Let us write $i$ for $j$, as is customary in mathematics (in distinction with electrical engineering). Let $T:=\varphi$ and $Z:=z=e^{iT}$, where $T\sim\mathcal U[-a,a]$, for some real $a>0$. We have to find the pdf of $Z$ with respect to the uniform distribution, say $\mu$, on the unit circle $C$ in $\mathbb C$.
The measure $\mu$ on $C$ is the push-forward measure of the uniform distribution $\mathcal U(-\pi,\pi]$ under the map $(-\pi,\pi]\ni s\mapsto e^{is}\in C$. So, $\mu$ is determined by the identity \begin{equation*} \int_C g\,d\mu=\frac1{2\pi}\int_{-\pi}^\pi g(e^{is})\,ds \end{equation*} for all nonnegative Borel functions $g$ on $C$.
The pdf (say $p$) of $Z=e^{iT}$ with respect to $\mu$ is then determined by the identity \begin{equation*} Eh(e^{iT})=\int_C hp\,d\mu \end{equation*} for all nonnegative Borel functions $h$ on $C$, which can be rewritten as \begin{equation*} \frac1{2a}\int_{-a}^a h(e^{it})\,dt=\frac1{2\pi}\int_{-\pi}^\pi h(e^{is})p(e^{is})\,ds. \tag{1} \end{equation*} Next, introduce \begin{equation*} K:=K_a:=\Big\lceil\frac{a/\pi-1}2\Big\rceil, \end{equation*} so that $(2K-1)\pi<a\le(2K+1)\pi$. Then \begin{align*} \int_{-a}^a h(e^{it})\,dt&=\sum_{k=1-K}^{K-1}\int_{(2k-1)\pi}^{(2k+1)\pi} h(e^{it})\,dt \\ & +\int_{(2K-1)\pi}^a h(e^{it})\,dt+\int_{-a}^{(1-2K)\pi} h(e^{it})\,dt \\ &=(2(K-1)+1)_+\int_{-\pi}^\pi h(e^{is})\,ds \\ & +\int_{-\pi}^\pi h(e^{is})I\{\arg e^{is}<a-2K\pi\}\,ds \\ &+\int_{-\pi}^\pi h(e^{is})I\{\arg e^{is}>2K\pi-a\}\,ds, \end{align*} where $w_+:=\max(0,w)$ for real $w$, $I$ denotes the indicator, and $\arg e^{is}:=s$ for $s\in(-\pi,\pi]$. Looking back at (1), we now conclude that \begin{equation*} p(c)=\frac\pi a\,\big( (2K_a-1)_++I\{\arg c<a-2K_a\pi\}+I\{\arg c>2K_a\pi-a\}\big) \end{equation*} for ($\mu$-almost all) $c\in C$.