Problem 8
Question
If \(z_{1}\) and \(z_{2}\) are two non-zero complex numbers such that \(\left|z_{1}+z_{2}\right|=\left|z_{1}\right|+\left|z_{2}\right|\), then \(\arg \left(z_{1}\right)-\arg \left(z_{2}\right)\) is equal to (a) \(-\pi\) (b) \(-\pi / 2\) (c) \(\pi / 2\) (d) 0
Step-by-Step Solution
Verified Answer
The argument difference is 0, so (d) is correct.
1Step 1: Understand the Problem
We need to find the difference between the arguments of two complex numbers \(z_{1}\) and \(z_{2}\), given that the magnitude condition \(\left|z_{1} + z_{2}\right| = \left|z_{1}\right| + \left|z_{2}\right|\) holds.
2Step 2: Analyze the Magnitude Condition
The condition \(\left|z_{1} + z_{2}\right| = \left|z_{1}\right| + \left|z_{2}\right|\) implies that \(z_{1}\) and \(z_{2}\) have the same direction, meaning they are collinear and can be expressed as positive multiples of a common direction vector. This occurs if and only if both vectors point in the same direction.
3Step 3: Explore Geometric Interpretation
The magnitudes add up when \(z_{1}\) and \(z_{2}\) point in the same or directly opposite direction. If they point in the same direction, the angle between them is zero. If they point in opposite directions, the angle between them is \(\pi\).
4Step 4: Interpret Argument Difference
For the condition to hold, and the magnitudes to simply add without reducing the net magnitude, \(\arg(z_1)\) and \(\arg(z_2)\) must be such that their vectors point in the same direction. Hence, \(\arg z_1 - \arg z_2 = 0\).
5Step 5: Conclusion
Given the information, the angles must differ by zero for their magnitudes to add without a change in net distance apart from zero.
Key Concepts
Argument of Complex NumbersMagnitude of Complex NumbersCollinearity of Vectors
Argument of Complex Numbers
The argument of a complex number, often referred to as the angle, is a crucial concept in understanding the position of complex numbers on the complex plane. Each complex number can be represented as a point or a vector from the origin in a two-dimensional space, where the horizontal axis represents the real part, and the vertical axis represents the imaginary part.
The argument of a complex number, denoted as \( \arg(z) \), is the angle the vector makes with the positive real axis. It's measured in radians, and for a complex number \( z = a + bi \), the argument \( \theta \) can be found using the arctangent function:
In the context of the exercise, the condition \( \arg(z_1) - \arg(z_2) = 0 \) tells us that the complex numbers \( z_1 \) and \( z_2 \) are pointing in the same direction. Thus, their arguments are equal.
The argument of a complex number, denoted as \( \arg(z) \), is the angle the vector makes with the positive real axis. It's measured in radians, and for a complex number \( z = a + bi \), the argument \( \theta \) can be found using the arctangent function:
- \( \theta = \tan^{-1}(\frac{b}{a}) \)
In the context of the exercise, the condition \( \arg(z_1) - \arg(z_2) = 0 \) tells us that the complex numbers \( z_1 \) and \( z_2 \) are pointing in the same direction. Thus, their arguments are equal.
Magnitude of Complex Numbers
The magnitude, or the modulus, of a complex number measures how far the number is from the origin in the complex plane. For a complex number \( z = a + bi \), its magnitude is calculated as:
In our exercise, it's crucial to note that the equality \( \left|z_1 + z_2\right| = \left|z_1\right| + \left|z_2\right| \) occurs under a very specific condition. This condition tells us that the vectors \( z_1 \) and \( z_2 \) are in the same line, meaning they are collinear. When two vectors point in the same direction or are directly opposite to each other, their magnitudes mathematically add or subtract perfectly due to their directional alignment.
This property is leveraged in the problem to determine the relations and difference of their arguments.
- \( |z| = \sqrt{a^2 + b^2} \)
In our exercise, it's crucial to note that the equality \( \left|z_1 + z_2\right| = \left|z_1\right| + \left|z_2\right| \) occurs under a very specific condition. This condition tells us that the vectors \( z_1 \) and \( z_2 \) are in the same line, meaning they are collinear. When two vectors point in the same direction or are directly opposite to each other, their magnitudes mathematically add or subtract perfectly due to their directional alignment.
This property is leveraged in the problem to determine the relations and difference of their arguments.
Collinearity of Vectors
Collinearity in vectors means that vectors lie along the same straight line. They can be expressed as scalar multiples of one another. For complex numbers viewed as vectors on the complex plane, collinearity implies that their direction vectors align perfectly, either in the same direction or opposite directions.
A critical take-away from the exercise is the condition \( \left|z_1 + z_2\right| = \left|z_1\right| + \left|z_2\right| \). This shows that \( z_1 \) and \( z_2 \) must be pointing in the same direction for their magnitudes to add perfectly without cancellation through vector subtraction.
A critical take-away from the exercise is the condition \( \left|z_1 + z_2\right| = \left|z_1\right| + \left|z_2\right| \). This shows that \( z_1 \) and \( z_2 \) must be pointing in the same direction for their magnitudes to add perfectly without cancellation through vector subtraction.
- If they point in the same direction, the difference in their arguments is zero.
- If they point in opposite directions, the difference would be \( \pi \) (although not the case here).
Other exercises in this chapter
Problem 8
If \(\omega\) is one cube root of unity, then prove that \(\left(1-\omega+\omega^{2}\right)^{5}+\left(1+\omega-\omega^{2}\right)^{5}=32\)
View solution Problem 8
If \(1, \omega, \omega^{2}\) are the cube roots of unity, then prove that \((1-\omega)^{3}-\left(1+\omega^{2}\right)^{3}=0 .\)
View solution Problem 9
Find the cube roots of \(27 .\)
View solution Problem 9
If \(z\) and \(\omega\) are two non-zero complex numbers such that \(|\mathrm{z}|=|\omega|\) and \(\arg (z)+\arg\) \((\omega)=\pi\), then \(z\) is equal to (a)
View solution