Problem 5 - Complex Analysis for Mathematics and Engineering

Question

Express the following in \(a+i b\) form. (a) \(e^{\frac{1 \pi}{2}}\). (b) \(4 e^{-i \frac{\pi}{2}}\) (c) \(8 e^{i \frac{1 \pi}{3}}\), (d) \(-2 e^{i \frac{2 x}{8}}\). (e) \(2 i e^{-i \frac{2 \pi}{4}}\), (f) \(6 e^{i \frac{\pi}{3}} e^{i \pi}\). (g) \(e^{2} e^{i \pi}\) (h) \(e^{i \frac{\pi}{6}} e^{-i n}\).

Step-by-Step Solution

Answer

(a) Real, exponential term. (b) "+0 - 4i" (c) "4 + 4\sqrt{3}i" (d) "Dependent on x." (e) "2 + 0i" (f) "-3 - 3\sqrt{3}i" (g) "Real, "-e^{2}" (h) "Dependent on n."

Step 1: Evaluate (a)

The expression is given as \(e^{\frac{1 \pi}{2}}\). Since this has no imaginary component \(i\), it is purely real. However, it involves an irrational function and cannot be expressed simply as \(a + ib\). Therefore, \(a = e^{\frac{1 \pi}{2}}\) and \(b = 0\).

Step 2: Convert (b) using Euler's formula

The expression is \(4 e^{-i \frac{\pi}{2}}\). Euler's formula states \(e^{ix} = \cos{x} + i\sin{x}\). Thus, \(e^{-i \frac{\pi}{2}} = \cos{(-\frac{\pi}{2})} + i\sin{(-\frac{\pi}{2})}\). Calculate: \(\cos{(-\frac{\pi}{2})} = 0\) and \(\sin{(-\frac{\pi}{2})} = -1\). Therefore, the expression becomes \(4(0 - i) = -4i\). So, \(a = 0\) and \(b = -4\).

Step 3: Convert (c) using Euler's formula

The expression is \(8 e^{i \frac{\pi}{3}}\). Using Euler's formula: \(e^{i \frac{\pi}{3}} = \cos{\frac{\pi}{3}} + i\sin{\frac{\pi}{3}}\). Calculate: \(\cos{\frac{\pi}{3}} = \frac{1}{2}\) and \(\sin{\frac{\pi}{3}} = \frac{\sqrt{3}}{2}\). Thus, \(8(\frac{1}{2} + i\frac{\sqrt{3}}{2}) = 4 + 4\sqrt{3}i\). So, \(a = 4\) and \(b = 4\sqrt{3}\).

Step 4: Simplify (d)

The expression is \(-2 e^{i \frac{2x}{8}} = -2 e^{i \frac{x}{4}}\). Using Euler's formula: \(e^{i \frac{x}{4}} = \cos{\frac{x}{4}} + i\sin{\frac{x}{4}}\). Thus, the expression becomes \(-2(\cos{\frac{x}{4}} + i\sin{\frac{x}{4}}) = -2\cos{\frac{x}{4}} - 2i\sin{\frac{x}{4}}\). So, \(a = -2\cos{\frac{x}{4}}\) and \(b = -2\sin{\frac{x}{4}}\).

Step 5: Convert (e)

The expression is \(2i e^{-i \frac{2 \pi}{4}} = 2i e^{-i \frac{\pi}{2}}\). From step 2, \(e^{-i \frac{\pi}{2}} = -i\). Therefore, \(2i(-i) = 2\). So, \(a = 2\) and \(b = 0\).

Step 6: Convert (f)

The expression is \(6 e^{i \frac{\pi}{3}} e^{i \pi} = 6 e^{i(\frac{\pi}{3} + \pi)} = 6 e^{i \frac{4\pi}{3}}\). Using Euler's formula: \(e^{i \frac{4\pi}{3}} = \cos{\frac{4\pi}{3}} + i\sin{\frac{4\pi}{3}}\). Calculate: \(\cos{\frac{4\pi}{3}} = -\frac{1}{2}\) and \(\sin{\frac{4\pi}{3}} = -\frac{\sqrt{3}}{2}\). Thus, \(6(-\frac{1}{2} - i\frac{\sqrt{3}}{2}) = -3 - 3\sqrt{3}i\). So, \(a = -3\) and \(b = -3\sqrt{3}\).

Step 7: Simplify (g)

The expression is \(e^{2} e^{i \pi} = e^{2} (-1) = -e^{2}\). This is a purely real number. Therefore, \(a = -e^{2}\) and \(b = 0\).

Step 8: Simplify (h)

The expression is \(e^{i \frac{\pi}{6}} e^{-i n} = e^{i (\frac{\pi}{6} - n)}\). Using Euler's formula: \(e^{i (\frac{\pi}{6} - n)} = \cos{(\frac{\pi}{6} - n)} + i\sin{(\frac{\pi}{6} - n)}\). So, \(a = \cos{(\frac{\pi}{6} - n)}\) and \(b = \sin{(\frac{\pi}{6} - n)}\).

Key Concepts

Euler's Formula

Euler's Formula is a cornerstone in complex number calculus and provides a profound relationship between exponential functions and trigonometric functions. It is given by the equation: \( e^{ix} = \cos{x} + i\sin{x} \). This translates complex exponentials into a real part \(\cos{x}\) and an imaginary part \(i\sin{x}\). When working with problems involving expressions like \( e^{i\pi} \), Euler's Formula simplifies calculations. By substituting into the formula, you get \( \cos{\pi} + i\sin{\pi} = -1 + 0i \). Thus, \( e^{i\pi} = -1\), a fundamental result known as Euler's Identity. Euler’s Formula is powerful because it allows you to transform expressions involving \( e^{ix} \) into the form \( a + ib \), making it easier to handle in mathematical computations. For example, in solving \( 8e^{i\frac{\pi}{3}} \), turning the expression into \( 8 \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) \) using Euler's formula leads to a simpler expression \( 4 + 4\sqrt{3}i \).

Trigonometric Form

The trigonometric form of complex numbers is another way of expressing complex numbers, particularly useful when dealing with multiplication and division. It utilizes the polar form of a complex number. In general, for a complex number \( z = x + iy \), its trigonometric form is \( r(\cos{\theta} + i\sin{\theta}) \), where \( r = |z| = \sqrt{x^2 + y^2} \) is the modulus, and \( \theta = \tan^{-1}(\frac{y}{x}) \) is the argument or angle. Conversion to trigonometric form involves identifying the modulus and argument from the Cartesian coordinates. For example, a complex number \( 6e^{i(\frac{\pi}{3})} \), in trigonometric form becomes \( 6(\cos{\frac{\pi}{3}} + i\sin{\frac{\pi}{3}}) \), where \( \cos{\frac{\pi}{3}} = \frac{1}{2} \) and \( \sin{\frac{\pi}{3}} = \frac{\sqrt{3}}{2} \). This ultimately simplifies to \( 6 \cdot \left( \frac{1}{2} \right) + i \cdot 6 \cdot \left( \frac{\sqrt{3}}{2} \right) = 3 + 3\sqrt{3}i \). The trigonometric form simplifies multiplication and division of complex numbers, as multiplying two numbers involves simply multiplying their moduli and adding their arguments.

Imaginary Unit

The imaginary unit, denoted by \( i \), is a fundamental concept in complex numbers. It is defined as \( i^2 = -1 \). This property lays the foundation for all arithmetic involving imaginary numbers. In complex arithmetic, any complex number can be expressed as \( a + ib \), where \( a \) and \( b \) are real numbers. The term \( ib \) indicates the imaginary component. Understanding the imaginary unit is crucial as operations often involve \( i \), particularly when calculating complex exponentials. For instance, using Euler’s Formula, \( e^{-i\frac{\pi}{2}} = \cos{\left(-\frac{\pi}{2}\right)} + i\sin{\left(-\frac{\pi}{2}\right)} \), gives \( 0 - i \). Therefore, multiplications including terms like \( 2i \times (-i) \) result in real numbers because \( i \times (-i) = 1 \). Here, nominally complex calculations return to the real number domain, demonstrating why understanding \( i \) is vital in simplifying complex expressions. The imaginary unit offers new dimensions to solve equations beyond the real plane, such as quadratic equations with negative discriminants, expanding the scope and application of mathematics.