Q47 / 160 · TS EAMCET 2021 (06 Aug Shift 2) · Mathematics
If $Z=x+i y$ is a complex number and $\sqrt{x^2-2 x+8}+(x+4) i=y(2+i)$, then $Z$ is equal to
- A$\frac{-28}{9}-\frac{16}{9} i$
- Correct answerB$-2+2 i$
- C$\frac{2}{3}-\frac{2}{3} i$
- D$\frac{-2}{5}-\frac{2 i}{5}$
Solution
Given, $\sqrt{x^2-2 x+8}+(x+4) i=y(2+i)$
On equating real and imaginary parts,
$\begin{aligned}
& \sqrt{x^2-2 x+8}=2 y \\
& \Rightarrow \quad x^2-2 x+8=4 y^2 \\
& \text { and } \\
& x+4=y \\
& \Rightarrow \quad x=y-4 \\
& \therefore \quad(y-4)^2-2(y-4)+8=4 y^2 \\
& \Rightarrow \quad 3 y^2+10 y-32=0 \\
& \Rightarrow \quad(y-2)(3 y+16)=0 \\
& \Rightarrow \quad y=2, \frac{-16}{3} \\
& \therefore \quad x=-2, \frac{-28}{3} \\
& \text { Hence, } z=-2+2 i \text { or } \frac{-28}{3}-\frac{16}{3} i \\
&
\end{aligned}$
Hence, the correct answer is option: 'B'.