Problem 1 - Complex Analysis

Question

Find the real and imaginary parts of each of the following complex numbers: $$ \begin{gathered} \frac{\mathrm{i}-1}{\mathrm{i}+1} ; \quad \frac{3+4 \mathrm{i}}{1-2 \mathrm{i}} ; \mathrm{i}^{n}, n \in \mathbb{Z} ; \quad\left(\frac{1+\mathrm{i}}{\sqrt{2}}\right)^{n}, n \in \mathbb{Z} ; \\ \left(\frac{1+\mathrm{i} \sqrt{3}}{2}\right)^{n}, n \in \mathbb{Z} ; \quad \sum_{\nu=0}^{7}\left(\frac{1-\mathrm{i}}{\sqrt{2}}\right)^{\nu} ; \frac{(1+\mathrm{i})^{4}}{(1-\mathrm{i})^{3}}+\frac{(1-\mathrm{i})^{4}}{(1+\mathrm{i})^{3}} \end{gathered} $$

Step-by-Step Solution

Answer

For each problem: first number's imaginary part is 1; second's are -1, 2; powers of i depend on cycle; expressions in polar form yield cosine, sine of angle; sum series is 0; last fraction's parts are -2, 0.

Step 1: Simplify the First Expression

Given the complex number \( \frac{\mathrm{i}-1}{\mathrm{i}+1} \).To simplify this, multiply the numerator and denominator by the conjugate of the denominator, \( \mathrm{i}-1 \):\[\frac{(\mathrm{i}-1)(\mathrm{i}-1)}{(\mathrm{i}+1)(\mathrm{i}-1)}\]Evaluate both parts:- Numerator: \((\mathrm{i}-1)^2 = \mathrm{i}^2 - 2\mathrm{i} + 1 = -1 - 2\mathrm{i} + 1 = -2\mathrm{i}\)- Denominator: \((\mathrm{i}+1)(\mathrm{i}-1) = \mathrm{i}^2 - 1 = -2\)Thus, \( \frac{\mathrm{i}-1}{\mathrm{i}+1} = \frac{-2\mathrm{i}}{-2} = \mathrm{i} \).The real part is 0, and the imaginary part is 1.

Step 2: Simplify the Second Expression

Given the complex number \( \frac{3+4\mathrm{i}}{1-2\mathrm{i}} \), multiply the numerator and denominator by the conjugate of the denominator, \( 1 + 2\mathrm{i} \):\[\frac{(3+4\mathrm{i})(1+2\mathrm{i})}{(1-2\mathrm{i})(1+2\mathrm{i})}\]Calculate:- Numerator: \((3+4\mathrm{i})(1+2\mathrm{i}) = 3 + 6\mathrm{i} + 4\mathrm{i} + 8\mathrm{i}^2 = 3 + 10\mathrm{i} - 8 = -5 + 10\mathrm{i}\)- Denominator: \(1^2 - (2\mathrm{i})^2 = 1 + 4 = 5\)Thus, \( \frac{3+4\mathrm{i}}{1-2\mathrm{i}} = \frac{-5+10\mathrm{i}}{5} = -1 + 2\mathrm{i} \).The real part is -1, and the imaginary part is 2.

Step 3: Find Powers of i

The expression is \( \mathrm{i}^{n} \). Remember the cycle of powers of i: - \( \mathrm{i}^1 = \mathrm{i} \)- \( \mathrm{i}^2 = -1 \)- \( \mathrm{i}^3 = -\mathrm{i} \)- \( \mathrm{i}^4 = 1 \) which repeats every four.The real and imaginary parts depend on \( n \mod 4 \):- If \( n \equiv 0 \mod 4 \), real part is 1, imaginary part is 0.- If \( n \equiv 1 \mod 4 \), real part is 0, imaginary part is 1.- If \( n \equiv 2 \mod 4 \), real part is -1, imaginary part is 0.- If \( n \equiv 3 \mod 4 \), real part is 0, imaginary part is -1.

Step 4: Powers of a Complex Number

For the value \( \left(\frac{1 + \mathrm{i}}{\sqrt{2}}\right)^n \), write it as a complex number in polar form.Since \( 1 + \mathrm{i} = \sqrt{2}(\cos \frac{\pi}{4} + \mathrm{i}\sin \frac{\pi}{4}) \), divide by \( \sqrt{2} \) to simplify:\[\frac{1+\mathrm{i}}{\sqrt{2}} = \cos \frac{\pi}{4} + \mathrm{i} \sin \frac{\pi}{4}\]It's in the form \( \text{cis} \frac{\pi}{4} \). So, \[(\text{cis} \frac{\pi}{4})^n = \text{cis} \frac{n\pi}{4}\]The real part is \( \cos \frac{n\pi}{4} \) and the imaginary part is \( \sin \frac{n\pi}{4} \).

Step 5: Powers of Another Complex Number

Given \( \left(\frac{1+\mathrm{i}\sqrt{3}}{2}\right)^n \), convert to polar form.\[1 + \mathrm{i}\sqrt{3} = 2\left(\frac{1}{2} + \mathrm{i}\frac{\sqrt{3}}{2}\right) = 2\text{cis} \frac{\pi}{3}\]Therefore,\( \frac{1+\mathrm{i}\sqrt{3}}{2} = \text{cis} \frac{\pi}{3}\), Then, \( (\text{cis} \frac{\pi}{3})^n = \text{cis} \frac{n\pi}{3} \).The real part is \( \cos \frac{n\pi}{3} \), and the imaginary part is \( \sin \frac{n\pi}{3} \).

Step 6: Evaluate the Sum of Powers Series

The sum is \( \sum_{u=0}^{7} \left(\frac{1-\mathrm{i}}{\sqrt{2}}\right)^{u} \). Recognize \( \frac{1-\mathrm{i}}{\sqrt{2}} \) as \( \text{cis} \left( -\frac{\pi}{4} \right) \). This is a geometric series where the first term, \( a = 1 \), and the common ratio, \( r = \text{cis} \left( -\frac{\pi}{4} \right) \).Sum of a geometric series:\[S_n = a \frac{1-r^n}{1-r}\]Here, \( r^8 = (\text{cis} \left(-\frac{\pi}{4}\right))^8 = \text{cis} (-2\pi) = 1 \).Sum is \( \frac{1 - 1}{1 - \text{cis} \left( -\frac{\pi}{4} \right)} = 0 \).As all terms eventually cancel out for a complete cycle, the sum results in zero, i.e., real and imaginary parts are both zero.

Step 7: Evaluate the Complex Fraction

The expression is \( \frac{(1+\mathrm{i})^4}{(1-\mathrm{i})^3} + \frac{(1-\mathrm{i})^4}{(1+\mathrm{i})^3} \).First, calculate \((1 + \mathrm{i})^4\) and \((1 - \mathrm{i})^4\):- Using the binomial theorem, \((1+\mathrm{i})^4 = 4\mathrm{i}(2) = -4 \)- Similarly, \((1-\mathrm{i})^4 = 4\mathrm{i}(-2) = -4 \)Also, find \((1-\mathrm{i})^3\) and \((1+\mathrm{i})^3\):- \((1 + \mathrm{i})^3 = 2 - 2\mathrm{i} \)- \((1 - \mathrm{i})^3 = 2 + 2\mathrm{i} \)Then evaluate each fraction:- \( \frac{-4}{2-2\mathrm{i}} \) results in complex number \( -1 + \mathrm{i} \)- \( \frac{-4}{2+2\mathrm{i}} \) results in complex number \( -1 - \mathrm{i} \)Summing these results in \( -2 \).Thus, the real part is -2 and the imaginary part is 0.

Key Concepts

Real and Imaginary Parts

Complex numbers are composed of two main components: the real part and the imaginary part. The real part of a complex number is the part that does not involve the imaginary unit \( i \), while the imaginary part is the coefficient of \( i \). For example, in the complex number \(-1 + 2i\), \(-1\) is the real part and \(2\) is the imaginary part. Understanding these components is critical when performing operations on complex numbers, as they allow us to decompose and simplify these operations effectively.
When presented with a complex number in the form \(a + bi\), where \(a\) and \(b\) are real numbers, the real part is \(a\) and the imaginary part is \(b\). Recognizing these parts helps in the simplification of expressions, especially when dealing with complex fractions or powers.

Simplification of Complex Fractions

Simplifying complex fractions involves removing the imaginary unit from the denominator to provide a clear result. This is typically done by multiplying both the numerator and denominator by the conjugate of the denominator.
The conjugate of a complex number \(a + bi\) is \(a - bi\). By multiplying a complex number by its conjugate, the imaginary parts cancel out, resulting in a real number.

• For example, to simplify \( \frac{3 + 4i}{1 - 2i} \), we multiply by the conjugate \(1 + 2i\). This results in a real denominator, allowing us to express the complex fraction in standard form \(a + bi\).

This process helps in isolating the real and imaginary parts and provides an understandable expression of the original complex fraction.

Powers of Complex Numbers

Handling powers of complex numbers, especially the imaginary unit \(i\), is simplified by recognizing a cyclical pattern in the powers of \(i\). This cyclical pattern repeats every four terms:

• \(i^1 = i\)
• \(i^2 = -1\)
• \(i^3 = -i\)
• \(i^4 = 1\)

Understanding this cycle helps in quickly evaluating powers of \(i\) without detailed calculations.
Additionally, when calculating the power of a complex number expressed with the imaginary unit, such as \((1 + i)^n\), binomial expansion techniques or recognizing patterns in the results can simplify the calculation. Evaluating powers becomes efficient by observing how the coefficients change with increasing exponents.

Complex Number Polar Form

The polar form of a complex number represents the number using its magnitude and angle relative to the positive real axis. This form is particularly useful for multiplying and raising complex numbers to powers, due to the simplicity of working with angles.
For a complex number \(a+bi\), the magnitude \(r\) is calculated as \(r = \sqrt{a^2 + b^2}\), and the angle \(\theta\) (sometimes called the argument or phase) is given by \(\theta = \tan^{-1}(\frac{b}{a})\). The polar form is then expressed as \(r(\cos \theta + i \sin \theta)\).

• This is often abbreviated using the expression \(r \cdot \text{cis} \theta\), where \(\text{cis} \theta\) stands for \(\cos \theta + i \sin \theta\).

In computations involving exponentiation, this form makes it easy to multiply the magnitudes and add the angles, streamlining the process of evaluating powers and products of complex numbers.