SATHEE: Complex Numbers 1 Question 1

Complex Numbers 1 Question 1

1. 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 (2019 Main, 12 April II)

(a) $n = 20$ and $Re (z) = - 10$

(b) $n = 40$ and $Re (z) = 10$

(d) $n = 20$ and $Re (z) = 10$

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Answer:

Correct Answer: 1. (c)

Solution:

Let $z = x + 10 i$ , as $Im (z) = 10$ (given).

Since $z$ satisfies,

$\begin{aligned} \frac{2 z - n}{2 z + n} & = 2 i - 1, n \in N, \\ ∴ (2 x + 20 i - n) & = (2 i - 1) (2 x + 20 i + n) \\ \Rightarrow (2 x - n) + 20 i & = (- 2 x - n - 40) + (4 x + 2 n - 20) i \end{aligned}$

On comparing real and imaginary parts, we get

$\begin{aligned} 2 x - n & = - 2 x - n - 40 and 20 = 4 x + 2 n - 20 \\ \Rightarrow & 4 x & = - 40 and 4 x + 2 n = 40 \\ \Rightarrow & x & = - 10 and - 40 + 2 n = 40 \Rightarrow n = 40 \\ So, n & = 40 and x = Re (z) = - 10 \end{aligned}$

Complex Numbers 1 Question 1

1. 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 (2019 Main, 12 April II)

Answer:

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Complex Numbers 1 Question 1

1. Let z∈C with Im(z)=10 and it satisfies 2z−n2z+n=2i−1 for some natural number n, then (2019 Main, 12 April II)

Answer:

Engineering Preparation

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NCERT Exercise Video Solution

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1. 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 (2019 Main, 12 April II)