(i) Solve: x+iy+2−3i=i(5−i)(3+4i)x+iy+2-3i=i(5-i)(3+4i)x+iy+2−3i=i(5−i)(3+4i) Solution First multiply (5−i)(3+4i):(5−i)(3+4i)=15+20i−3i−4i2=15+17i+4=19+17iNow multiply by i:i(19+17i)=19i+17i2=−17+19iEquate real and imaginary parts:x+iy+2−3i=−17+19i(x+2)+i(y−3)=−17+19ix+2=−17⇒x=−19y−3=19⇒y=22\begin{aligned} & \boxed{\text{First multiply }(5-i)(3+4i):}\\ \\ & (5-i)(3+4i)=15+20i-3i-4i^2 \\ &=15+17i+4=19+17i \\ \\ & \boxed{\text{Now multiply by }i:}\\ \\ & i(19+17i)=19i+17i^2=-17+19i \\ \\ & \boxed{\text{Equate real and imaginary parts:}}\\ \\ & x+iy+2-3i=-17+19i \\ & (x+2)+i(y-3)=-17+19i \\ \\ & x+2=-17 \Rightarrow x=-19 \\ & y-3=19 \Rightarrow y=22 \end{aligned}First multiply (5−i)(3+4i):(5−i)(3+4i)=15+20i−3i−4i2=15+17i+4=19+17iNow multiply by i:i(19+17i)=19i+17i2=−17+19iEquate real and imaginary parts:x+iy+2−3i=−17+19i(x+2)+i(y−3)=−17+19ix+2=−17⇒x=−19y−3=19⇒y=22 x=−19, y=22\boxed{x=-19,\ y=22}x=−19, y=22