দেওয়া আছে, $z = 8e^{i\pi/3} = 8(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}) = 8(\frac{1}{2} + i\frac{\sqrt{3}}{2}) = 4 + 4\sqrt{3}i$। তাহলে, $$ \begin{aligned} |e^{iz}| &= |e^{i(4+4\sqrt{3}i)}| \\ &= |e^{4i - 4\sqrt{3}}| \\ &= |e^{4i} \cdot e^{-4\sqrt{3}}| \\ &= |e^{4i}| \cdot |e^{-4\sqrt{3}}| \\ &= 1 \cdot e^{-4\sqrt{3}} \\ &= e^{-4\sqrt{3}} \end{aligned} $$ যেহেতু $|e^{i\theta}| = 1$।