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Accedmychevron_right11thchevron_rightmathchevron_rightComplex Numberschevron_rightExercise 1.2

(iii)

Solve:

x2+i+y3−i=4+5i\frac{x}{2+i}+\frac{y}{3-i}=4+5i

Solution

Rationalize each fraction:x2+i=x(2−i)(2+i)(2−i)=x(2−i)5y3−i=y(3+i)(3−i)(3+i)=y(3+i)10x(2−i)5+y(3+i)10=4+5iMultiply by 10:2x(2−i)+y(3+i)=40+50i(4x−2xi)+(3y+yi)=40+50i(4x+3y)+i(−2x+y)=40+50iEquate real and imaginary parts:4x+3y=40(1)−2x+y=50(2)From (2): y=50+2xSubstitute in (1): 4x+3(50+2x)=404x+150+6x=40⇒10x=−110⇒x=−11y=50+2(−11)=28\begin{aligned} & \boxed{\text{Rationalize each fraction:}}\\ \\ & \frac{x}{2+i}=\frac{x(2-i)}{(2+i)(2-i)}=\frac{x(2-i)}{5} \\ & \frac{y}{3-i}=\frac{y(3+i)}{(3-i)(3+i)}=\frac{y(3+i)}{10} \\ \\ & \frac{x(2-i)}{5}+\frac{y(3+i)}{10}=4+5i \\ \\ & \boxed{\text{Multiply by }10:}\\ \\ & 2x(2-i)+y(3+i)=40+50i \\ & (4x-2xi)+(3y+yi)=40+50i \\ & (4x+3y)+i(-2x+y)=40+50i \\ \\ & \boxed{\text{Equate real and imaginary parts:}}\\ \\ & 4x+3y=40 \quad (1)\\ & -2x+y=50 \quad (2)\\ \\ & \text{From (2): }y=50+2x \\ & \text{Substitute in (1): }4x+3(50+2x)=40 \\ & 4x+150+6x=40 \Rightarrow 10x=-110 \Rightarrow x=-11 \\ & y=50+2(-11)=28 \end{aligned} x=−11, y=28\boxed{x=-11,\ y=28}
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