Complex Numbers 1 Question 1
1. Let $z \in C$ with $\operatorname{Im}(z)=10$ and it satisfies $\frac{2 z-n}{2 z+n}=2 i-1$ for some natural number $n$, then (2019 Main, 12 April II)
(a) $n=20$ and $\operatorname{Re}(z)=-10$
(b) $n=40$ and $\operatorname{Re}(z)=10$
(c) $n=40$ and $\operatorname{Re}(z)=-10$
(d) $n=20$ and $\operatorname{Re}(z)=10$
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Answer:
Correct Answer: 1. (c)
Solution:
- Let $z=x+10 i$, as $\operatorname{Im}(z)=10$ (given).
Since $z$ satisfies,
$$ \begin{aligned} \frac{2 z-n}{2 z+n} & =2 i-1, n \in N, \\ \therefore \quad(2 x+20 i-n) & =(2 i-1)(2 x+20 i+n) \\ \Rightarrow(2 x-n)+20 i & =(-2 x-n-40)+(4 x+2 n-20) i \end{aligned} $$
On comparing real and imaginary parts, we get
$$ \begin{array}{rlrl} & & 2 x-n & =-2 x-n-40 \text { and } 20=4 x+2 n-20 \\ \Rightarrow & & 4 x & =-40 \text { and } 4 x+2 n=40 \\ \Rightarrow & & x & =-10 \text { and }-40+2 n=40 \Rightarrow n=40 \\ & \text { So, } \quad n & =40 \text { and } x=\operatorname{Re}(z)=-10 \end{array} $$
