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4-1.Complex numbers
hard
Let $z \in C$ with $Im(z) = 10$ and it satisfies $\frac{{2z - n}}{{2z + n}} = 2i - 1$ for some natural number $n$. Then
A
$n = 40$ and $Re(z) = 10$
B
$n = 20$ and $Re(z) = 10$
C
$n = 40$ and $Re(z) = -10$
D
$n = 20$ and $Re(z) = -10$
(JEE MAIN-2019)
Solution
Let $z=x+10 i$
given $\frac{2 z-n}{2 z+n}=2 i-1$
$\Rightarrow \frac{2(x+10 i)-n}{2(x+10 i)+n}=2 i-1$
$\Rightarrow(2 x-n)+20 i=(2 i-1)[(2 x+n)+20 i]$
Comparing real and imaginary part
$\Rightarrow 2 x-n=2(-20)-(2 x+n)$ and $20=2(2 x+n)-20$
$\Rightarrow 2 x-n=-40-2 x-n$ and $20=4 x+2 n-20$
$\Rightarrow 4 x=-40$ and $4 x+2 n=40$
$\Rightarrow x=-10$ and $-40+2 n=40$
$\Rightarrow n=40$
$\Rightarrow n=40$ and $\operatorname{Re}(z)=-10$
Standard 11
Mathematics
