Question

qual o valor de m de modo que o quociente 3+mi sobre 2-1 seja um numero imaginario puro

Answer 1

$\mathrm{\dfrac{3+mi}{2-i}=a+bi\ \to\ \dfrac{3+mi}{2-i}.\dfrac{2+i}{2+i}=a+bi}\\\\\\ \mathrm{\dfrac{6+3i+2mi+mi^2}{4+2i-2i-i^2}=a+bi\ \to\ \dfrac{6+(3+2m)i+m(-1)}{4-(-1)}=a+bi}\\\\\\ \mathrm{\dfrac{(6-m)+(3+2m)i}{5}=a+bi\ \to\ \bigg(\dfrac{6-m}{5}\bigg)+\bigg(\dfrac{3+2m}{5}\bigg)i=a+bi}\\\\\\ \mathrm{*\ Como\ \Re{(Z)}=0\ \to\ a=0\ \to\ \bigg(\dfrac{6-m}{5}\bigg)=0}\\\\\\ \mathrm{6-m=0\ \to\ \boxed{\mathbf{m=6}}}$