Let $$z = x + iy$$

$$\therefore$$ $$|z| = \sqrt {{x^2} + {y^2}} $$

Given, $$|z| = 3$$

$$\therefore$$ $$\sqrt {{x^2} + {y^2}} = 3$$

$$ \Rightarrow {x^2} + {y^2} = 9 = {3^2}$$

This represent a circle with center at (0, 0) and radius = 3

Now, given

$$\arg (z - 1) - \arg (z + 1) = {\pi \over 4}$$

$$ \Rightarrow \arg (x + iy - 1) - \arg (x + iy + 1) = {\pi \over 4}$$

$$ \Rightarrow \arg (x - 1 + iy) - \arg (x + 1 + iy) = {\pi \over 4}$$

$$ \Rightarrow {\tan ^{ - 1}}\left( {{y \over {x - 1}}} \right) - {\tan ^{ - 1}}\left( {{y \over {x + 1}}} \right) = {\pi \over 4}$$

$$ \Rightarrow {\tan ^{ - 1}}\left( {{{{y \over {x - 1}} - {y \over {x + 1}}} \over {1 + {y \over {x - 1}} \times {y \over {x + 1}}}}} \right) = {\pi \over 4}$$

$$ \Rightarrow {\tan ^{ - 1}}\left( {{{{{xy + y - xy + y} \over {{x^2} - 1}}} \over {{{{x^2} - 1 + {y^2}} \over {{x^2} - 1}}}}} \right) = {\pi \over 4}$$

$$ \Rightarrow {\tan ^{ - 1}}\left( {{{xy + y - xy + y} \over {{x^2} - 1 + {y^2}}}} \right) = {\pi \over 4}$$

$$ \Rightarrow {{2y} \over {{x^2} - 1 + {y^2}}} = \tan \left( {{\pi \over 4}} \right)$$

$$ \Rightarrow 2y = {x^2} + {y^2} - 1$$

$$ \Rightarrow {x^2} + {y^2} - 2y - 1 = 0$$

$$ \Rightarrow {x^2} + {(y - 1)^2} = {(\sqrt 2 )^2}$$

This represent a circle with center at (0, 1) and radius $$\sqrt 2 $$.

From diagram you can see both the circles do not cut anywhere.