Question

if z-1/z-i is purely imaginary, find the locus of z​

Answer 1

Appropriate Question :-

• If z is a complex number and if (z - 1)/(z + i) is purely imaginary, find the locus of z.

Basic Concept Used :-

Purely Imaginary :-

• A complex number z = x + iy is purely imaginary iff Real part of z = 0.

Solution :-

Since, z is a complex number,

So,

• Let z = x + iy

Consider,

$\rm :\longmapsto\:\dfrac{z - 1}{z - i}$

$\rm \: = \: \:\dfrac{x + iy - 1}{x + iy - i}$

$\rm \: = \: \:\dfrac{x - 1 + iy }{x + i(y - 1)}$

$\rm \: = \: \:\dfrac{x - 1 + iy }{x + i(y - 1)} \times \dfrac{x - i(y - 1)}{x - i(y - 1)}$

$\rm \: = \: \:\dfrac{ {x}^{2} - x + iyx - i(y - 1)(x - 1) - {i}^{2}y(y - 1)}{ {x}^{2} - {i}^{2} {(y - 1)}^{2} }$

$\rm \: = \: \:\dfrac{ {x}^{2} - x + iyx - i(xy - x - y + 1) + y(y - 1)}{ {x}^{2} + {(y - 1)}^{2} }$

$\rm \: = \: \:\dfrac{ {x}^{2} - x + {y}^{2} - y + i(yx - xy + x + y - 1)}{ {x}^{2} + {(y - 1)}^{2} }$

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

Since,

$\rm :\longmapsto\:\dfrac{z - 1}{z - i} \: is \: purely \: imaginary.$

$\bf\implies \:Re\bigg(\dfrac{z - 1}{z - i} \bigg) = 0$

$\rm :\implies\:\dfrac{ {x}^{2}+{y}^{2} - x - y}{ {x}^{2}+{(y - 1)}^{2}} = 0$

$\bf\implies \: {x}^{2} + {y}^{2} - x - y = 0$

$\bf\implies \:Locus \: is \: a \: circle \: whose \: centre \: is \: \bigg(\dfrac{1}{2} ,\dfrac{1}{2} \bigg)$

and

$\rm :\longmapsto\:radius \: = \sqrt{ {\bigg(\dfrac{1}{2}\bigg) }^{2} + {\bigg(\dfrac{1}{2}\bigg) }^{2} }$

$\rm \: = \: \: \sqrt{\dfrac{1}{4} + \dfrac{1}{4} }$

$\rm \: = \: \: \sqrt{\dfrac{1 + 1}{4} }$

$\rm \: = \: \: \sqrt{\dfrac{2}{4} }$

$\rm \: = \: \: \sqrt{\dfrac{1}{2}}$

$\rm \: = \: \:\dfrac{1}{ \sqrt{2} }$