Let $$z=x+iy$$

$$\left| {z - 1} \right| = \left| {z + 1} \right|{\left( {x - 1} \right)^2} + {y^2}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {x + 1} \right)^2} + {y^2}$$

$$ \Rightarrow {\mathop{\rm Re}\nolimits} \,z = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow x = 0$$

$$\left| {z - 1} \right| = \left| {z - i} \right|{\left( {x - 1} \right)^2} + {y^2}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {x^2} + {\left( {y - 1} \right)^2}$$

$$ \Rightarrow x = y$$

$$\left| {z + 1} \right| = \left| {z - i} \right|{\left( {x + 1} \right)^2} + {y^2}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {x^2} + {\left( {y - 1} \right)^2}$$

Only $$(0,0)$$ will satisfy all conditions.

$$ \Rightarrow $$ Number of complex number $$z=1$$

Given $$S = \left\{ {\left( {x,y} \right):y = x + 1\,\,} \right.\,$$

and $$\,\,\,\left. {0 < x < 2} \right\}$$

As $$\,\,\,\,x \ne x + 1\,\,\,$$

for any $$\,\,\,x \in \left( {0,2} \right) \Rightarrow \left( {x,x} \right) \notin S$$

$$\therefore$$ $$S$$ is not reflexive.

Hence $$S$$ in not an equivalence relation.

Also $$\,\,\,T = \left\{ {x,\left. y \right)} \right.:x - y$$ is an integer $$\left. {} \right\}$$

as $$x - x = 0$$ is an integer $$\forall x \in R$$

$$\therefore$$ $$T$$ is reflexive.

If $$x-y$$ is an integer then $$y-x$$ is also an integer

$$\therefore$$ $$T$$ is symmetric

If $$x-y$$ is an integer and $$y - z$$ is an integer then

$$(x-y)+(y-z)=x-z$$ is also an integer.

$$\therefore$$ $$T$$ is transitive

Hence $$T$$ is an equivalence relation

$$\left( {{1 \over {i - 1}}} \right) = {1 \over { - i - 1}} = {{ - 1} \over {i + 1}}$$

$$z$$ lies on or inside the circle with center $$(-4,0)$$ and radius $$3$$ units.



From the Argand diagram maximum value of $$\left| {z + 1} \right|$$ is $$6$$