Respuesta :
Given the following complex number
[tex]z=4-6i[/tex]We will find the cube root of the complex number using the following formula:
[tex]^3\sqrt{z}=\sqrt[3]{|z|}*(cos\text{ }\frac{\theta+2\pi k}{3}+i*sin\text{ }\frac{\theta+2\pi k}{3})[/tex]The formula is called De Moivre's theorem of the nth root
We have substituted n = 3
So, first, we will convert the given number from the rectangular form to the polar form
[tex]\begin{gathered} |z|=\sqrt{4^2+6^2}\approx7.211 \\ \theta=tan^{-1}\frac{-6}{4}=303.7\degree \end{gathered}[/tex]Substitute the magnitude and the angle and k = 0, 1, 2
So, there are 3 cubic roots of the given number as follows:
[tex]\begin{gathered} k=0\rightarrow z_1=\sqrt[3]{7.211}(cos\frac{303.7}{3}+i*sin\frac{303.7}{3})=1.932(cos101.23+i*sin101.23) \\ \\ k=1\rightarrow z_2=\sqrt[3]{7.211}(cos\frac{303.7+2\pi}{3}+i*sin\frac{303.7+2\pi}{3})=1.932(cos221.23+i*sin221.23) \\ \\ k=2\rightarrow z_3=\sqrt[3]{7.211}(cos\frac{303.7+4\pi}{3}+i*sin\frac{303.7+4\pi}{3})=1.932(cos341.23+i*sin341.23) \end{gathered}[/tex]