Question

If $z=x+iy$and $|2z+1|=|z-3|$ , then show that $3x^2+3y^2+10x=8$


please help..moderators​

Answer 1

Given :-

$\rm :\longmapsto\:z = x + iy$

$\rm :\longmapsto\: |2z + 1| = |z - 3|$

To Show :-

$\rm :\longmapsto\: {3x}^{2} + {3y}^{2} + 10x = 8$

Basic Concept Used :-

Modulus of a complex number

Let us consider a complex number z = x + iy, then Modulus of z is represented as |z| and is given by

$\rm :\longmapsto\: |z| = \sqrt{ {x}^{2} + {y}^{2} }$

Solution :-

Given that

$\rm :\longmapsto\:z = x + iy$

and

$\rm :\longmapsto\: |2z + 1| = |z - 3|$

On substituting the value of z = x + iy, we get

$\rm :\longmapsto\: |2(x + iy) + 1| = |x + iy - 3|$

$\rm :\longmapsto\: |2x + 2iy + 1| = |x + iy - 3|$

$\rm :\longmapsto\: |(2x + 1)+ 2iy| = |(x - 3) + iy|$

$\rm :\longmapsto\: \sqrt{ {(2x + 1)}^{2} + {(2y)}^{2} } = \sqrt{ {(x - 3)}^{2} + {y}^{2} }$

On squaring both sides, we get

$\rm :\longmapsto\: {(2x + 1)}^{2} + {(2y)}^{2} = {(x - 3)}^{2} + {y}^{2}$

$\rm :\longmapsto\: {4x}^{2} + 1 + 4x + {4y}^{2} = {x}^{2} + 9 - 6x + {y}^{2}$

$\bf\implies \: {3x}^{2} + {3y}^{2} - 10x = 8$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

If z is a complex number, then

$\rm :\longmapsto\: |z| = | \overline{z}|$

$\rm :\longmapsto\: |z \overline{z}| = { |z| }^{2}$

$\rm :\longmapsto\: |z_1z_2| = |z_1| |z_2|$

$\rm :\longmapsto\:\overline{z_1 + z_2} \: = \: \overline{z_1} + \overline{z_2}$

$\rm :\longmapsto\:\overline{z_1 - z_2} \: = \: \overline{z_1} - \overline{z_2}$

$\rm :\longmapsto\:\overline{z_1z_2} \: = \: \overline{z_1} \: \: \: \overline{z_2}$