Question

Let $z\in C$ with $\text{Im}\left(z\right)=10$ and it satisfies
$\frac{2z-n}{2z+n}=2i–1$ for some natural number $n$. Then

• A. $n=20$ and $\text{Re}\left(z\right)=10$
• B. $n=20$ and $\text{Re}\left(z\right)=-10$
• C. $n=40$ and $\text{Re}\left(z\right)=10$
• D. $n=40$ and $\text{Re}\left(z\right)=-10$

Answer 1

The correct option is D $n=40$ and $\text{Re}\left(z\right)=-10$

Let $z=x+10i$
$\Rightarrow \frac{2(x+10i)-n}{2(x+10i)+n}=2i-1\Rightarrow (2x-n)+20i=(2i-1)\left[\right(2x+n)+20i]$
On comparing real and imaginary part, we have -
$2x-n=2(-20)-(2x+n)\text{and}20=2(2x+n)-20\Rightarrow 2x-n=-40-2x-n\text{and}20=4x+2n-20\Rightarrow 4x=-40\text{and}4x+2n=40\Rightarrow x=-10\text{and}-40+2n=40\Rightarrow n=40\Rightarrow n=40\text{and}Re\left(z\right)=-10$