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Chapter 3: Problem 197

Let \(z, w\) be complex numbers such that \(\bar{z}+i \bar{w}=0\) and \(\arg z w=\pi\). Then \(\arg z\) equals \([2004]\) (A) \(\frac{\pi}{4}\) (B) \(\frac{5 \pi}{4}\) (C) \(\frac{3 \pi}{4}\) (D) \(\frac{\pi}{2}\)

Short Answer

Expert verified
The argument \(\arg z = \frac{3\pi}{4}\) (Option C).

Step by step solution

01

Analyze the First Equation

Given the equation \(\bar{z} + i\bar{w} = 0\). This implies that \(\bar{z} = -i\bar{w}\). Taking the complex conjugate of both sides gives \(z = i w\). This tells us that the complex numbers \(z\) and \(w\) are related by multiplication by the imaginary unit \(i\).
02

Analyze the Argument of zw

We know that \(\arg(zw) = \pi\). If \(z = re^{i\theta_1}\) and \(w = se^{i\theta_2}\), then \(zw = rse^{i(\theta_1 + \theta_2)}\). The argument condition \(\theta_1 + \theta_2 = \pi\) holds based on the argument property.
03

Substitute Relations into Argument Equation

Based on Step 1, substitute \(z = iw\) into the argument condition. Then if \(w = se^{i\theta_2}\), we have \(z = is e^{i\theta_2} = se^{i(\theta_2 + \frac{\pi}{2})}\), which gives \(\theta_1 = \theta_2 + \frac{\pi}{2}\).
04

Solve the Argument Equation

Using the relation from Step 2, substitute \(\theta_1 = \theta_2 + \frac{\pi}{2}\) into the equation \(\theta_1 + \theta_2 = \pi\). This yields \(\theta_2 + \frac{\pi}{2} + \theta_2 = \pi\).
05

Simplify and Solve

Simplify the equation from Step 4: \(2\theta_2 + \frac{\pi}{2} = \pi\). Subtract \(\frac{\pi}{2}\) from both sides to get \(2\theta_2 = \frac{\pi}{2}\). Dividing both sides by 2 gives \(\theta_2 = \frac{\pi}{4}\). Substituting back, \(\theta_1 = \frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4}\).
06

Conclusion: Determine \(\arg z\)

The argument of \(z\) ends up being \(\arg z = \frac{3\pi}{4}\), which corresponds to option \(C\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Argument of Complex Numbers
The argument of a complex number is an important concept that helps us understand the direction or orientation of the complex number on the complex plane. A complex number, usually expressed in the form \(z = a + bi\), can also be written in polar form as \(z = r e^{i\theta}\), where \(r\) is the magnitude, and \(\theta\) is the argument.
The argument \(\theta\) represents the angle the line joining the origin to the point \((a, b)\) makes with the positive x-axis.
  • It is measured in radians.
  • The argument can be adjusted to be within any interval of \(2\pi\), usually \([0, 2\pi)\) or \((-\pi, \pi]\).
Understanding the argument is crucial in determining solutions to problems involving orientations, like the given exercise where we deduced \(\arg(zw) = \pi\), indicating the product lies along the negative real axis.
Complex Conjugate
The complex conjugate of a complex number is another essential concept, often used for simplifying complex expressions and solving complex equations. Given a complex number \(z = a + bi\), its conjugate is denoted \(\bar{z}\) and is defined as \(\bar{z} = a - bi\).
  • The conjugate flips the sign of the imaginary component while keeping the real component unchanged.
  • Multiplying a complex number by its conjugate yields a real number: \(z \cdot \bar{z} = a^2 + b^2\).
In the exercise, we used the property that \(\bar{z} + i\bar{w} = 0\) to establish a relationship with the imaginary unit \(i\), showing how complex conjugation helps develop solutions to equations involving imaginary components.
Multiplication of Complex Numbers
Multiplying complex numbers involves combining both their magnitudes and arguments. If \(z = r_1 e^{i\theta_1}\) and \(w = r_2 e^{i\theta_2}\), then their product is given by \(zw = r_1 r_2 e^{i(\theta_1 + \theta_2)}\).
  • The magnitudes multiply as \(r_1 \times r_2\).
  • The arguments add up: \(\theta_1 + \theta_2\).
This rule simplifies many computations and helps in visually predicting the position of the resulting product in the complex plane.
In our exercise's context, knowing that \(\arg(zw) = \pi\) allowed us to find \(\theta_1 + \theta_2 = \pi\). This was key in determining the specific argument of the number \(z\).

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Most popular questions from this chapter

If \(z_{1}, z_{2}, z_{3}\) are the vertices of an equilateral triangle in the complex plane and \(z_{0}\) is the centroid, then (A) \(\frac{1}{z_{1}-z_{2}}+\frac{1}{z_{2}-z_{3}}+\frac{1}{z_{3}-z_{1}}=0\) (B) \(\left(z_{1}-z_{2}\right)^{2}+\left(z_{2}-z_{3}\right)^{2}+\left(z_{3}-z_{1}\right)^{2}=0\) (C) \(z_{1}^{2}+z_{2}^{2}+z_{3}^{2}=3 z_{0}^{2}\) (D) \(z_{1}^{2}+z_{2}^{2}+z_{3}^{2}=z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}\)

If \(z_{1}, z_{2}, z_{3}\) are complex numbers such that \(\left|z_{1}\right|=\left|z_{2}\right|=\) \(\left|z_{3}\right|=\left|\frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}}\right|=1\), then \(\left|z_{1}+z_{2}+z_{3}\right|\) is (A) equal to 1 (B) less than 1 (C) greater than 3 (D) equal to 3

The roots of the equation \(x^{6}+x^{3}+1=0\) are \(\cos \left(\frac{p \pi}{9}\right) \pm i \sin \left(\frac{p \pi}{9}\right)\), where \(p=\) (A) 2 (B) 8 (C) 14 (D) 20

If \(z\) satisfies \(|z+1|<|z-2|\), and \(\omega=3 z+2+i\), then (A) \(|\omega+1|<|\omega-8|\) (B) \(|\omega+1|<|\omega-7|\) (C) \(\omega+\bar{\omega}>7\) (D) \(|\omega+5|<|\omega-4|\)

If \(\omega\) is an imaginary cube root of unity, then \((1+\omega\) \(\left.-\omega^{2}\right)^{7}\) equals: (A) \(128 \omega\) (B) \(-128 \omega\) (C) \(128 \omega^{2}\) (D) \(-128 \omega^{2}\)
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