Let S=z=x+iy:(2z-3i)/(4z+2i), is, a, real, number.

Question

Let $S={z=x+iy:\frac{2z-3i}{4z+2i}\, is\, a\, real\, number}$. Then which of the following is NOT correct?

Accepted Answer

Given that $z=x+i y$

$$
\begin{aligned}
& \text { then } \frac{2 z-3 i}{4 z+2 i}=\frac{2(x+i y)-3 i}{4(x+i y)+2 i} \\
& =\frac{2 x+i(2 y-3)}{4 x+i(4 y+2)} \times \frac{4 x-i(4 y+2)}{4 x-i(4 y+2)} \\
& =\frac{8 x^2+(2 y-3)(4 y+2)}{(4 x)^2+(4 y+2)^2}+i\left(\frac{4 x(2 y-3)-2 x(4 y+2)}{(4 x)^2+(4 y+2)^2}\right)
\end{aligned}
$$

Since, $\frac{2 z-3 i}{4 z+2 i}$ is Real $\Rightarrow \operatorname{Img}\left(\frac{2 z-3 i}{4 z+2 i}\right)=0$

$$
\begin{aligned}
& \Rightarrow 4 x(2 y-3)-2 x(4 y+2)=0 \\
& \Rightarrow 2 x(4 y-6-4 y-2)=0 \\
& \Rightarrow 2 x(-8)=0 \Rightarrow x=0
\end{aligned}
$$

Also, $(4 x)^2+(4 y+2)^2 \neq 0 \Rightarrow y+x^2+y^2 \neq \frac{-1}{4}$

If $x=0$, then $y \neq-\frac{1}{2}$

Solutions · 1