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If \( \alpha \) is a root of the equation \( x^2 + x + 1 = 0 \) and \( \displaystyle \sum_{k=1}^{n}\left(\alpha^k+\frac{1}{\alpha^k}\right)^2 = 20 \), then \( n \) is equal to ________________.

Two number \( k_1 \) and \( k_2 \) are randomly chosen from the set of natural numbers. Then, the probability that the value of \( i^{k_1} + i^{k_2} \) \( (i = \sqrt{-1}) \) is non-zero, equals:

If \( \alpha+i\beta \) and \( \gamma+i\delta \) are the roots of \( x^{2}-(3-2i)x-(2i-2)=0 \), \( i=\sqrt{-1} \), then \( \alpha\gamma+\beta\delta \) is equal to:

Let integers \( a,b\in[-3,3] \) be such that \( a+b\ne 0 \). Then the number of all possible ordered pairs \( (a,b) \), for which \( \left|\dfrac{z-a}{\,z+b\,}\right|=1 \) and , where \( \omega \) and \( \omega^2 \) are the roots of \( x^2+x+1=0 \), is equal to ________.

Let \( z_1, z_2 \) and \( z_3 \) be three complex numbers on the circle \( |z| = 1 \) with \( \arg(z_1) = -\pi/4 \), \( \arg(z_2) = 0 \) and \( \arg(z_3) = \pi/4 \).
If \( \left| z_1\overline{z_2} + z_2\overline{z_3} + z_3\overline{z_1} \right|^{2} = \alpha + \beta \sqrt{2}, \alpha, \beta \in \mathbb{Z} \), then the value of \( \alpha^{2} + \beta^{2} \) is:

Let the product of \( \omega_1 = (8 + i) \sin \theta + (7 + 4i)\cos \theta \) and \( \omega_2 = (i + 8i)\sin \theta + (4 + 7i)\cos \theta \) be \( \alpha + i\beta \), \( i = \sqrt{-1} \). Let \( p \) and \( q \) be the maximum and the minimum values of \( \alpha + \beta \) respectively. Then \( p + q \) is equal to

Let \( |z_1 - 8 - 2i| \le 1 \) and \( |z_2 - 2 + 6i| \le 2 \), \( z_1, z_2 \in \mathbb{C} \). Then the minimum value of \( |z_1 - z_2| \) is:

Let \( \alpha, \beta \) be the roots of the equation \( x^2 - ax - b = 0 \) with \( \operatorname{Im}(\alpha) < \operatorname{Im}(\beta) \). Let \( P_n = \alpha^n - \beta^n \). If \( P_3 = -5\sqrt{7}\,i \), \( P_4 = -3\sqrt{7}\,i \), \( P_5 = 11\sqrt{7}\,i \) and \( P_6 = 45\sqrt{7}\,i \), then \( |\alpha^4 + \beta^4| \) is equal to __________.

Let \( O \) be the origin, the point \( A \) be \( z_1 = \sqrt{3} + 2\sqrt{2}\,i \), the point \( B(z_2) \) be such that \( \sqrt{3}\,|z_2| = |z_1| \) and \( \arg(z_2) = \arg(z_1) + \frac{\pi}{6} \). Then:

Let \( A = \left\{ \theta \in [0, 2\pi] : 1 + 10\,Re\left( \frac{2\cos\theta + i\sin\theta}{\cos\theta - 3i\sin\theta} \right) = 0 \right\}. \) Then \( \displaystyle \sum_{\theta \in A} \theta^{2} \) is equal to:

Among the statements

(\( \mathrm{S1} \)) : The set \( \left\{ z \in \mathbb{C} - \{-i\} : |z| = 1 \text{ and } \frac{z-i}{z+i} \text{ is purely real} \right\} \) contains exactly two elements, and

(\( \mathrm{S2} \)) : The set \( \left\{ z \in \mathbb{C} - \{-1\} : |z| = 1 \text{ and } \frac{z-1}{z+1} \text{ is purely imaginary} \right\} \) contains infinitely many elements.

If \( \alpha \) and \( \beta \) are the roots of the equation \( 2z^2 - 3z - 2i = 0 \), where \( i = \sqrt{-1} \), then \( 16\,. \mathrm{Re}\!\left( \frac{\alpha^{19}+\beta^{19} + \alpha^{11} + \beta^{11}}{\alpha^{15} + \beta^{15}} \right) \cdot \mathrm{Im}\!\left( \frac{\alpha^{19}+\beta^{19} + \alpha^{11} + \beta^{11}}{\alpha^{15} + \beta^{15}} \right) \) is equal to:

The number of complex numbers \( z \), satisfying \( |z| = 1 \) and \(\displaystyle \left| \frac{z}{\overline{z}} + \frac{\overline{z}}{z} \right| = 1 \), is:

If \( z_1, z_2, z_3 \in \mathbb{C} \) are the vertices of an equilateral triangle, whose centroid is \( z_0 \) then \( \displaystyle\sum_{k=1}^{3}\left(z_k - z_0\right)^2 \) is equal to:

Let $\left| \frac{\bar{z} - i}{2\bar{z} + i} \right| = \frac{1}{3}, z \in C$, be the equation of a circle with center at $C$. If the area of the triangle, whose vertices are at the points $(0, 0)$, $C$ and $(\alpha, 0)$ is $11$ square units, then $\alpha^2$ equals:

Let \( z \) be a complex number such that \( |z| = 1 \). If \( \dfrac{2 + k^2 {z}}{k + \bar{z}} = kz, \; k \in \mathbb{R} \), then the maximum distance of \( k + i k^2 \) from the circle \( |z - (1 + 2i)| = 1 \) is:

If the locus of \( z\in \mathbb{C} \), such that \( Re\!\left(\dfrac{z-1}{2z+i}\right)+Re\!\left(\dfrac{\overline{z}-1}{2\overline z-i}\right)=2 \), is a circle of radius \( r \) and centre \( (a,b) \), then \( \dfrac{15ab}{r^2} \) is equal to:

Let the curve \( z(1+i)+\bar{z}(1-i)=4 , \; z \in \mathbb{C} \), divide the region \( |z-3|\le 1 \) into two parts of areas \( \alpha \) and \( \beta \). Then \( |\alpha-\beta| \) equals:

Let \( z \in \mathbb{C} \) be such that \( \frac{z^2 + 3i}{z - 2 + i} = 2 + 3i \). Then the sum of all possible values of \( z^2 \) is:

← PreviousQuestion 19 of 19

JEE Main 2025 3rd April Morning Shift

Let \( z \in \mathbb{C} \) be such that \( \frac{z^2 + 3i}{z - 2 + i} = 2 + 3i \). Then the sum of all possible values of \( z^2 \) is: