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Chapter 23: Problem 15

If \(z=x+j y\), determine the locus defined by \(\arg (z-1)=\frac{\pi}{6}\) If \(\arg (z-1)=\frac{\pi}{6}\), then \(\arg (x+j y-1)=\frac{\pi}{6}\) i.e. $$ \arg [(x-1)+j y]=\frac{\pi}{6} $$ In Fig. 25.3, \(\theta=\tan ^{-1}\left(\frac{y}{x}\right)\) i.e. \(\arg z=\tan ^{-1}\left(\frac{y}{x}\right)\) Hence, in this example, \(\tan ^{-1}\left(\frac{y}{x-1}\right)=\frac{\pi}{6}\) i.e. \(\frac{y}{x-1}=\tan \frac{\pi}{6}=\tan 30^{\circ}=\frac{1}{\sqrt{3}}\)Thus, if \(\frac{y}{x-1}=\frac{1}{\sqrt{3}}\), then \(y=\frac{1}{\sqrt{3}}(x-1)\) Hence, the locus of \(\arg (z-1)=\frac{\pi}{6}\) is a straight line \(y=\frac{1}{\sqrt{3}} x-\frac{1}{\sqrt{3}}\)

Short Answer

Expert verified
The locus is the line \( y = \frac{1}{\sqrt{3}} x - \frac{1}{\sqrt{3}} \).

Step by step solution

01

Rewrite the Complex Number

Rewrite the given complex number in terms of real and imaginary components: \[ z = x + jy \]. We are given \( \arg(z - 1) = \frac{\pi}{6} \).
02

Substitute and Simplify

Substitute \( z - 1 = (x - 1) + jy \) into the argument expression. This becomes: \[ \arg((x - 1) + jy) = \frac{\pi}{6} \].
03

Relate to Tangent Function

The argument of a complex number \( a + jb \) is given by \( \tan^{-1}\left(\frac{b}{a}\right) \). Substitute into this formula: \[ \tan^{-1}\left(\frac{y}{x-1}\right) = \frac{\pi}{6} \].
04

Solve for \( y \) in terms of \( x \)

Recognize that \( \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \). Thus, \[ \frac{y}{x-1} = \frac{1}{\sqrt{3}} \], resulting in \[ y = \frac{1}{\sqrt{3}}(x-1) \].
05

Express as a Linear Equation

Substitute \( y = \frac{1}{\sqrt{3}}(x-1) \) to obtain a linear equation: \[ y = \frac{1}{\sqrt{3}}x - \frac{1}{\sqrt{3}} \]. This equation represents a straight line, which is the locus.

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Key Concepts

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

Locus
In mathematics, understanding the concept of a locus is quite essential, especially when dealing with equations and geometric figures. A locus refers to a set of points satisfying a particular condition or a group of conditions. In the context of complex numbers, identifying a locus helps map out positions in the complex plane that fulfill specific characteristics.
For example, in this exercise, we explore the locus of points defined by a complex argument condition: \( \arg(z - 1) = \frac{\pi}{6} \). By analysing this condition, we derived a line in the xy-plane. Here, the line derived from the condition's constraints forms our locus.
This helps us visualize how a simple constraint on complex numbers can translate into a geometric form, reinforcing the strong link between algebra and geometry. Learning to identify loci across different constraint types enriches your problem-solving toolbox.
Argument of Complex Numbers
The argument of a complex number is akin to its directional angle in the complex plane. When we have a complex number represented as \( z = x + jy \), its argument is the angle the line connecting this point to the origin makes with the positive x-axis.
In mathematical terms, for a complex number represented as \( a + jb \), the argument is given by \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \). This places the complex number on a certain path, or direction, within the complex plane.
In this exercise, realizing that \( \arg(z - 1) = \frac{\pi}{6} \) guides us to calculate the slope and characteristics of the line, as transforming arguments directly hints at the orientation or 'tilt' of the locus line. Recognizing how to manipulate the argument of complex numbers can significantly aid in complex plane navigation, providing clarity on how equations evolve with angle conditions.
Equation of a Line
The equation of a line represents a fundamental geometric shape in coordinate systems, and it showcases the relationship between different variables in a linear manner. In two-dimensional space, a line is often represented by a linear equation of the form \( y = mx + b \), where \( m \) is the slope, and \( b \) is the y-intercept.
In the context of this exercise, with the argument condition \( \arg(z - 1) = \frac{\pi}{6} \), we deduced a linear equation from this. The derivation involves setting the tangent of the argument to \( \frac{1}{\sqrt{3}} \), which transforms into the line equation \( y = \frac{1}{\sqrt{3}}x - \frac{1}{\sqrt{3}} \).
Understanding how complex number conditions can redefine a plane into linear equations enhances our capacity to decipher and interpret complex to geometric translations smoothly, proving invaluable in diverse mathematical and practical applications.

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Most popular questions from this chapter

Determine, in polar form, \(\ln (3+j 4)\) $$ \begin{aligned} \ln (3+j 4) &=\ln [5 \angle 0.927]=\ln \left[5 \mathrm{e}^{j 0.927}\right] \\ &=\ln 5+\ln \left(\mathrm{e}^{j 0.927}\right) \\ &=\ln 5+j 0.927 \\ &=1.609+j 0.927 \\ &=\mathbf{1 . 8 5 7} \angle \mathbf{2 9 . 9 5}^{\circ} \text { or } \mathbf{1 . 8 5 7} \angle \mathbf{0 . 5 2 3} \end{aligned} $$

Determine the locus defined by arg \(z=\frac{\pi}{4}\), given that \(z=x+j y\) In Fig. \(23.3\) above, \(\quad \theta=\tan ^{-1}\left(\frac{y}{x}\right)\) where \(\theta\) is called the argument and is written as \(\arg z=\tan ^{-1}\left(\frac{y}{x}\right)\) Figure 23.4] Hence, in this example, \(\tan ^{-1}\left(\frac{y}{x}\right)=\frac{\pi}{4}\) i.e. \(\frac{y}{x}=\tan \frac{\pi}{4}=\tan 45^{\circ}=1\) Thus, \(\quad\) if \(\frac{y}{x}=1\), then \(y=x\) Figure \(23.5\) Hence, the locus (or path) of \(\arg z=\frac{\pi}{4}\) is a straight line \(y=x\) (with \(y>0\) ) as shown in Fig. 23.5.

Change \(6 \mathrm{e}^{2-j 3}\) into \((a+j b)\) form. \(6 \mathrm{e}^{2-j 3}=\left(6 \mathrm{e}^{2}\right)\left(\mathrm{e}^{-j 3}\right)\) by the laws of indices \(=6 \mathrm{e}^{2} \angle-3 \mathrm{rad}\left(\right.\) or \(\left.6 \mathrm{e}^{2} \angle-171.89^{0}\right)\) in polar form $$ \begin{aligned} &=6 \mathrm{e}^{2}[\cos (-3)+j \sin (-3)] \\ &=(-43.89-j 6.26) \text { in }(a+j b) \text { form } \end{aligned} $$

Determine the locus defined by \(|z-2|=3\), given that \(z=x+j y\) If \(\begin{aligned} z=x+j y, \quad \text { then } \quad|z-2|=|x+j y-2| \\\ &=|(x-2)+j y|=3 \end{aligned}\) On the Argand diagram shown in Figure \(23.3\), \(|z|=\sqrt{x^{2}+y^{2}}\) Hence, in this case, \(|z-2|=\sqrt{(x-2)^{2}+y^{2}}=3\) from which, \((x-2)^{2}+y^{2}=3^{2}\) From Chapter \(14,(x-a)^{2}+(y-b)^{2}=r^{2}\) is a circle, with centre \((a, b)\) and radius \(r\). Hence, \((x-2)^{2}+y^{2}=3^{2}\) is a circle, with centre \((2,0)\) and radius 3 The locus of \(|z-2|=3\) is shown in Figure 23.6.

Change \((3-j 4)\) into (a) polar form, (b) exponential form. (a) \((3-j 4)=5 \angle-53.13^{\circ}\) or \(5 \angle-0.927\) in polar form (b) \(\quad(3-j 4)=5 \angle-0.927=5 \mathrm{e}^{-j 0.927}\) in exponential form
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