Answer:
Step-by-step explanation:
second
let r(cos theta+i sin theta)=-3+8i
[tex]r cos~ \theta =-3\\r ~sin~\theta=8\\square and add\\r^2=9+64=73\\r=\sqrt{73} \\divide\\tan \theta=-\frac {8}{3}\\sin \theta ~is ~positive~and cos ~\theta~is~negative.\\so \theta ~lies ~in~second ~quadrant.[/tex]
