Complex Numbers
Complex Number in Iota Form
Complex Numbers_PYQ
Grade 11
Question:
Let $z \in \mathbb{C}$ with $\text{Im}(z) = 10$ and it satisfies $\dfrac{2z - n}{2z + n} = 2i - 1$ for some natural number $n$, then
$n = 20$ and $\text{Re}(z) = -10$
$n = 40$ and $\text{Re}(z) = 10$
$n = 40$ and $\text{Re}(z) = -10$
$n = 20$ and $\text{Re}(z) = 10$
Step-by-Step Solution
Key Concept: Equating real and imaginary parts separately after cross-multiplication reduces a complex equation to a 2×2 real system.
**Step 1: Set up with z = x + 10i**
Let $z = x + 10i$. Cross-multiplying: $2z - n = (2i - 1)(2z + n)$.
**Step 2: Expand and equate real parts**
$2x - n + 20i = (2i-1)(2x+n+20i)$. Real part of RHS: $-(2x+n) - 40 = -(2x+n+40)$. So $2x - n = -(2x+n) - 40 \Rightarrow 4x = -40 \Rightarrow x = -10$, i.e., $\text{Re}(z) = -10$.
**Step 3: Equate imaginary parts to find n**
Imaginary part of RHS: $2(2x+n) - 20$. Setting equal to $20$: $2(2(-10)+n) = 40 \Rightarrow 2n - 40 = 40 \Rightarrow n = 40$.
Correct Answer: 3