Problem 77
Question
If \(|z+\bar{z}|+|z-\bar{z}|=8\), then \(z\) lies on (A) a circle (B) a straight line (C) a square (D) None of these
Step-by-Step Solution
Verified Answer
(C) a square
1Step 1: Express z and bar(z) in terms of a and b
Let the complex number \( z = a + bi \), where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit. The conjugate of \( z \) is \( \bar{z} = a - bi \).
2Step 2: Calculate the expression z + bar(z) and z - bar(z)
Compute \( z + \bar{z} = (a + bi) + (a - bi) = 2a \) and \( z - \bar{z} = (a + bi) - (a - bi) = 2bi \).
3Step 3: Evaluate the absolute values
Determine \( |z + \bar{z}| = |2a| = 2|a| \) and \( |z - \bar{z}| = |2bi| = 2|b| \).
4Step 4: Apply the given equation
We are given that \( |z+\bar{z}| + |z-\bar{z}| = 8 \). Substitute the expressions from Step 3: \( 2|a| + 2|b| = 8 \).
5Step 5: Simplify the equation
Divide the entire equation by 2 to simplify: \( |a| + |b| = 4 \).
6Step 6: Interpret the geometrical representation
The equation \( |a| + |b| = 4 \) represents a geometric figure in the coordinate plane. It describes a square centered at the origin with vertices at \( (4,0), (-4,0), (0,4), (0,-4) \).
7Step 7: Conclude the figure on which z lies
Since the equation represents a square, \( z \) lies on a square.
Key Concepts
Geometric Interpretation of Complex NumbersModulus of Complex NumbersComplex Conjugate
Geometric Interpretation of Complex Numbers
Complex numbers are fascinating because they combine real numbers with imaginary numbers to create a two-dimensional number system. This enables them to be interpreted geometrically in the Argand plane.
Here, any complex number \( z = a + bi \) can be visualized as a point \((a, b)\) or a vector starting from the origin \((0, 0)\) to \((a, b)\). This geometric representation makes complex numbers very intuitive, as they can directly illustrate operations like addition and subtraction.
The original exercise shows that if \(|z+\bar{z}| + |z-\bar{z}| = 8\), the solutions form a geometric figure, which is found to be a square on the plane. This is derived from interpreting the equation in the context of complex numbers' geometry, demonstrating complex numbers' ability to represent sophisticated geometric shapes.
Here, any complex number \( z = a + bi \) can be visualized as a point \((a, b)\) or a vector starting from the origin \((0, 0)\) to \((a, b)\). This geometric representation makes complex numbers very intuitive, as they can directly illustrate operations like addition and subtraction.
The original exercise shows that if \(|z+\bar{z}| + |z-\bar{z}| = 8\), the solutions form a geometric figure, which is found to be a square on the plane. This is derived from interpreting the equation in the context of complex numbers' geometry, demonstrating complex numbers' ability to represent sophisticated geometric shapes.
Modulus of Complex Numbers
The modulus of a complex number provides its distance from the origin in the complex plane, akin to finding the magnitude of a vector. For a complex number \( z = a + bi \), the modulus is defined as \( |z| = \sqrt{a^2 + b^2} \).
This concept is integral when evaluating expressions like \(|z + \bar{z}|\) and \(|z - \bar{z}|\). In our exercise, the modulus helps simplify the problem. We find that \( |z + \bar{z}| = 2|a| \) and \( |z - \bar{z}| = 2|b| \), leveraging the real and imaginary parts factored by two.
The equation \( |a| + |b| = 4 \) highlights a geometric condition related to these moduli, reinforcing the notion that moduli provide critical insights into the distances and orientations of complex numbers on the complex plane.
This concept is integral when evaluating expressions like \(|z + \bar{z}|\) and \(|z - \bar{z}|\). In our exercise, the modulus helps simplify the problem. We find that \( |z + \bar{z}| = 2|a| \) and \( |z - \bar{z}| = 2|b| \), leveraging the real and imaginary parts factored by two.
The equation \( |a| + |b| = 4 \) highlights a geometric condition related to these moduli, reinforcing the notion that moduli provide critical insights into the distances and orientations of complex numbers on the complex plane.
Complex Conjugate
Complex conjugates are pairs of complex numbers that have the same real part but opposite imaginary parts. For a complex number \( z = a + bi \), the conjugate is \( \bar{z} = a - bi \).
The key property of complex conjugates is that their product is always a real number, specifically \( z \bar{z} = (a + bi)(a - bi) = a^2 + b^2 \). This reveals that the conjugate mechanism neutralizes the imaginary component.
In the exercise, the operations \( z + \bar{z} = 2a \) and \( z - \bar{z} = 2bi \) simplify the problem by isolating these components. Understanding conjugates simplifies solution pathways and allows insights into symmetry and balance within complex numbers.
Therefore, complex conjugates are a powerful tool in manipulating complex numbers, offering paths to simplifying calculations and exploring geometric interpretations.
The key property of complex conjugates is that their product is always a real number, specifically \( z \bar{z} = (a + bi)(a - bi) = a^2 + b^2 \). This reveals that the conjugate mechanism neutralizes the imaginary component.
In the exercise, the operations \( z + \bar{z} = 2a \) and \( z - \bar{z} = 2bi \) simplify the problem by isolating these components. Understanding conjugates simplifies solution pathways and allows insights into symmetry and balance within complex numbers.
Therefore, complex conjugates are a powerful tool in manipulating complex numbers, offering paths to simplifying calculations and exploring geometric interpretations.
Other exercises in this chapter
Problem 75
If \(z^{4}=(z-1)^{4}\), then the roots are represented in the argand plane by the points that are (A) collinear (B) concyclic (C) vertices of a parallelogram (D
View solution Problem 76
The maximum value of \(|z|\) when \(z\) satisfies the condition \(\left|z+\frac{2}{z}\right|=2\) is (A) \(\sqrt{3}-1\) (B) \(\sqrt{3}+1\) (C) \(\sqrt{3}\) (D) \
View solution Problem 78
The complex number which satisfies the equation \(z+\sqrt{2}|z+1|+i=0\) is (A) \(2-i\) (B) \(-2-i\) (C) \(2+i\) (D) \(-2+i\)
View solution Problem 79
\(\tan \left[i \log \frac{a-i b}{a+i b}\right]\) is equal to (A) \(\frac{2 a b}{a^{2}+b^{2}}\) (B) \(\frac{a^{2}-b^{2}}{2 a b}\) (C) \(\frac{2 a b}{a^{2}-b^{2}}
View solution