin{array}{ccc}{3} & {0} & {0} \\ {1} & {-\omega^{2}-1} & {\omega^{2}} \\ {1} & {\omega^{2}} & {\omega}\end{array}\right|=3(-1-\omega-\omega)=-3 \mathrm{z} \Rightarrow \mathrm{k}=-\mathrm{z}$