Solution
$\mathrm{Z}=\frac{\mathrm{i}-1}{\cos \frac{\pi}{3}+\mathrm{i} \sin \frac{\pi}{3}}=\frac{\mathrm{i}-1}{\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}}=\frac{2(\mathrm{i}-1)(1-\sqrt{3} \mathrm{i})}{(1+\sqrt{3} \mathrm{i})(1-\sqrt{3} \mathrm{i})}=\frac{2(\mathrm{i}+\sqrt{3} \mathrm{i}-1+\sqrt{3})}{1+3}$
\[
\begin{aligned}
& z=\frac{\sqrt{3}-1}{2}+\frac{\sqrt{3}+1}{2} i \Rightarrow|z|=\frac{1}{2} \sqrt{(\sqrt{3}-1)^{2}+(\sqrt{3}+1)^{2}}=\sqrt{2} \\
& \arg (z)=\tan ^{-1}\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)=75^{\circ}=\frac{5 \pi}{12} \Rightarrow z=\sqrt{2}\left(\cos \frac{5 \pi}{12}+i \sin \frac{5 \pi}{12}\right)
\end{aligned}
\]
.
