Let z be a complex number such that |z+2|=1 and lm((z+1)/(z+2))=(1)/(5).

Question

Let $z$ be a complex number such that $|z+2|=1$ and $lm(\frac{z+1}{z+2})=\frac{1}{5}$. Then the value of $|Re(\vec{z+2})|$ is

Accepted Answer

$$\begin{aligned}
& |z+2|=1 \
& \operatorname{Im}_m\left(\frac{z+1}{z+2}\right)=\frac{1}{5} \
& |\operatorname{Re}(\overline{z+2})|=?
\end{aligned}$$

Let $z=x+i y$

$$\begin{aligned}
& \because|z+2|=1 \Rightarrow(x+2)^2+y^2=1 \quad \ldots(1) \
& I_m\left(\frac{z+1}{z+2}\right)=\frac{1}{5} \Rightarrow I_m\left(\frac{x+i y+1}{x+i y+2}\right)=\frac{1}{5} \
& \Rightarrow I_m\left|\frac{[(x+1)+i y][(x+)-i y]}{(x+2)^2+y^2}\right|=\frac{1}{5}
\end{aligned}$$

$$\frac{y(x+2)-y(x+)}{(x+2)^2+y^2}=\frac{1}{5}\quad \text{.... (2)}$$

$$\Rightarrow y=\frac{1}{5}$$

Substituting in equation (1)

$$\begin{aligned}
& (x+2)^2+\frac{1}{25}=1 \
& (x+2)^2=\frac{24}{25} \
& \Rightarrow x=-2 \pm \frac{\sqrt{24}}{5} \
& |\operatorname{Re}(\overline{x+i y+2})| \
& =x+2= \pm \frac{\sqrt{24}}{5}=\frac{2 \sqrt{6}}{5}
\end{aligned}$$

Solutions · 1