Solution Given: z1=2+3i,z2=1−aiz_1=2+3i,\quad z_2=1-aiz1=2+3i,z2=1−ai Compute z1z2z_1z_2z1z2: z1z2=(2+3i)(1−ai)=2−2ai+3i−3ai2=2−2ai+3i+3a=(2+3a)+i(3−2a)\begin{aligned} z_1z_2&=(2+3i)(1-ai) \\ &=2-2ai+3i-3ai^2 \\ &=2-2ai+3i+3a \\ &=(2+3a)+i(3-2a) \end{aligned}z1z2=(2+3i)(1−ai)=2−2ai+3i−3ai2=2−2ai+3i+3a=(2+3a)+i(3−2a) So: Im(z1z2)=3−2a\operatorname{Im}(z_1z_2)=3-2aIm(z1z2)=3−2a Given Im(z1z2)=7\operatorname{Im}(z_1z_2)=7Im(z1z2)=7: 3−2a=7⇒−2a=4⇒a=−23-2a=7\Rightarrow -2a=4\Rightarrow a=-23−2a=7⇒−2a=4⇒a=−2 a=−2\boxed{a=-2}a=−2