If the locus of z C, such that Re((z-1)/(2z+i))+(z-12z-i)=2, is a circle of radius r and c…

Question

If the locus of z \in  \mathbb{C}, such that Re$(\frac{z-1}{2z+i})+(\frac{\vec{z}-1}{2\vec{z}-i})=2$, is a circle of radius r and center $(a,b)$, then $\frac{15ab}{{r}^{2}}$ is equal to :

Accepted Answer

$\operatorname{Re}\left(\frac{\mathrm{z}-1}{2 \mathrm{z}+\mathrm{i}}\right)+\operatorname{Re}\left(\frac{\overline{\mathrm{z}}-1}{2 \bar{z}-\mathrm{i}}\right)=2$

Here, $\frac{\mathrm{z}-1}{2 \mathrm{z}+\mathrm{i}}=\left(\frac{\overline{\bar{z}-1}}{2 \overline{\mathrm{z}}-\mathrm{i}}\right)=2$

$=\operatorname{Re}\left(\frac{z-1}{2 z+i}\right)+\operatorname{Re}\left(\overline{\frac{z-1}{2 z+i}}\right)=2$

$=2 \operatorname{Re}\left(\frac{z-1}{2 z+1}\right)=2 \Rightarrow \operatorname{Re}\left(\frac{z-1}{2 z+i}\right)=1$

$$\begin{aligned}
& \text { Let } z=x+i y \
& \operatorname{Re}\left(\frac{(x-1)+i y}{2 x+i(2 y+1)}\right)=1 \Rightarrow \operatorname{Re}\left[\frac{((x-1)+i y)(2 x-i(y+1)}{(2 x+i(2 y+1)(2 x-i(2 y+1))}\right]=1 \
& \Rightarrow \frac{2 x(x-1)+y(2 y+1)}{4 x^2+(2 y+1)^2}=1 \
& \Rightarrow 2 x^2-2 x+2 y^2+y=4 x^2+4 y^2+1+4 y \
& \Rightarrow 2 x^2+2 y^2+3 y+2 x+1=0 \
& \Rightarrow x^2+y^2+x+\frac{3}{2} y+\frac{1}{2}=0 \
& \text { centre }=\left(\frac{-1}{2}, \frac{-3}{4}\right), r=\sqrt{\frac{1}{4}+\frac{9}{16}-\frac{1}{2}}=\frac{\sqrt{5}}{4} \
& a=\frac{-1}{2}, b=\frac{-3}{4}, r^2=\frac{5}{16}
\end{aligned}$$

$15 \frac{\mathrm{ab}}{\mathrm{r}^2}=15 \times\left(\frac{-1}{2}\right) \times\left(\frac{-3}{4}\right) \times \frac{16}{5}=18$

Solutions · 1