Explanation

(b)
Since, $\left| \frac{z}{z - i/3} \right|$=1
$\Rightarrow \ \ \ 3|z| = |3z - i|$
$\Rightarrow \ \ \ 3|x + iy| = \left| 3(x + iy) - i \right|$ [put $z = x + iy$]
$\Rightarrow \ \ \ \ \ \ 3\sqrt{x^{2} + y^{2}} = \sqrt{(3x)^{2} + (3y - 1)^{2}}$
$\Rightarrow \ \ 9x^{2} + 9y^{2} = 9x^{2} + 9y^{2} + 1 - 6y$
$\Rightarrow \ \ \ \ y = \frac{1}{6}$
Which shows that $z$ lies on a straight line.