Question

Let $f:C\to C$ be a function defined as $f\left(z\right)=\frac{z+i}{z-i}$ for all complex numbers $z\ne i$ and $z_n=f\left(z_{n-1}\right)$ for all $n\in N$. If $z_0=K+i$ and $z_{2020}=1+2020i$, then the value of $\left(\frac{1}{K}\right)$ is

Answer 1

$f\left(z\right)=\frac{z+i}{z-i}$
$f\left(f\right(z\left)\right)=\frac{\frac{z+i}{z-i}+1}{\frac{z+i}{z-i}-i}$
$=\frac{z+i+zi+1}{z+i-zi-1}$
$=\frac{(z+1)(i+1)}{(z-1)(1-i)}\times \frac{(1+i)}{(1+i)}$
$=\frac{(z+1)(i+1{)}^2}{(z-1)\times 2}$
$\Rightarrow f_2\left(z\right)=\left(\frac{z+1}{z-1}\right)i\left[\text{Let}{f}_2\right(z)=f(f\left(z\right)\left)\right]$

$f_3\left(z\right)=f\left(f\right(f\left(z\right)\left)\right)=\frac{\frac{z+1}{z-1}i+i}{\frac{z+1}{z-1}i-i}$
$=\frac{(z+1)+(z-1)}{(z+1)-(z-1)}$
$=z$

Similarly, $f_4\left(z\right)=\frac{z+i}{z-i}=f\left(z\right)$
$\Rightarrow f_{n+3}\left(z\right)=f_n\left(z\right)...\left(1\right)$

Now, $z_n=f\left(z_{n-1}\right)$
$\Rightarrow z_n=f\left(f\right(z_{n-2}\left)\right)=f_2\left(z_{n-2}\right)$
$\Rightarrow z_n=f\left(f\right(f\left(z_{n-3}\right)\left)\right)=f_3\left(z_{n-3}\right)$
$⋮$
$\Rightarrow z_n=f_{n-1}\left(z\right)...\left(2\right)$
From $\left(1\right)$ and $\left(2\right),$ we get
$\Rightarrow z_{n+3}=z_n$

Now, $z_{2020}=z_{3\times 673+1}$
$\Rightarrow 1+2020i=z_1$
$\Rightarrow 1+2020i=\frac{z_0+i}{z_0-i}$
$\Rightarrow 1+2020i=\frac{K+i+i}{K+i-i}$
$\Rightarrow 1+2020i=1+\frac{2}{K}i$
Comparing real and imaginarry part we get
$\Rightarrow \frac{2}{K}=2020$
$\Rightarrow \frac{1}{K}=1010$