The problem asks us to find a point \(z_2 = x + iy\) such that a given point \(z_1 = 1 + i\) is the reflection of \(z_2\) in the line defined by the equation \(i\bar{z} - iz = 5\). Reflection involves understanding the geometric relationship between the original point, the reflecting line, and the reflected point.

Understanding the Reflecting Line

The given equation of the line is \(i\bar{z} - iz = 5\). To understand this line geometrically, let's substitute \(z = x + iy\) and its conjugate \(\bar{z} = x - iy\).

Substituting these into the equation:

\(i(x - iy) - i(x + iy) = 5\)

\(ix - i^2y - ix - i^2y = 5\)

Since \(i^2 = -1\), this simplifies to:

\(ix - (-1)y - ix - (-1)y = 5\)

\(ix + y - ix + y = 5\)

\(2y = 5\)

\(y = \frac{5}{2}\)

So, the reflecting line is the horizontal line \(y = \frac{5}{2}\) in the complex plane.

Reflection in a Horizontal Line

When a point \(z_2 = x + iy\) (corresponding to point \((x, y)\)) is reflected in a horizontal line \(y = c\), the reflected point \(z_1 = x' + iy'\) (corresponding to point \((x', y')\)) has the following properties:

• The x-coordinate remains unchanged: \(x' = x\).
• The y-coordinate is transformed such that the midpoint of the y-coordinates is \(c\). That is, \(\frac{y + y'}{2} = c\), which means \(y' = 2c - y\).

Applying Reflection to Find \(z_2\)

We are given that \(z_1 = 1 + i\) is the reflection of \(z_2 = x + iy\) in the line \(y = \frac{5}{2}\). The point \(z_1 = 1 + i\) corresponds to the coordinate \((1, 1)\). The point \(z_2 = x + iy\) corresponds to the coordinate \((x, y)\). The reflecting line is \(y = \frac{5}{2}\), so \(c = \frac{5}{2}\).

Using the reflection properties:

• The x-coordinate of the reflected point \(z_1\) must be equal to the x-coordinate of the original point \(z_2\). The x-coordinate of \(z_1 = 1 + i\) is 1. The x-coordinate of \(z_2 = x + iy\) is \(x\). Therefore, \(x = 1\).
• The y-coordinate of the reflected point \(z_1\) is related to the y-coordinate of the original point \(z_2\) and the line constant \(c\) by \(y' = 2c - y\). The y-coordinate of \(z_1 = 1 + i\) is 1. The y-coordinate of \(z_2 = x + iy\) is \(y\). The line constant is \(c = \frac{5}{2}\). Therefore, \(1 = 2\left(\frac{5}{2}\right) - y\).

Now, let's solve for \(y\):

\(1 = 5 - y\)

\(y = 5 - 1\)

\(y = 4\)

So, the coordinates of the point \(z_2\) are \((x, y) = (1, 4)\). This corresponds to the complex number \(z_2 = 1 + 4i\).

Verification

Let's check if \(z_1 = 1 + i\) is indeed the reflection of \(z_2 = 1 + 4i\) in the line \(y = 5/2\).

• The midpoint of the segment connecting \((1, 1)\) and \((1, 4)\) is \(\left(\frac{1+1}{2}, \frac{1+4}{2}\right) = \left(\frac{2}{2}, \frac{5}{2}\right) = \left(1, \frac{5}{2}\right)\). The y-coordinate of the midpoint is \(5/2\), which lies on the line \(y = 5/2\). This confirms the midpoint property.
• The segment connecting \((1, 1)\) and \((1, 4)\) is a vertical line segment. The reflecting line \(y = 5/2\) is a horizontal line. A vertical segment is perpendicular to a horizontal line. This confirms the perpendicularity property.

Both properties hold true, confirming that \(z_2 = 1 + 4i\) is the correct point.

Step-by-Step Solution Summary

1. Identify the given points: \(z_1 = 1+i\) and \(z_2 = x+iy\).
2. Identify the reflecting line equation: \(i\bar{z} - iz = 5\).
3. Convert the line equation to its Cartesian form \(y=c\) by substituting \(z = x+iy\). The line is \(y = 5/2\).
4. Recall the properties of reflection across a horizontal line \(y=c\): the x-coordinate is unchanged, and the y-coordinate \(y'\) of the reflected point relates to the original y-coordinate \(y\) by \(y' = 2c - y\).
5. Apply these properties using \(z_1 = 1+i\) as the reflected point \((x', y') = (1, 1)\) and \(z_2 = x+iy\) as the original point \((x, y)\) and \(c = 5/2\).
6. From the x-coordinate property, \(x = x'\), so \(x = 1\).
7. From the y-coordinate property, \(y' = 2c - y\), so \(1 = 2(5/2) - y\).
8. Solve the equation \(1 = 5 - y\) for \(y\), which gives \(y = 4\).
9. The point \(z_2\) is \(x + iy = 1 + 4i\).

Revision Table: Key Concepts for Reflection

Additional Information: Complex Numbers and Lines

The equation of a line in the complex plane can be represented in various forms. The general form is often given as \(a\bar{z} + \bar{a}z = c\), where \(a\) is a complex number related to the normal vector of the line and \(c\) is a real constant. In our case, the equation \(i\bar{z} - iz = 5\) can be rewritten.

Comparing \(i\bar{z} - iz = 5\) with \(a\bar{z} + \bar{a}z = c\):

Here, \(a = i\) and \(\bar{a} = -i\). The constant is \(c = 5\). So the form is \(i\bar{z} + (-i)z = 5\), which matches the given equation.

The normal vector to the line \(a\bar{z} + \bar{a}z = c\) is proportional to \(a\). In this case, \(a = i\), which corresponds to the direction \((0, 1)\) in the Cartesian plane. A line with a normal vector in the direction \((0, 1)\) is a horizontal line, which aligns with our finding that the line is \(y = 5/2\).

Reflection in a line \(a\bar{z} + \bar{a}z = c\) maps a point \(z\) to \(z'\) such that \(\frac{z' + z}{2}\) is on the line and \((z' - z)\) is parallel to \(a\). However, for simple horizontal or vertical lines like \(y=c\) or \(x=c\), using the Cartesian coordinate geometry approach (as demonstrated above) is often more intuitive and straightforward.

For a horizontal line \(y=c\), the direction \(a\) is purely imaginary (like \(i\) or \(-i\)). For a vertical line \(x=c\), the direction \(a\) is purely real (like \(1\) or \(-1\)). Our line \(y = 5/2\) is horizontal, which is consistent with the normal vector being in the imaginary direction (\(a=i\)).

The final answer is \(1 + 4i\).