Answer:
[tex]3\sqrt{6}+3\sqrt{2}i[/tex] and [tex]-3\sqrt{6}-3\sqrt{2}i[/tex]
Step-by-step explanation:
For a complex number [tex]z=a+bi[/tex] in rectangular form, polar form is [tex]z=r[cos(\theta)+isin(\theta)][/tex] where [tex]r=\sqrt{a^2+b^2}[/tex] and [tex]\theta=tan^{-1}(\frac{b}{a})[/tex]:
[tex]r=\sqrt{a^2+b^2}\\\\r=\sqrt{(36)^2+(36\sqrt{3})^2}\\\\r=\sqrt{1296+3888}\\\\r=\sqrt{5184}\\\\r=72[/tex]
[tex]\theta=tan^{-1}(\frac{b}{a})\\ \\\theta=tan^{-1}(\frac{36\sqrt{3}}{36})\\ \\\theta=tan^{-1}(\sqrt{3})\\\\\theta=\frac{\pi}{3}[/tex]
Thus, our complex number in polar form is [tex]z=72[cos(\frac{\pi}{3})+isin(\frac{\pi}{3})][/tex]. This will help us determine the square roots of the complex number using the formula [tex]\sqrt[n]{r}\biggr[cis\bigr({\frac{\theta+2\pi k}{n})\biggr][/tex] for [tex]k=0,1,2,\: ... \:,n-1[/tex]. Note that [tex]cis(\theta)=cos(\theta)+isin(\theta)[/tex]:
1st square root for k = 2-1 = 1
[tex]\sqrt[n]{r}\biggr[cis\bigr({\frac{\theta+2\pi k}{n})\biggr][/tex]
[tex]\sqrt[2]{72}\biggr[cis\bigr({\frac{\frac{\pi}{3}+2\pi(1)}{2})\biggr][/tex]
[tex]6\sqrt{2}\biggr[cis\bigr({\frac{\frac{\pi}{3}+2\pi}{2})\biggr][/tex]
[tex]6\sqrt{2}\biggr[cis\bigr(\frac{\pi}{6}+\pi)\biggr][/tex]
[tex]6\sqrt{2}\biggr[cis\bigr(\frac{7\pi}{6})\biggr][/tex]
[tex]6\sqrt{2}\biggr[cos(\frac{7\pi}{6})+isin(\frac{7\pi}{6})\biggr]\\ \\6\sqrt{2}(-\frac{\sqrt{3}}{2}-\frac{1}{2}i)\\\\-3\sqrt{6}-3\sqrt{2}i[/tex]
2nd square root for k = 1-1 = 0
[tex]\sqrt[6]{72}\biggr[cis\bigr({\frac{\frac{\pi}{3}+2\pi(0)}{2}\bigr)\biggr][/tex]
[tex]6\sqrt{2}\biggr[cis(\frac{\pi}{6} )\biggr][/tex]
[tex]6\sqrt{2}[cos(\frac{\pi}{6})+isin(\frac{\pi}{6})]\\ \\6\sqrt{2}(\frac{\sqrt{3}}{2}+\frac{1}{2})\\\\3\sqrt{6}+3\sqrt{2}i[/tex]
Therefore, the square roots of [tex]36+36\sqrt{3}i[/tex] are [tex]3\sqrt{6}+3\sqrt{2}i[/tex] and [tex]-3\sqrt{6}-3\sqrt{2}i[/tex].
