We have,
$$
|a|=1 \Rightarrow \arg (a)=\theta
$$


$$
\begin{aligned}
& \therefore \quad a=\cos \theta+i \sin \theta \\
& \therefore\left(\frac{1+i z}{1-i z}\right)^4=a=\left(\frac{1+i z}{1-i z}\right)^4=\cos \theta+i \sin \theta \\
& =\left(\frac{1+i z}{1-i z}\right)=(\cos \theta+i \sin \theta)^{1 / 4} \\
& =\frac{1+i z}{1-i z}=\cos \left(\frac{2 k \pi+\theta}{4}\right)+i \sin \left(\frac{2 k \pi+\theta}{4}\right) \\
& k=0,1,2,3
\end{aligned}
$$
Apply componendo and dividendo.
$$
\begin{gathered}
\frac{1}{i z}=\frac{\cos \left(\frac{2 k \pi+\theta}{4}\right)+i \sin \left(\frac{2 k \pi+\theta}{4}\right)+1}{\cos \left(\frac{2 k \pi+\theta}{4}\right)+i \sin \left(\frac{2 k \pi+\theta}{4}\right)-1} \\
\Rightarrow \quad z=\frac{\cos \left(\frac{2 k \pi+\theta}{4}\right)+i \sin \left(\frac{2 k \pi+\theta}{4}\right)-1}{i\left[\cos \left(\frac{2 k \pi+\theta}{4}\right)+i \sin \left(\frac{2 k \pi+\theta}{4}\right)+1\right]} \\
z=\frac{-1+\cos \left(\frac{2 k \pi+\theta}{4}\right)+i \sin \left(\frac{2 k \pi+\theta}{4}\right)}{i\left(1+\cos \frac{2 k \pi+\theta}{4}+i \sin \frac{2 k \pi+\theta}{4}\right)}
\end{gathered}
$$


$$
\begin{aligned}
& z=\frac{\left[\begin{array}{r}
-2 \sin ^2\left(\frac{2 k \pi+\theta}{8}\right)+i 2 \sin \left(\frac{2 k \pi+\theta}{8}\right) \\
\left.\cos \left(\frac{2 k \pi+\theta}{8}\right)\right]
\end{array}\right]}{\left[\begin{array}{r}
2 \cos ^2 \frac{2 k \pi+\theta}{8}+i 2 \sin \left(\frac{2 k \pi+\theta}{8}\right) \\
\cos \left(\frac{2 k \pi+\theta}{8}\right)
\end{array}\right]} \\
& z=\frac{i \sin \left(\frac{2 k \pi+\theta}{8}\right)\left[\begin{array}{c}
\cos \frac{2 k \pi+\theta}{8} \\
+i \sin \frac{2 k \pi+\theta}{8}
\end{array}\right]}{i \cos \left(\frac{2 k \pi+\theta}{8}\right)\left[\begin{array}{c}
\cos \frac{2 k \pi+\theta}{8} \\
+i \sin \left(\frac{2 k \pi+\theta}{8}\right)
\end{array}\right]} \\
&
\end{aligned}
$$


$$
\Rightarrow \quad z=\tan \left(\frac{2 k \pi+\theta}{8}\right) \Rightarrow k=0,1,2,3
$$