Question

Let z ∈ C with Im(z) = 10 and it satisfies \(\frac{{2z - n}}{{2z + n}} = 2i - 1\) for some natural number n. Then:
1. n = 20 and Re(z) = -10
2. n = 40 and Re(z) = 10
3. n = 40 and Re(z) = -10
4. n = 20 and Re(z) = 10

Answer 1

Correct Answer - Option 3 : n = 40 and Re(z) = -10

From question, ‘z’ belongs to complex number.

z = x + iy

From question, Im(z) = 10,

⇒ z = x + 10i

From question,

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

Now,

\(\Rightarrow \frac{{2\left( {x + 10i} \right) - n}}{{2\left( {x + 10i} \right) + n}} = - 1 + 2i\) 

\(\Rightarrow \frac{{\left( {2x - n} \right) + 20i}}{{\left( {2x + n} \right) + 20i}} = - 1 + 2i\) 

⇒ (2x – n) + 20i = (-1 + 2i)((2x + n) + 20i)

⇒ (2x – n) + 20i = -2x – n – 20i + 4xi + 2ni + 40i²

∵ [i² = 1]

⇒ (2x – n) + 20i = -2x – n + (-20 + 4x + 2n)i – 40

⇒ (2x – n) + 20i = (-2x – n – 40) + (-20 + 4x + 2n)i

On equating real part on both sides,

⇒ 2x – n = -2x – n – 40

⇒ 4x = -40

⇒ x = -10

∴ Re(z) = -10

On equating imaginary part on both sides,

⇒ 20 = -20 + 4x + 2n

⇒ 40 = 4(-10) + 2n

⇒ 2n = 80

∴ n = 40