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, #-1+2i=sqrt5e^(ialpha)#, where #alpha=arctan(-2)#
and #3-i=sqrt10e^(ibeta)#, where #beta=arctan(-1/3)#
Hence #(-1+2i)*(3-i)=sqrt50e^(i(alpha+beta))#
As #tan(alpha+beta)=(tanalpha+tanbeta)/(1-tanalpha*tanbeta)# or
#tan(alpha+beta)=((-2)-1/3)/(1-(-2)*(-1/3))=(-7/3)/(1-2/3)=(-7/3)/(1/3)=-7#
Hence #(-1+2i)*(3-i)=sqrt50e^(i(arctan(-7)))#
or #(-1+2i)*(3-i)=sqrt50[cosarctan(-7)+isinarctan(-7)]#
