Problem 38
Question
By the quadratic formula, \(e^{z}=-\frac{1}{2}+\frac{\sqrt{3}}{2} i\) or \(e^{z}=-\frac{1}{2}-\frac{\sqrt{3}}{2} i .\) Hence $$z=\ln \left(-\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)=\left(\frac{2 \pi}{3}+2 n \pi\right) i \quad \text { or } \quad z=\ln \left(-\frac{1}{2}-\frac{\sqrt{3}}{2} i\right)=\left(\frac{4 \pi}{3}+2 n \pi\right) i$$
Step-by-Step Solution
Verified Answer
\(z=\frac{2\pi i}{3}+2n\pi i\) or \(z=\frac{4\pi i}{3}+2n\pi i\).
1Step 1: Analyze Given Exponential Equation
Recognize that we have two separate solutions given by the exponential equation, either \(e^{z} = -\frac{1}{2} + \frac{\sqrt{3}}{2} i\) or \(e^{z} = -\frac{1}{2} - \frac{\sqrt{3}}{2} i\). These expressions involve complex numbers in polar form.
2Step 2: Convert Exponential Form to Logarithmic Form
We need to find \(z\) by converting the exponential expressions into logarithmic form: \(z = \ln(-\frac{1}{2} + \frac{\sqrt{3}}{2} i)\) and \(z = \ln(-\frac{1}{2} - \frac{\sqrt{3}}{2} i)\).
3Step 3: Express the Complex Number in Polar Form
Note that \(-\frac{1}{2} + \frac{\sqrt{3}}{2} i\) represents a point on the unit circle at an angle of \(\frac{2\pi}{3}\) radians (120 degrees). Similarly, \(-\frac{1}{2} - \frac{\sqrt{3}}{2} i\) is at \(\frac{4\pi}{3}\) radians (240 degrees).
4Step 4: Use the Logarithm of a Complex Number Formula
The logarithm of a complex number given by \(re^{i\theta}\) is \(\ln|r| + i\theta\). Here \(|r| = 1\), so the logarithm reduces to simply \(z = i\theta\). So, \(z = i\left(\frac{2\pi}{3} + 2n\pi\right)\) for the first expression, and \(z = i\left(\frac{4\pi}{3} + 2n\pi\right)\) for the second, where \(n\) is an integer.
Key Concepts
Exponential FunctionsLogarithmic FunctionsPolar Coordinates
Exponential Functions
Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. They are commonly used in various fields, such as finance, physics, and biology, to describe growth and decay processes. In the realm of complex numbers, exponential functions become particularly interesting. When dealing with complex exponents, we use Euler's formula:
This interplay between exponential functions and trigonometry is foundational to understanding complex analysis.
- The formula states that for any real number \( \theta \), \( e^{i\theta} = \cos(\theta) + i\sin(\theta) \).
- This converts complex exponential equations into trigonometric form, which can be graphically represented as a point on the unit circle.
This interplay between exponential functions and trigonometry is foundational to understanding complex analysis.
Logarithmic Functions
Logarithmic functions are the inverses of exponential functions. They play an essential role in calculating the continuous growth rates or comprehending the natural scales involved in phenomena. In complex analysis, logarithms stretch beyond real numbers and help in determining the angle or phase of a complex number. For a complex number in polar form, \( re^{i\theta} \), its logarithm is expressed as:
- \( \ln(z) = \ln|r| + i\theta \)
- where \(|r|\) is the magnitude (or modulus) of the complex number, and \(\theta\) is the angle (or argument).
Polar Coordinates
Polar coordinates are a two-dimensional coordinate system where each point is determined by a distance from a reference point and an angle from a reference direction. This system is especially useful in representing complex numbers because each complex number corresponds naturally to a radius and angle.
- The polar form of a complex number is \( z = re^{i\theta} \), where \(r\) is the modulus and \(\theta\) is the argument.
- This form simplifies operations like multiplication or division, reducing them to addition or subtraction of angles.
- Moreover, polar coordinates shine in visualizing complex numbers as points or vectors originating from the origin.
Other exercises in this chapter
Problem 36
\(x^{2}-y^{2}-4 x+(-2 x y-4 y) i=0+0 i\) implies \(x^{2}-y^{2}-4 x=0\) and \(y(-2 x-4)=0 .\) If \(y=0\) then \(x(x-4)=0\) and so \(z=0\) and \(z=4 .\) If \(-2 x
View solution Problem 37
\(|10+8 i|=\sqrt{164}\) and \(|11-6 i|=\sqrt{157}\). Hence \(11-6 i\) is closer to the origin.
View solution Problem 38
We have $$(\cos \theta+i \sin \theta)^{3}=\cos 3 \theta+i \sin 3 \theta$$ Also $$\begin{aligned} (\cos \theta+i \sin \theta)^{3} &=\cos ^{3} \theta+3 \cos ^{2}
View solution Problem 38
$$\left|\frac{1}{2}-\frac{1}{4} i\right|=\frac{\sqrt{5}}{4} \text { and }\left|\frac{2}{3}+\frac{1}{6} i\right|=\frac{\sqrt{17}}{6}, \text { since } \frac{\sqrt
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