We first covert them into trigonometric form. In this form #a+bi=r(costheta+isintheta)# or #a+bi=r*e^(itheta)#, where #r=sqrt(a^2+b^2)# and #theta=arctan(b/a)#
Hence, #7i+1=1+7isqrt50e^(ialpha)#, where #alpha=arctan7#
and #-3i+1=1-3i=sqrt10e^(ibeta)#, where #beta=arctan(-3)#
Hence #(7i+1)/(-3i+1)=sqrt5e^(i(alpha-beta))#
As #tan(alpha-beta)=(tanalpha-tanbeta)/(1+tanalpha*tanbeta)# or
#tan(alpha+beta)=(7-(-3))/(1+7*(-3))=10/(-20)=-1/2#
Hence #(7i+1)/(-3i+1)=sqrt5e^(i(arctan(-1/2)))#
or #(7i+1)/(-3i+1)=sqrt5[cosarctan(-1/2)+isinarctan(-1/2)]#
