JEE Mains · Maths · STD 11 - 4.1 complex nubers
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 \(Re(z) = -10\)
Answer & Solution
Correct Answer
(C) \(n = 40\) and \(Re(z) = -10\)
Step-by-step Solution
Detailed explanation
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 \(20=2(2 x+n)-20\)…
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