Problem 108
Question
Let \(\bar{b} z+b \bar{z}=c, b \neq 0\), be a line in the complex plane, where \(\bar{b}\) is the complex conjugate of \(b .\) If a point \(z_{1}\) is the reflection of a point \(z_{2}\) through the line, then \(\bar{z}_{1} b+z_{2} \bar{b}=\) (A) \(4 c\) (B) \(2 c\) (C) \(_{C}\) (D) None of these
Step-by-Step Solution
Verified Answer
Option (B) \( 2c \).
1Step 1: Express line equation
The given equation of the line is \( \bar{b}z + b\bar{z} = c \)."
2Step 2: Reflect a point in a line
If a point \( z_1 \) is the reflection of \( z_2 \) across the line, the midpoint \( M \) between \( z_1 \) and \( z_2 \) should be on the line. Therefore, \( M = \frac{z_1 + z_2}{2} \) satisfies \( \bar{b}M + b\bar{M} = c \)."
3Step 3: Substitute midpoint into line equation
Substitute \( M = \frac{z_1 + z_2}{2} \) into the line equation: \[ \bar{b} \left(\frac{z_1 + z_2}{2}\right) + b \left(\frac{\bar{z}_1 + \bar{z}_2}{2}\right) = c \]. Simplifying, we obtain \( \frac{\bar{b}(z_1 + z_2) + b(\bar{z}_1 + \bar{z}_2)}{2} = c \)."
4Step 4: Solve for conjugate sums
Multiply through by 2 and rearrange terms to isolate \( \bar{z}_1 b + z_2 \bar{b} \):\[ \bar{b} z_1 + \bar{b} z_2 + b \bar{z}_1 + b \bar{z}_2 = 2c \].Since the conjugates reflect symmetrically, rearranging gives \( \bar{z}_1 b + z_2 \bar{b} = 2c \).
Key Concepts
Line Equation in Complex NumbersReflection Point in GeometryComplex Conjugate Properties
Line Equation in Complex Numbers
Understanding line equations in the complex plane requires a shift from traditional 2D Cartesian coordinates to our friend, the complex number. A complex number, denoted by \( z = x + yi \), has a real part \( x \) and an imaginary part \( y \). The beauty of complex numbers is that they inherently hold both horizontal and vertical dimensions.
To describe a line in this plane, we leverage complex conjugates. The equation \( \bar{b}z + b\bar{z} = c \), where \( \bar{b} \) is the conjugate of complex \( b \), effectively behaves as a linear equation.
Here's how it works:
To describe a line in this plane, we leverage complex conjugates. The equation \( \bar{b}z + b\bar{z} = c \), where \( \bar{b} \) is the conjugate of complex \( b \), effectively behaves as a linear equation.
Here's how it works:
- Each term \( \bar{b}z \) and \( b\bar{z} \) contributes to forming what can be interpreted as a 'slope'.
- These terms use the complex conjugate, which flips the sign of the imaginary part, thereby serving the role of y-interchange.
Reflection Point in Geometry
Reflecting a point across a line in the complex plane is an elegant process reminiscent of finding reflections in regular Euclidean geometry, but with a complex twist. When a point \( z_1 \) is the reflection of a point \( z_2 \) across a line, the midpoint between these two points lies exactly on the line.
The midpoint \( M \) is calculated using \( M = \frac{z_1 + z_2}{2} \). This midpoint ensures symmetrical distances from the line.
In terms of steps:
The midpoint \( M \) is calculated using \( M = \frac{z_1 + z_2}{2} \). This midpoint ensures symmetrical distances from the line.
In terms of steps:
- Verify the midpoint is on the line by substituting \( M \) into the line equation.
- Ensure the equation \( \bar{b}M + b\bar{M} = c \) holds true to confirm symmetry.
- Use this midpoint relationship to express how \( z_1 \) and \( z_2 \) are linked through the line.
Complex Conjugate Properties
The complex conjugate, notated as \( \bar{z} \) for a complex number \( z = x + yi \), has an intriguing property: it flips the imaginary part, converting \( yi \) to \(-yi \). This subtle transformation becomes quite powerful in geometric interpretations.
Here’s a quick breakdown:
Here’s a quick breakdown:
- The conjugate maintains the modulus \( |z| \) of the complex number, effectively preserving its 'magnitude'.
- The conjugate operation reflects the complex number across the real axis—this reflection is crucial in determining the geometric positions of points.
- Properties like \( z\bar{z} = x^2 + y^2 \) offer tools for exploring and simplifying expressions involving reflection without losing generality.
Other exercises in this chapter
Problem 105
If \(|z-i|=1\) and \(\arg (z)=\theta, \theta \in\left(0, \frac{\pi}{2}\right)\), then the value of \(\cot \theta-\frac{2}{z}\) is equal to (A) 0 (B) \(i\) (C) \
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If \(i=\sqrt{-1}\), then \(4+5\left(-\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)^{334}+3\left(-\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)^{365}\) is equal to (A) \(1-i
View solution Problem 109
Let \(z_{1}\) and \(z_{2}\) be roots of the equation \(z^{2}+p z+q=0\), where the coefficients \(p\) and \(q\) may be complex numbers. Let \(A\) and \(B\) repre
View solution Problem 111
If \(z_{1}, z_{2}, z_{3}\) are complex numbers such that \(\left|z_{1}\right|=\left|z_{2}\right|=\) \(\left|z_{3}\right|=\left|\frac{1}{z_{1}}+\frac{1}{z_{2}}+\
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