If $\frac{(1+i) x-i}{2+i}+\frac{(1+2 i) y+i}{2-i}=1$, then $(x, y)$ is equal to
If $\frac{(1+i) x-i}{2+i}+\frac{(1+2 i) y+i}{2-i}=1$, then $(x, y)$ is equal to
- $\left(\frac{7}{3}, \frac{-7}{15}\right)$
- $\left(\frac{7}{3}, \frac{7}{15}\right)$
- $\left(\frac{7}{5}, \frac{-7}{15}\right)$
- $\left(\frac{7}{5}, \frac{7}{15}\right)$
Solution
$
\begin{aligned}
& \frac{(1+i) x-i}{2+i}+\frac{(1+2 i) y+i}{2-i}=1 \\
& \Rightarrow \frac{[(1+i) x-i](2-i)}{\left(4-i^2\right)}+\frac{[(1+2 i) y+i](2+i)}{\left(4-i^2\right)}=1 \\
& \Rightarrow \frac{2(1+i) x-2 i-i(1+i) x+i^2}{4+1} \\
& +\frac{2(1+2 i) y+2 i+i(1+2 i) y+i^2}{(4+1)}=1 \\
& \Rightarrow \frac{\left(2+2 i-i-i^2\right) x-2 i+i^2}{5} \\
& +\frac{\left(4 i+2+i+2 i^2\right) y+2 i+i^2}{5}=1 \\
& \Rightarrow \frac{(2+i+1) x-2 i-1}{5}+\frac{(5 i+2-2) y+2 i-1}{5}=1 \\
& \Rightarrow(3+i) x-2 i-1+(5 i) y+2 i-1=5 \\
& \Rightarrow \quad(3+i) x+5 i y=7 \\
& \Rightarrow \quad 3 x+i x+5 i y-7=0 \\
& \Rightarrow \quad(3 x-7)+(x+5 y) i=0+i 0 \\
&
\end{aligned}
$
On comparing, we get
$
\begin{gathered}
3 x-7=0 \\
\Rightarrow \quad x=\frac{7}{3} \text { and } x+5 y=0 \\
y=\frac{-7}{15}
\end{gathered}
$
Hence, $(x, y)=\left(\frac{7}{3}, \frac{-7}{15}\right)$
Asked in: AP EAMCET 2013
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