Explanation

(c)
$\left| \mathcal{\ z -}iRe\left( \mathcal{z} \right) \right|\mathcal{= |z -}Im\left( \mathcal{z} \right)|$
If $\mathcal{z =}x + iy$
Then $|\ x + iy - ix| = |x + iy - y|$
$\sqrt{x^{2} + (y - x)^{2}} = \sqrt{(x - y)^{2} + y^{2}}$
or $x^{2} = y^{2}$
$\therefore\ \ x = \pm y$
$\Rightarrow \ \ Re\left( \mathcal{z} \right) = \pm Im\mathcal{(z)}$
$\Rightarrow \ \ Re\left( \mathcal{z} \right) + Im\left( \mathcal{z} \right) = 0$
and Re $\mathcal{(z) -}Im\left( \mathcal{z} \right) = 0$