if `(2z-n)/(2z+n)=2i-1,n in N in N` and `IM(z)=10`, then

Options

A. n = 20 and Re (z) = -10
B. n = 40 and Re (z) = 10
C. n = 40 and Re (z) = -10
D. n = 20 and Re (z) = 10

Answer

To solve the problem, we start with the equation:

\[ \frac{2z - n}{2z + n} = 2i - 1 \]

where \( n \in \mathbb{N} \) and \( \text{IM}(z) = 10 \).

### Step 1: Express \( z \) in terms of its real and imaginary parts
Let \( z = x + iy \), where \( x \) is the real part and \( y \) is the imaginary part. Given that \( \text{IM}(z) = 10 \), we have:

\[ z = x + 10i \]

### Step 2: Substitute \( z \) into the equation
Substituting \( z \) into the equation gives:

\[ \frac{2(x + 10i) - n}{2(x + 10i) + n} = 2i - 1 \]

This simplifies to:

\[ \frac{2x + 20i - n}{2x + 20i + n} = 2i - 1 \]

### Step 3: Separate the numerator and denominator
Now, we can rewrite the left-hand side:

\[ \frac{(2x - n) + 20i}{(2x + n) + 20i} \]

### Step 4: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:

\[ (2x - n) + 20i = (2i - 1)((2x + n) + 20i) \]

Expanding the right-hand side:

\[ (2i - 1)(2x + n) + (2i - 1)(20i) \]

Calculating \( (2i - 1)(20i) \):

\[ = 40i - 20 \]

Thus, we have:

\[ (2x - n) + 20i = (2i - 1)(2x + n) + 40i - 20 \]

### Step 5: Expand and combine like terms
Expanding \( (2i - 1)(2x + n) \):

\[ = (4xi + 2ni - 2x - n) \]

Putting it all together:

\[ (2x - n) + 20i = (-2x - n + 40) + (4x + 2n)i \]

### Step 6: Equate real and imaginary parts
From the real parts:

\[ 2x - n = -2x - n + 40 \]

From the imaginary parts:

\[ 20 = 4x + 2n \]

### Step 7: Solve the equations
From the real part equation:

\[ 2x - n = -2x - n + 40 \implies 4x = 40 \implies x = -10 \]

From the imaginary part equation:

\[ 20 = 4(-10) + 2n \implies 20 = -40 + 2n \implies 2n = 60 \implies n = 30 \]

### Conclusion
Thus, we find:

- The real part of \( z \) is \( x = -10 \)
- The value of \( n \) is \( 30 \)

### Final Answer
- Real part of \( z \): \( -10 \)
- Value of \( n \): \( 30 \)