Answer: The roots are

\(z=2\cos\frac{\pi}{9}+\mathrm{i}\left(2\sin\frac{\pi}{9}-5\right)\),

\(z=2\cos\frac{7\pi}{9}+\mathrm{i}\left(2\sin\frac{7\pi}{9}-5\right)\),

\(z=2\cos\frac{13\pi}{9}+\mathrm{i}\left(2\sin\frac{13\pi}{9}-5\right)\).

Let \(w=z+5\mathrm{i}\). Then

\(w^3=4+4\sqrt{3}\,\mathrm{i}\).

Write the right-hand side in polar form.

Its modulus is

\(|4+4\sqrt{3}\,\mathrm{i}|=\sqrt{4^2+(4\sqrt{3})^2}=\sqrt{16+48}=8\).

Its argument is

\(\arg(4+4\sqrt{3}\,\mathrm{i})=\tan^{-1}\!\left(\frac{4\sqrt{3}}{4}\right)=\tan^{-1}(\sqrt{3})=\frac{\pi}{3}\).

So

\(4+4\sqrt{3}\,\mathrm{i}=8\left(\cos\frac{\pi}{3}+\mathrm{i}\sin\frac{\pi}{3}\right)\).

Hence

\(w^3=8\left(\cos\frac{\pi}{3}+\mathrm{i}\sin\frac{\pi}{3}\right)\).

The cube roots have modulus \(\sqrt[3]{8}=2\), and arguments

\(\frac{\frac{\pi}{3}+2k\pi}{3}=\frac{\pi}{9}+\frac{2k\pi}{3}\qquad (k=0,1,2).\)

So the three values of \(w\) are

\(w_1=2\left(\cos\frac{\pi}{9}+\mathrm{i}\sin\frac{\pi}{9}\right),\)

\(w_2=2\left(\cos\frac{7\pi}{9}+\mathrm{i}\sin\frac{7\pi}{9}\right),\)

\(w_3=2\left(\cos\frac{13\pi}{9}+\mathrm{i}\sin\frac{13\pi}{9}\right).\)

Since \(z=w-5\mathrm{i}\),

\(z_1=2\cos\frac{\pi}{9}+\mathrm{i}\left(2\sin\frac{\pi}{9}-5\right),\)

\(z_2=2\cos\frac{7\pi}{9}+\mathrm{i}\left(2\sin\frac{7\pi}{9}-5\right),\)

\(z_3=2\cos\frac{13\pi}{9}+\mathrm{i}\left(2\sin\frac{13\pi}{9}-5\right).\)

These are in the required form \(r\cos\theta+\mathrm{i}(r\sin\theta-5)\), with \(r=2\).