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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 Re $z$ =1, then it is necessary that

Option 1)
$\left | \beta \right |=1\;$

Option 2)
$\; \; \beta \in (1,\infty )\;$

Option 3)
$\; \beta \in (0,1)\;$

Option 4)
$\; \beta \in (-1,0)$

Answers (2)

best_answer

As we learnt in

Definition of Complex Number -

$z=x+iy, x,y\epsilon R$ & i²=-1

- wherein

Real part of z = Re (z) = x & Imaginary part of z = Im (z) = y

$z^{2}+\alpha z+\beta=0$

Let z = 1 + iy

So that $(1+iy)^{2}+\alpha(1+iy)+\beta=0$

$\Rightarrow\ \;1-y^{2}+i2y+\alpha+i\alpha y+\beta=0$

$\therefore\ \;(1-y^{2}+\alpha + \beta)+i(2+\alpha)y=0$

$\therefore\ \; \alpha=-2$ and $1-y^{2}-2+\beta=0$

$\Rightarrow\ \;y^{2}=\beta-1>0$

$\therefore\ \; \beta-1>0$

$\beta \epsilon(1,\; \infty)$

Correct option is 2.

Option 1)

$\left | \beta \right |=1\;$

This an incorrect option.

Option 2)

$\; \; \beta \in (1,\infty )\;$

This the correct option.

Option 3)

$\; \beta \in (0,1)\;$

This an incorrect option.

Option 4)

$\; \beta \in (-1,0)$

This an incorrect option.

Posted by

prateek

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