For a complex number $$z = x + iy$$, it's conjugate $$\overline z = x - iy$$. Now z is purely real when $$y = 0$$.

When $$y = 0$$ then $$z = x + i \times (0) = x$$ and $$\overline z = x - i \times (0) = x$$

$$\therefore$$ $$z = \overline z $$ when z is purely real.

Now given, $$w = {{2 + 3z + 4{z^2}} \over {2 - 3z + 4{z^2}}}$$ is real

$$\therefore$$ $$w = \overline w $$

$$ \Rightarrow {{2 + 3z + 4{z^2}} \over {2 - 3z + 4{z^2}}} = \left( {\overline {{{2 + 3z + 4{z^2}} \over {2 - 3z + 4{z^2}}}} } \right)$$

$$ \Rightarrow {{2 + 3z + 4{z^2}} \over {2 - 3z + 4{z^2}}} = {{2 + 3\overline z + 4{{(\overline z )}^2}} \over {2 - 3\overline z + 4{{(\overline z )}^2}}}$$

$$ \Rightarrow 4 - 6\overline z + 8{(\overline z )^2} + 6z - 9z\overline z + 12z{(\overline z )^2} + 8{(\overline z )^2} - 12{z^2}\overline z + 16{z^2}{(\overline z )^2} = 4 + 6\overline z + 8{(\overline z )^2} - 6z - 9z\overline z - 12z{(\overline z )^2} + 8{z^2} + 12{z^2}\overline z + 16{z^2}{(\overline z )^2}$$

$$ \Rightarrow - 6\overline z + 6z + 12z{(\overline z )^2} + 12{z^2}\overline z = 6\overline z - 6z - 12z{(\overline z )^2} + 12{z^2}\overline z $$

$$ \Rightarrow 6(z - \overline z ) + 12z\overline z (\overline z - z) = 6(\overline z - z) + 12z\overline z (z - \overline z )$$

$$ \Rightarrow 12(z - \overline z ) + 24(\overline z - z)(z\overline z ) = 0$$

$$ \Rightarrow 12(z - \overline z )[1 - 2z\overline z ] = 0$$

$$ \Rightarrow 12(x + iy - (x - iy))[1 - 2|z{|^2}] = 0$$ [as $$|z{|^2} = z\overline z $$]

$$ \Rightarrow 12 \times 2iy[1 - 2|z{|^2}] = 0$$

$$ \Rightarrow 24iy[1 - 2|z{|^2}] = 0$$

$$\therefore$$ $$y = 0$$ or $$1 - 2|z{|^2} = 0$$

$$y = 0$$ not possible as given z is a complex number with non-zero imaginary part.

$$\therefore$$ $$1 - 2|z{|^2} = 0$$

$$ \Rightarrow |z{|^2} = {1 \over 2} = 0.5$$