Answer:
[tex]z_1=(5\sqrt{2})^{1/3}cis(\frac{7\pi}{12})[/tex]
Step-by-step explanation:
To find the roots of the complex number you use the following formula:
[tex]z=re^{i\theta}\\\\z_k= (r)^{1/n}[cos(\frac{\theta+2\pi k}{n})+sin(\frac{\theta+2\pi k}{n})];\ \ k=0,1,2..n-1[/tex] (1)
in this case the polar number in polar form is:
[tex]r=\sqrt{5^2+5^2}=5\sqrt{2}\\\\\theta=tan^{-1}(\frac{-5}{5})=-45\°[/tex]
By replacing in (1) you obtain:
[tex]z_0=(5\sqrt{2})^{1/3}cis(\frac{-\pi/4+0}{3})\\\\z_1=(5\sqrt{2})^{1/3}cis(\frac{-\pi/4+2\pi}{3})=(5\sqrt{2})^{1/3}cis(\frac{7\pi}{12})\\\\z_2=(5\sqrt{2})^{1/3}cis(\frac{-\pi/4+4\pi}{3})=(5\sqrt{2})^{1/3}cis(\frac{15\pi}{12})[/tex]
hence, you have:
h. 52‾√3cis(7π12)
