Question

for any purely imaginary number such that img(z)<0 then arg(z) is

Answer 1

for any purely imaginary no. Re(z)=0
therefore in z=x+iy x=0 and y=-y (according to the question)
tan alpha= -y/0=infinity
alpha= π/2=90°
since, z lies in y' axis
hence, theta=3π/2=270°
therefore arg(z)=270°(ans)