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

(C) $n=40$ and $\text{Re}(z)=-10$

(A) $n=40$ and $\text{Re}(z)=10$

(B) $n=20$ and $\text{Re}(z)=10$

(D) $n=20$ and $\text{Re}(z)=-10$

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Complex Numbers

Let

z \in C

with

Im (z) = 10

and it satisfies

\frac{2 z - n}{2 z + n} = 2 i - 1

for some natural number

n

. Then :

Answer: (C)

Clear the fraction in

\frac{2 z - n}{2 z + n} = 2 i - 1

2 z - n = (2 i - 1) (2 z + n)

. Expanding and collecting,

4 z = 4 i z + 2 in

, so

z = \frac{in}{2 ( 1 - i )} = \frac{in ( 1 + i )}{4} = \frac{n ( i - 1 )}{4}

. Hence

Re (z) = - \frac{n}{4}

and

Im (z) = \frac{n}{4}

. Since

Im (z) = 10

n = 40

, and then

Re (z) = - 10

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Single correct option