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Accedmychevron_right11thchevron_rightmathchevron_rightComplex Numberschevron_rightExercise 1.2

(ii)

Solve:

(x+iy)(1−i)=(2−3i)(−5+5i)−35i(x+iy)(1-i)=(2-3i)(-5+5i)-\frac{3}{5}i

Solution

First multiply (2−3i)(−5+5i):(2−3i)(−5+5i)=−10+10i+15i−15i2=−10+25i+15=5+25iSubtract 35i:5+25i−35i=5+(25−35)i=5+1225iExpand left side:(x+iy)(1−i)=x−xi+iy−iyi=x−xi+iy−yi2=(x+y)+i(y−x)Equate real and imaginary parts:x+y=5(1)y−x=1225(2)(1)+(2): 2y=5+1225=1475 ⇒ y=14710x=5−y=5−14710=−9710\begin{aligned} & \boxed{\text{First multiply }(2-3i)(-5+5i):}\\ \\ & (2-3i)(-5+5i)=-10+10i+15i-15i^2 \\ &=-10+25i+15=5+25i \\ \\ & \boxed{\text{Subtract }\frac{3}{5}i:}\\ \\ & 5+25i-\frac{3}{5}i=5+\left(25-\frac{3}{5}\right)i=5+\frac{122}{5}i \\ \\ & \boxed{\text{Expand left side:}}\\ \\ & (x+iy)(1-i)=x-xi+iy-iyi \\ &=x-xi+iy-yi^2 \\ &=(x+y)+i(y-x) \\ \\ & \boxed{\text{Equate real and imaginary parts:}}\\ \\ & x+y=5 \quad (1)\\ & y-x=\frac{122}{5} \quad (2)\\ \\ & (1)+(2):\ 2y=5+\frac{122}{5}=\frac{147}{5}\ \Rightarrow\ y=\frac{147}{10} \\ & x=5-y=5-\frac{147}{10}=-\frac{97}{10} \end{aligned} x=−9710, y=14710\boxed{x=-\frac{97}{10},\ y=\frac{147}{10}}
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