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Step-by-Step Solution
Step 1: Express z in terms of x and y
Let the complex number z be written as
z = x + i y .
Step 2: Use the given modulus condition
The condition
\left|\frac{z + i}{z - 3i}\right| = 1
implies that the distance of the point z from -i is the same as the distance of z from 3i . Geometrically, this suggests that z lies on the perpendicular bisector of the line segment joining -i and 3i .
If we write z = x + i y , then
|z + i| = |x + i(y+1)| and |z - 3i| = |x + i(y-3)|.
The equality
|x + i(y+1)| = |x + i(y-3)|
implies y+1 = -(y-3) or y+1 = y-3 when we consider their imaginary parts squared and real parts squared to be equal. Solving properly shows that y = 1.
Step 3: Substitute y = 1 into z
Hence, z = x + i.
Step 4: Express \omega in terms of x and y
We have
\omega = z \overline{z} \;-\; 2z \;+\; 2.
Since z = x + i, its conjugate is \overline{z} = x - i. Therefore,
\[
z \overline{z} = (x + i)(x - i) = x^2 - i^2 = x^2 + 1.
\]
Hence,
\[
\omega = (x^2 + 1) - 2(x + i) + 2 = x^2 + 1 - 2x - 2i + 2.
\]
So,
\[
\omega = x^2 - 2x + 3 - 2i.
\]
Step 5: Find the minimum value of Re( \omega )
The real part of \omega is
\mathrm{Re}(\omega) = x^2 - 2x + 3.
We can rewrite this as
\[
x^2 - 2x + 3 = (x - 1)^2 + 2.
\]
This expression attains its minimum value when (x - 1)^2 = 0, i.e. x = 1.
Step 6: Substitute x = 1 to find the corresponding \omega
When x = 1, we have
z = 1 + i.
Then
\[
\omega = (1 + i)(1 - i) - 2(1 + i) + 2.
\]
First, compute (1 + i)(1 - i) = 1 - i^2 = 1 + 1 = 2. Thus,
\[
\omega = 2 - 2(1 + i) + 2 = 2 - 2 - 2i + 2 = 2 - 2i.
\]
Factor out 2:
\[
\omega = 2 (1 - i).
\]
Step 7: Express \omega in polar form
Observe that
1 - i = \sqrt{2} e^{-i \pi/4},
so
\[
\omega = 2 \sqrt{2} e^{-i \pi/4}.
\]
Step 8: Determine when \omega^n is real
Consider
\[
\omega^n = \left(2 \sqrt{2} e^{-i \pi/4}\right)^n = 2^n (\sqrt{2})^n \, e^{-\,i\,n\,\pi/4} = 2^{n + \frac{n}{2}} \, e^{-\,i\,n\,\pi/4} = 2^{\frac{3n}{2}} \, e^{-\,i\,\frac{n\pi}{4}}.
\]
For \omega^n to be real, the exponential part
e^{-\,i\,\frac{n\pi}{4}}
must be real. This happens if
\[
-\frac{n\pi}{4} = k\pi \quad \text{for some integer } k,
\]
which simplifies to
\[
\frac{n}{4} = -k \quad \Rightarrow \quad n = 4 |k|.
\]
Since n \in \mathbb{N}, the smallest positive value of n for which \omega^n is real is
\[
n = 4.
\]
Final Answer
\boxed{4}
