Question

Let $z \in C$ with $\text{Im}(z) = 10$ and it satisfies $\frac{2z-n}{2z+n} = 2i - 1$, $i = \sqrt{-1}$ for some natural number $n$, then}

• A. n = 20 and $\text{Re}(z) = -10$
• B. n = 40 and $\text{Re}(z) = -10$
• C. n = 40 and $\text{Re}(z) = 10$
• D. n = 20 and $\text{Re}(z) = 10$

Hint:

When solving equations involving complex numbers, always express the complex number in the form $x+iy$ and then equate the real and imaginary parts on both sides of the equation. Remember that $i^2 = -1$.

Answer 1

The Correct Option is B

Step 1: Express $z$ in terms of its real and imaginary parts. We are given that $\text{Im}(z) = 10$. Let $z = x + iy$, where $x = \text{Re}(z)$ and $y = \text{Im}(z)$. Thus, we have $z = x + 10i$.
Step 2: Substitute $z$ into the given equation. The given equation is $\frac{2z-n}{2z+n} = 2i - 1$. Substitute $z = x + 10i$ into the equation: \[ \frac{2(x + 10i) - n}{2(x + 10i) + n} = 2i - 1 \] \[ \frac{(2x - n) + 20i}{(2x + n) + 20i} = -1 + 2i \]
Step 3: Cross-multiply and simplify the equation. Multiply both sides by $((2x + n) + 20i)$: \[ (2x - n) + 20i = (-1 + 2i)((2x + n) + 20i) \] Expand the right side: \[ (2x - n) + 20i = -1(2x + n) - 1(20i) + 2i(2x + n) + 2i(20i) \] \[ (2x - n) + 20i = (-2x - n) - 20i + (4x + 2n)i + 40i^2 \] Since $i^2 = -1$: \[ (2x - n) + 20i = (-2x - n) - 20i + (4x + 2n)i - 40 \] Group the real and imaginary terms on the right side: \[ (2x - n) + 20i = (-2x - n - 40) + (-20 + 4x + 2n)i \]
Step 4: Equate the real and imaginary parts. For two complex numbers to be equal, their real parts must be equal, and their imaginary parts must be equal. Equating the real parts: \[ 2x - n = -2x - n - 40 \] Equating the imaginary parts: \[ 20 = -20 + 4x + 2n \]
Step 5: Solve the system of equations for $x$ and $n$. From the real parts equation: \[ 2x - n = -2x - n - 40 \] Add $2x + n$ to both sides: \[ 4x = -40 \] \[ x = -10 \] So, $\text{Re}(z) = -10$. Now, substitute $x = -10$ into the imaginary parts equation: \[ 20 = -20 + 4(-10) + 2n \] \[ 20 = -20 - 40 + 2n \] \[ 20 = -60 + 2n \] Add 60 to both sides: \[ 80 = 2n \] \[ n = 40 \]
Step 6: State the final answer. We found $x = -10$ and $n = 40$. Therefore, $\text{Re}(z) = -10$ and $n = 40$.