Write the expression in Euler's formula 
$a=2+2i$ ,  $b=5e^{i\frac{\pi}{3}}$
such that $$\frac{b^5}{a^3}=Re^{\theta i}$$
find: $R$ and $\theta$

$$R=\sqrt{2^2+2^2}=\sqrt{8}$$,
$$tan^{-1}=\frac{2}{2}\rightarrow \theta=\frac{\pi}{4}$$
$$a=\sqrt{8}e^{\frac{\pi}{4}i}, a^3=({\sqrt{8}e^{\frac{\pi}{4}i}})^3=(\sqrt{8})^3e^{\frac{3\pi}{4}i}$$
$$b^5=5^5e^{\frac{5\pi}{3} i}$$
$$\frac{b^5}{a^3}=\frac{5^5e^{\frac{5\pi}{3} i}}{(\sqrt{8})^3e^{\frac{3\pi}{4}i}}$$
$$\frac{b^5}{a^3}=\color{red}{5^5-(\sqrt{8})^3}*e^{(\frac{5\pi}{3}-\frac{3\pi}{4})i}=(5^5-(\sqrt{8})^3)*e^{\frac{11}{12}\pi i}$$
$$R=5^5-(\sqrt{8})^3, \theta=\frac{11}{12}\pi$$
I can not find my mistake
 A: Write $a = 2\sqrt 2 e^{\pi/4 i}$ Then
$$\frac {b^5}{a^3} = \frac{5^5e^{5i\pi/3}}{(\sqrt 8)^3 e^{3/4\pi i}} = \frac{5^5}{(\sqrt 8)^3} e^{ (5/3 - 3/4) i \pi} = \frac{5^5}{(\sqrt 8)^3} e^{ \frac {11}{12} i \pi}$$
A: $$\frac{\left(5e^{\frac{\pi}{3}i}\right)^5}{\left(2+2i\right)^3}=$$
$$\frac{\left(5e^{\frac{\pi}{3}i}\right)^5}{\left(|2+2i|e^{\arg(2+2i)i}\right)^3}=$$
$$\frac{\left(5e^{\frac{\pi}{3}i}\right)^5}{\left(\sqrt{2^2+2^2}e^{\tan^{-1}\left(\frac{2}{2}\right)i}\right)^3}=$$
$$\frac{\left(5e^{\frac{\pi}{3}i}\right)^5}{\left(\sqrt{8}e^{\tan^{-1}\left(1\right)i}\right)^3}=$$
$$\frac{\left(5e^{\frac{\pi}{3}i}\right)^5}{\left(2\sqrt{2}e^{\frac{\pi}{4}i}\right)^3}=$$
$$\frac{5^5e^{5\cdot\frac{\pi}{3}i}}{2^3\left(\sqrt{2}\right)^3e^{3\cdot\frac{\pi}{4}i}}=$$
$$\frac{3125e^{\frac{5\pi}{3}i}}{8\left(2\sqrt{2}\right)e^{\frac{3\pi}{4}i}}=$$
$$\frac{3125e^{\frac{5\pi}{3}i}}{16\sqrt{2}e^{\frac{3\pi}{4}i}}=$$
$$\frac{3125}{16\sqrt{2}}e^{\left(\frac{5\pi}{3}-\frac{3\pi}{4}\right)i}$$
$$\frac{3125}{16\sqrt{2}}e^{\frac{11\pi}{12}i}$$
