If $z_r=\cos \frac{2r\pi }{4}+i\sin \frac{2r\pi }{4}$ where r = 0, 1, 2, 3 then $\frac{z_0+z_1}{z_2+z_3}$ equals 

Given $Z_r=\cos \frac{2r\pi }{4}+i\sin \frac{2r\pi }{4}r=0,1,2,3$

i.e., z,, r = 0, 1, 2, 3 are the 4th roots of unity. 

$\begin{array}{l}\therefore z_0=1,z_1=i,z_2=-1\text{ and }{z}_3-i\\ \therefore \frac{z_0+z_1}{z_2+z_3}=\frac{1+i}{-1-i}=\frac{-(1+i)}{(1+i)}=-1\end{array}$