Answer:
Second option.
Step-by-step explanation:
We will use the euler's formula to solve this question
[tex]\text{Solution: }\\\\\text{From the euler's formula,}\\\\e^{i\theta}=\cos\theta+i\sin\theta[/tex]
[tex]\text{Replacing value of }\theta\text{ by 80}^\circ\text{ and }320^\circ\text{ one by one,}[/tex]
[tex]\cos80^\circ+i\sin80^\circ=e^{80i}[/tex]
[tex]\text{And, }\cos320^\circ+i\sin320^\circ=e^{320i}[/tex]
[tex]\text{Now,}[/tex]
[tex]\dfrac{z_1}{z_2}=\dfrac{36(\cos320^\circ+i\sin320^\circ)}{12(\cos80^\circ+i\sin80^\circ)}=\dfrac{36e^{320i}}{12e^{80i}}=3e^{320i-80i}=3e^{240i}[/tex]
[tex]\text{or, }\dfrac{z_1}{z_2}=3(\cos240^\circ+i\sin240^\circ)[/tex]
[tex]=3\{\cos(180^\circ+60^\circ)+i\sin(180^\circ+60^\circ)\}[/tex]
[tex]=3\{\text{-}\cos60^\circ-i\sin60^\circ\}\hspace{2cm}[\ \because \cos(180+\theta)=-\cos\theta[/tex]
[tex]\sin(180+\theta)=-\sin\theta][/tex]
[tex]=3\ \Big\{-\dfrac{1}{2}-i\dfrac{\sqrt3}{2}\Big\}[/tex]
[tex]\therefore\ \dfrac{z_1}{z_2}=-\dfrac{3}{2}-\dfrac{3\sqrt3}{2}i[/tex]
