If z₁, z₂ are two complex numbers satisfying $\left|\frac{{\mathrm{z}}_1 - 3{\mathrm{z}}_2}{3 - {\mathrm{z}}_1\overline{{\mathrm{z}}_2}}\right| = 1, \left|{\mathrm{z}}_1\right| \ne 3$, then $\left|{\mathrm{z}}_2\right|$ is equal to

• 1

• 2

• 3

• 4

The value of $\sum _{\mathrm{n} = 0}^{\infty }{\left(\frac{2\mathrm{i}}{3}\right)}^{\mathrm{n}}$ is

• $\frac{9 + 6\mathrm{i}}{13}$

• $\frac{9 - 6\mathrm{i}}{13}$

• 9 + 6i

• 9 - 6i

The roots of the equation x - 3x - 2 = 0 are

• - 1, - 1, 2

• - 1, 1, - 2

• - 1, 2, - 3

• - 1, - 1, - 2

$\mathrm{If} \mathrm{\alpha }, \mathrm{\beta }, \mathrm{\gamma } \mathrm{are} \mathrm{the} \mathrm{roots} \mathrm{of} {\mathrm{x}}^3 + 2{\mathrm{x}}^2 - 3\mathrm{x} - 1 = 0 \mathrm{then} {\mathrm{\alpha }}^{- 2} + {\mathrm{\beta }}^{- 2} + {\mathrm{\gamma }}^{- 2} =$

• 12

• 13

• 14

• 15

215.

If ${\mathrm{\alpha }}_1, {\mathrm{\alpha }}_2, {\mathrm{\alpha }}_3$ respectively denote the moduli of the complex number - i, $\frac{1}{3}$(1 + i) and - 1 + i, 3 then their increasing order is

• ${\mathrm{\alpha }}_1, {\mathrm{\alpha }}_2, {\mathrm{\alpha }}_3$

• ${\mathrm{\alpha }}_3, {\mathrm{\alpha }}_2, {\mathrm{\alpha }}_1$

• ${\mathrm{\alpha }}_2, {\mathrm{\alpha }}_1, {\mathrm{\alpha }}_3$

• ${\mathrm{\alpha }}_3, {\mathrm{\alpha }}_1, {\mathrm{\alpha }}_2$

$\mathrm{If} \mathrm{\alpha } \mathrm{is} \mathrm{a} \mathrm{non}-\mathrm{real} \mathrm{root} \mathrm{of} {\mathrm{x}}^6 = 1, \mathrm{then} \frac{{\mathrm{\alpha }}^5 + {\mathrm{\alpha }}^{3 } + \mathrm{\alpha } + 1}{{\mathrm{\alpha }}^2 + 1} \mathrm{is} \mathrm{equal} \mathrm{to},$

• ${\mathrm{\alpha }}^2$

• 0

• $- {\mathrm{\alpha }}^2$

• $\mathrm{\alpha }$

The difference between two roots of the equation x³ - 13x² + 15x + 189 = 0 is 2. Then  the roots of the equation are :

• - 3, 5, 7

• - 3, - 7, - 9

• 3, - 5, 7

• - 3, - 7, 9

$$\begin{gathered} \mathrm{If} \mathrm{\alpha }, \mathrm{\beta }, \mathrm{\gamma } \mathrm{are} \mathrm{the} \mathrm{roots} \mathrm{of} \mathrm{the} \mathrm{equation} {\mathrm{x}}^3 - 6{\mathrm{x}}^2 + 11\mathrm{x} + 6 = 0, \\ \mathrm{then} \sum _{}{\mathrm{\alpha }}^2\mathrm{\beta } + \sum _{}{\mathrm{\alpha \beta }}^2 \mathrm{is} \mathrm{equal} \mathrm{to}\hspace{0.17em}: \end{gathered}$$

• 80

• 84

• 90

• - 84

The locus of the point z = x + iy satisfying the equation

$\left|\frac{\mathrm{z} - 1}{\mathrm{z} + 1}\right| = 1 \mathrm{is} \mathrm{given} \mathrm{by}$ :

• x = 0

• y = 0

• x = y

• x + y = 0

The equation of the locus of z such that $\left|\frac{\mathrm{z} + \mathrm{i}}{\mathrm{z} - \mathrm{i}}\right| = 2$, where z= x + iy is a complex number, is

• $3{\mathrm{x}}^2 + 3{\mathrm{y}}^2 + 10\mathrm{y} - 3 = 0$

• $3{\mathrm{x}}^2 + 3{\mathrm{y}}^2 + 10\mathrm{y} + 3 = 0$

• $3{\mathrm{x}}^2 - 3{\mathrm{y}}^2 - 10\mathrm{y} - 3 = 0$

• ${\mathrm{x}}^2 + {\mathrm{y}}^2 - 5\mathrm{y} + 3 = 0$