Question

$\tt{ \underline{ \underline\bold{For \: IIT \: Aisprants}}} \\ \\ \tt{let \: \alpha \: and \: \beta \: be \: real. \: z \: be \: a \: complex \: number.} \\ \tt{if \: {z}^{2} + \alpha \: z + \beta = 0 \: has \: two \: distinct \: roots} \\ \tt{on \: the \: line \: re \: z = 1 \: then \: \beta \: belongs \: to....}$
​

Answer 1

$\red {\huge {\underline{ \frak{Your \: answEr : }}}}$

$\blue{\huge \implies \boxed{\beta \in(1 \: ,∞)}}$

$\red {\huge {\underline{ \frak{Explanation : }}}}$

$\green{\large{ \underline{\mathcal{\star \:Given : }}}}$

$\tt{\alpha \: \beta \: are \: real \: number} \\ z \: \tt{is \: a \: complex \: number}$

${z}^{2} + \alpha z + \beta = 0 \: \tt{has \: two \: distinct} \\ \tt{roots \: on} \: \: \mathcal{R_ez = 1}$

$\green{\large{ \underline{\mathcal{\star \: To \: Find : }}}}$

$\tt{value \: of \: \beta }$

$\green{\large{ \underline{\mathcal{\star \: Solution : }}}}$

$\blacksquare \: \underline\frak{Quick \: Facts}$

• If the roots are ax² + bx + c = 0 are Distinct and Complex, then Discriminent D < 0

• D = b² - 4ac

$\blacksquare \: \underline\frak{Now \: head \: to \: the \: Question}$

★ Roots of the Equation ;

$\huge z = \frac{ - \alpha ± \sqrt{ { \alpha }^{2} - 4 \beta } }{2}$

★ Since z is Complex ;

$\huge D = { \alpha }^{2} - 4 \beta < 0$

★ Also Given that ;

$\huge R_ez = 1 \: , \: \frac{ - \alpha }{2} = 1$

$\large\implies \alpha = - 2$

$\large \therefore 4 - 4 \beta < 0 \\ \: \: \: \: \: \beta > 1$

$\purple{\large \implies \boxed{\beta \in(1 \: ,∞)}}$

$\huge{\red{\ddot{\smile}}}$

Answer 2

Answer:

Step-by-step explanation:

z²+αz+β = 0

$z_1$+ $z_2$= -α

$z_1$  $z_2$= β

x=1

$z_1$ =1+ iy

$z_2$ =1− iy

now, 1²+y²=β

β∈(1,∞)