Solution: Note that \[ \gamma (t) = \begin{cases} 3e^{\iota t}, & t\in \left[ 0, \frac{\pi}{2} \right],\\[1ex] \iota \left( 3 + \frac{\pi}{2} - t \right), & t \in \left[ \frac{\pi}{2}, \frac{\pi}{2}+3 \right] ,\\[1ex] t - \frac{\pi}{2} - 3, & \left[ \frac{\pi}{2} + 3, \frac{\pi}{2} + 6 \right] . \end{cases} \] is same as \[ \eta (t) = \begin{cases} 3e^{\iota t}, & t\in \left[ 0, \frac{\pi}{2} \right],\\[1ex] \iota t, & t \in \left[ 3,0 \right] ,\\[1ex] t, & \left[ 0,3 \right] . \end{cases} \] Let \begin{align*} \gamma _1(t) & = 3e^{\iota t}, \quad t \in \left[ 0, \frac{\pi }{2} \right] \\[1ex] \gamma _2(t) & = \iota t, \quad t \in \left[3, 0 \right] \\[1ex] \gamma _3(t) & = t, \quad t \in [0,3]. \end{align*}

Thus, \begin{align*} \int_\gamma \vert z \vert \mathrm{d} z & = \int _{\gamma_1 \cup \gamma _2 \cup \gamma _3} \vert z \vert \mathrm{d}z \\ & = \int_{\gamma _1} \left\vert \gamma _1(t) \right\vert \gamma _1'(t) \mathrm{d} t + \int_{\gamma _2} \left\vert \gamma _2(t) \right\vert \gamma _2'(t) \mathrm{d} t \\ & \kern 2cm + \int_{\gamma _3} \left\vert \gamma _3(t) \right\vert \gamma _3'(t) \mathrm{d} t \\[1ex] & = \int_0^{\frac{\pi}{2}} 3 \cdot 3\iota e^{\iota t}\mathrm{d} t + \int_3^0 t \iota \mathrm{d} t + \int_0^3 t \mathrm{d} t \\[1ex] & = 9\iota \int_0^{\frac{\pi }{2}} (\cos t - \iota \sin t)\mathrm{d} t + (1-\iota ) \int _0^3 t\mathrm{d} t\\ & = 9\iota \left[ \sin t + \iota \cos t \right]_0^{\pi /2} + (1 - \iota ) \left[ \frac{t^2}{2} \right]_0^3 \\ & = -\frac{9}{2} + \frac{9}{2}\iota . \end{align*} Thus, \[ \textcolor{blue}{ \boxed{ \int_{\gamma }\vert z \vert \mathrm{d} z = -\frac{9}{2} + \frac{9}{2}\iota } } \]