The equation $\vert z - z_0 \vert = r$, denotes a circle with center $z_0$ and radius $r$.
To find the maximum value of $\arg z$, consider the triangle formed by the tangent lines from the origin and find the angle.
Solution: We take $z = x + \iota y$.
The locus of point $P$ satisfying the equation \( \vert z - 5 - 3\iota \vert = 3 \) is a circle centered at the point \( (5, 3) \) with a radius of \( 3 \).
The Cartesian equation of the locus can be obtained by squaring both sides of the equation:
\begin{align*}
\vert z - 5 - 3 \iota \vert = 3 & \implies \vert x + \iota y - 5 - 3 \iota \vert = 5 \\
& \implies \sqrt{(x-5)^2 + (y - 3)^2} = 3 \\
& \implies (x - 5)^2 + (y - 3)^2 = 9.
\end{align*}
To find the maximum value of $\arg z$, consider the following picture.
The maximum value of $\arg z$ occurs when the line between the origin and $P$ is a tangent to the circle. Now we need to find the angle $\theta $ in the figure. Note that both the triangles $\triangle OAC$ and $\triangle OBC$ are congruent (by RHS congruence criterion). Thus, both the angles are same, which is denoted by $\theta $. Use triangle $OBC$, to get
\begin{align*}
\tan \theta = \frac{BC}{OB} & = \frac{3}{5} \implies \theta \tan ^{-1} \left( \frac{3}{5} \right) .
\end{align*}
Thus, the maximum value of $\arg z$ is equals to $\tan ^{-1} \left( \frac{3}{5} \right)$.
