As #z_1=-3+2i#, we have #|z_1|=sqrt(3^2+2^2)=sqrt13#
and hence #z_1=-3+2i=sqrt13(-3/sqrt13+i2/sqrt13)#
= #sqrt13(cosalpha+isinalpha)#, where #cosalpha=-3/sqrt13# and #sinalpha=2/sqrt13#
Observe that #alpha# lies in #Q2# and #tanalpha=-2/3#
Similarly #z_2=4+3i#, we have #|z_1|=sqrt(4^2+3^2)=sqrt25=5#
and hence #z_2=4+3i=5(4/5+i3/5)#
= #5(cosbeta+isinbeta)#, where #cosbeta=4/5# and #sinbeta=3/5#
Observe that #beta# lies in #Q1# and #tanbeta=3/4#
Hence #z_1/z_2=(sqrt13(cosalpha+isinalpha))/(5(cosbeta+isinbeta))#
= #(sqrt13(cosalpha+isinalpha))/(5(cosbeta+isinbeta))xx(cosbeta-isinbeta)/(cosbeta-isinbeta)#
= #sqrt13/5xx((cosalphacosbeta+sinalphasinbeta)+i(sinalphacosbeta-cosalphasinbeta))/(cos^2beta+sin^2beta)#
= #sqrt13/5(cos(alpha-beta)+isin(alpha-beta))#
Now #tan(alpha-beta)=(tanalpha-tanbeta)/(1+tanalphatanbeta)#
= #(-2/3+3/4)/(1+(-2/3)xx(3/4))#
= #(1/12)/(1-2/4)=1/12xx4/2=1/6# and #alpha-beta=tan^(-1)(1/6)#
As #alpha# lies in #Q2#, #beta# lies in #Q1# and #tan(alpha-beta)>0#
Argand of #z_1/z_2# is in #Q1#
