JEE Main Complex Numbers Practice Questions With Solutions

Let $z$z be a complex number such that the real part of $\frac{z-2i}{z+2i}$\frac{z - 2 i}{z + 2 i} is zero. Then, the maximum value of $|z-(6+8i)|$\left|\right. z - \left(\right. 6 + 8 i \left.\right) \left|\right. is equal to

The sum of all possible values of $\theta \in [-\pi ,2\pi ]$\theta \in \left[\right. - \pi , 2 \pi \left]\right., for which $\frac{1+i\cos \theta }{1-2i\cos \theta }$\frac{1 + i cos ⁡ \theta}{1 - 2 i cos ⁡ \theta} is purely imaginary, is equal to :

Let $z$z be a complex number such that $|z+2|=1$\left|\right. z + 2 \left|\right. = 1 and $\mathrm{lm}\left(\frac{z+1}{z+2}\right)=\frac{1}{5}$lm \left(\right. \frac{z + 1}{z + 2} \left.\right) = \frac{1}{5}. Then the value of $|\mathrm{Re}(\overline{z+2})|$\left|\right. Re \left(\right. \overset{―}{z + 2} \left.\right) \left|\right. is

If the set $R=\{(a,b):a+5b=42,a,b\in \mathbb{N}\}$R = \left{\right. \left(\right. a , b \left.\right) : a + 5 b = 42 , a , b \in \mathbb{N} \left.\right} has $m$m elements and $\sum _{n=1}^m(1-i^{n!})=x+iy$\sum_{n = 1}^{m} \left(\right. 1 - i^{n !} \left.\right) = x + i y, where $i=\sqrt{-1}$i = \sqrt{- 1}, then the value of $m+x+y$m + x + y is

If $z_1,z_2$z_{1} , z_{2} are two distinct complex number such that $\left|\frac{z_1-2z_2}{\frac{1}{2}-z_1{\overline{z}}_2}\right|=2$\left|\right. \frac{z_{1} - 2 z_{2}}{\frac{1}{2} - z_{1} \bar{z}_{2}} \left|\right. = 2, then

Let $S_1=\{z\in \mathbf{C}:|z|\le 5\},S_2=\{z\in \mathbf{C}:\mathrm{Im}\left(\frac{z+1-\sqrt{3}i}{1-\sqrt{3}i}\right)\ge 0\}$S_{1} = \left{\right. z \in \mathbf{C} : \left|\right. z \left|\right. \leq 5 \left.\right} , S_{2} = \left{\right. z \in \mathbf{C} : Im \left(\right. \frac{z + 1 - \sqrt{3} i}{1 - \sqrt{3} i} \left.\right) \geq 0 \left.\right} and $S_3=\{z\in \mathbf{C}:\mathrm{Re}(z)\ge 0\}$S_{3} = \left{\right. z \in \mathbf{C} : Re \left(\right. z \left.\right) \geq 0 \left.\right}. Then the area of the region $S_1\cap S_2\cap S_3$S_{1} \cap S_{2} \cap S_{3} is :

Consider the following two statements :

Statement I: For any two non-zero complex numbers $z_1,z_2,(|z_1|+|z_2|)|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}|\le 2(\left|z_1\right|+\left|z_2\right|)\text{, and }$z_{1} , z_{2} , \left(\right. \left|\right. z_{1} \left|\right. + \left|\right. z_{2} \left|\right. \left.\right) \left|\right. \frac{z_{1}}{\left|\right. z_{1} \left|\right.} + \frac{z_{2}}{\left|\right. z_{2} \left|\right.} \left|\right. \leq 2 \left(\right. \left|\right. z_{1} \left|\right. + \left|\right. z_{2} \left|\right. \left.\right) ,\text{ and}\textrm{ }

Statement II : If $x,y,z$x , y , z are three distinct complex numbers and $\mathrm{a},\mathrm{b},\mathrm{c}$a , b , c are three positive real numbers such that $\frac{\mathrm{a}}{|y-z|}=\frac{\mathrm{b}}{|z-x|}=\frac{\mathrm{c}}{|x-y|}$\frac{a}{\left|\right. y - z \left|\right.} = \frac{b}{\left|\right. z - x \left|\right.} = \frac{c}{\left|\right. x - y \left|\right.}, then $\frac{{\mathrm{a}}^2}{y-z}+\frac{{\mathrm{b}}^2}{z-x}+\frac{{\mathrm{c}}^2}{x-y}=1$\frac{a^{2}}{y - z} + \frac{b^{2}}{z - x} + \frac{c^{2}}{x - y} = 1.

Between the above two statements,

The area (in sq. units) of the region $S=\{z\in \mathbb{C}:|z-1|\le 2;(z+\overline{z})+i(z-\overline{z})\le 2,\mathrm{lm}(z)\ge 0\}$S = \left{\right. z \in \mathbb{C} : \left|\right. z - 1 \left|\right. \leq 2 ; \left(\right. z + \bar{z} \left.\right) + i \left(\right. z - \bar{z} \left.\right) \leq 2 , lm \left(\right. z \left.\right) \geq 0 \left.\right} is

Let $\alpha$\alpha and $\beta$\beta be the sum and the product of all the non-zero solutions of the equation $(\overline{z}{)}^2+|z|=0,z\in C$\left(\right. \bar{z} \left.\right)^{2} + \left|\right. z \left|\right. = 0 , z \in C. Then $4({\alpha }^2+{\beta }^2)$4 \left(\right. \alpha^{2} + \beta^{2} \left.\right) is equal to :