Complex operation

 (5 - 3i)^1/3.

\(r = mod z = |z| = \sqrt{5^2+(-3)^2}=\sqrt{34}\)

\(5-3i=\sqrt{34}(\frac{5}{\sqrt{34}}-\frac{3i}{\sqrt{34}})\\ 4th\;quad\\ argument \;of\; z = arg(z) =\theta=2\pi-acos(\frac{5}{\sqrt{34}})\\ \theta\approx 5.7427658\;radians\\ 5-3i=\sqrt{34}\;e^{5.7427658i} \)
 

\(|z^{1/3}| = \sqrt{34}^{1/3}=\sqrt[6]{34}\)

\(1)  arg(z^{1/3})=5.7427658/3 = 1.914255\\ 2)  arg(z^{1/3}) = 1.914255 +\frac{2 \pi}{3} = 4.008650\\ 3)  arg(z^{1/3}) = 1.914255 +2*\frac{2 \pi}{3}= 6.103045\\\)

So the 3 cubed roots of are 5-3i  are

\(\sqrt[6]{34}*e^{1.914255i}\quad and \quad \sqrt[6]{34}*e^{4.008650i}\quad and \quad \sqrt[6]{34}*e^{6.103045i}\)

cos(1.914255) = -0.336745771286

sin(1.914255) = 0.94159560615

\(1st\;root=\sqrt[6]{34}\;(-0.336746\;+\;0.941596\;i)\)

etc

check the first root

(34^(1/6)*e^(1.914255*i))^3 = 4.9999975792700509-3.0000040345460377i      Near enough    

z = (5 - 3i)^(1/3)

Algebraic form:
z = 1.7707675-0.3224815i

Exponential form:
z = 1.7998922 × ei (-10°19'16″)

Trigonometric form:
z = 1.7998922 × (cos (-10°19'16″) + i sin (-10°19'16″))

Polar form:
r = |z| = 1.79989
φ = arg z = -10.32125° = -10°19'16″ = -0.05734π
 

THIS IS THE ACCURATE SOLUTION. THE FIRST ANSWER CALCULATED (5-3i)^1*3

Could you talk us through it please ?