If |z - 6|

Question

If |z - 6| < |z - 2|, its solution is given by (A) Re(z) > 0 (B) Re(z) < 0 (C) Re(z) > 4 (D) Re(z) < 4

Tardigrade Admin · Accepted Answer

Let z = a + ib Now, z - 6 = (a - 6) + ib and z - 2 = (a - 2) + ib ∴ | z- 6 | < | z -2| ⇒ (a - 6)2 + b2 < (a - 2)2 + b2 ⇒ (a -6)2 < (a - 2)2 ⇒ - 8a < - 36 + 4 ⇒ - 8a < - 32 ⇒ a > 4 ∴ Re(z) > 4.

Q. If $∣ z - 6∣ < ∣ z - 2∣$ , its solution is given by

$R e (z) > 0$

$R e (z) < 0$

$R e (z) > 4$

$R e (z) < 4$

Let $z = a + ib$
Now, $z - 6 = (a - 6) + ib$
and $z - 2 = (a - 2) + ib$
$∴ ∣ z - 6∣ < ∣ z - 2∣$
$\Rightarrow (a - 6)^{2} + b^{2} < (a - 2)^{2} + b^{2}$
$\Rightarrow (a - 6)^{2} < (a - 2)^{2}$
$\Rightarrow - 8 a < - 36 + 4$
$\Rightarrow - 8 a < - 32$
$\Rightarrow a > 4$
$∴ R e (z) > 4$ .

Q. If ∣z−6∣<∣z−2∣, its solution is given by

Re(z)>0

Re(z)<0

Re(z)>4

Re(z)<4