If $\mathrm{Z} = \mathrm{cos\alpha } + \mathrm{isin\alpha }, 0 <\hspace{0.17em}\mathrm{\alpha } <\hspace{0.17em}\frac{\mathrm{\pi }}{4}$ then $\left|\frac{1 + {\mathrm{Z}}^4}{1 - {\mathrm{Z}}^3}\right| =$

$$\begin{gathered} \mathrm{z} = \mathrm{cos\alpha } + \mathrm{isin\alpha } 0 <\hspace{0.17em}\mathrm{\alpha } <\hspace{0.17em}\frac{\mathrm{\pi }}{4} \\ \Rightarrow {\mathrm{z}}^4 = {(\mathrm{cos\alpha } + \mathrm{isin\alpha })}^4 = \cos 4\mathrm{\alpha } + \mathrm{isin}4\mathrm{\alpha } \\ \mathrm{and} {\mathrm{z}}^3 = {(\mathrm{cos\alpha } + \mathrm{isin\alpha })}^3 = \cos 3\mathrm{\alpha } + \mathrm{isin}3\mathrm{\alpha } \\ \Rightarrow \left|\frac{1 + {\mathrm{z}}^4}{1 - {\mathrm{z}}^3}\right| = \left|\frac{1 + \cos 4\mathrm{\alpha } + \mathrm{isin}4\mathrm{\alpha }}{1 - \cos 3\mathrm{\alpha } - \mathrm{isin}3\mathrm{\alpha }}\right| \\ = \left|\frac{2{\cos }^{2}2\mathrm{\alpha } + \mathrm{i}(2\sin 2\mathrm{\alpha } \cos 2\mathrm{\alpha })}{2{\sin }^2{\displaystyle \frac{3\mathrm{\alpha }}{2}} - \mathrm{i}(2\sin {\displaystyle \frac{3\mathrm{\alpha }}{2}} \cos {\displaystyle \frac{3\mathrm{\alpha }}{2}})}\right| \\ = \frac{2\cos 2\mathrm{\alpha } \left|\cos 2\mathrm{\alpha } + \mathrm{isin}2\mathrm{\alpha }\right|}{\left|2\mathrm{isin}{\displaystyle \frac{3\mathrm{\alpha }}{2}}\right|\left|\cos {\displaystyle \frac{3\mathrm{\alpha }}{2}} - \mathrm{isin}{\displaystyle \frac{3\mathrm{\alpha }}{2}}\right|} \\ = \frac{\cos 2\mathrm{\alpha }}{\sin 3{\displaystyle \frac{\mathrm{\alpha }}{2}}} \end{gathered}$$