Problem 1 - Complex Analysis

Question

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

Step-by-Step Solution

Answer

First complex number: imaginary=1; Second: real=-1, imaginary=2; \(\mathrm{i}^n\): dependent on \(n\mod4\); Others: result from polar forms or series.

Step 1: Simplifying \( \frac{\mathrm{i}-1}{\mathrm{i}+1} \)

The aim is to eliminate the imaginary part from the denominator by multiplying both the numerator and denominator by the conjugate of the denominator, which is \( \mathrm{i}-1 \). So, \( \frac{(\mathrm{i}-1)(\mathrm{i}-1)}{(\mathrm{i}+1)(\mathrm{i}-1)} = \frac{(\mathrm{i}-1)^2}{(\mathrm{i}^2-1^2)} \). Simplifying further: \((\mathrm{i}-1)^2 = \mathrm{i}^2 - 2\mathrm{i} + 1 = -1 - 2\mathrm{i} + 1 = -2\mathrm{i} \) and \( \mathrm{i}^2 - 1 = -1 - 1 = -2 \). Thus, \( \frac{-2\mathrm{i}}{-2} = \mathrm{i} \), which gives a real part of 0 and an imaginary part of 1.

Step 2: Simplifying \( \frac{3+4 \mathrm{i}}{1-2 \mathrm{i}} \)

First, multiply both the numerator and the denominator by the conjugate of the denominator \( 1+2\mathrm{i} \). Thus, \( \frac{(3+4\mathrm{i})(1+2\mathrm{i})}{(1-2\mathrm{i})(1+2\mathrm{i})} = \frac{3 + 6\mathrm{i} + 4\mathrm{i} + 8 \mathrm{i}^2}{1 - 4\mathrm{i}^2} \). With \( \mathrm{i}^2 = -1 \), this becomes \( \frac{3 + 10\mathrm{i} - 8}{1 + 4} = \frac{-5 + 10\mathrm{i}}{5} \). Dividing through by 5 gives us \( -1 + 2\mathrm{i} \). Thus, the real part is -1 and the imaginary part is 2.

Step 3: Simplifying \( \mathrm{i}^n \) for Integer \( n \)

The powers of \( \mathrm{i} \) cycle every 4 steps: \( \mathrm{i}^1 = \mathrm{i}, \mathrm{i}^2 = -1, \mathrm{i}^3 = -\mathrm{i}, \mathrm{i}^4 = 1 \) and repeats. Thus, for \( \mathrm{i}^n \), the imaginary part is: if \( n \mod 4 = 1 \) or 3, then the imaginary part is 1 or -1 respectively; otherwise, it is 0. The real part is -1 for \( n \mod 4 = 2 \), and 1 for \( n \mod 4 = 0 \); otherwise, it is 0.

Step 4: Simplifying \( \left(\frac{1+\mathrm{i}}{\sqrt{2}}\right)^n \)

Convert to polar form: \( \frac{1+\mathrm{i}}{\sqrt{2}} = e^{i\frac{\pi}{4}} \). Thus, \( \left(\frac{1+\mathrm{i}}{\sqrt{2}}\right)^n = e^{i\frac{n\pi}{4}} \). The real and imaginary components are \( \cos\left(\frac{n\pi}{4}\right) \) and \( \sin\left(\frac{n\pi}{4}\right) \) respectively.

Step 5: Simplifying \( \left(\frac{1+\mathrm{i} \sqrt{3}}{2}\right)^n \)

This complex number can also be converted to polar form as \( e^{i\frac{\pi}{3}} \) since its magnitude is 1 and angle \( \frac{\pi}{3} \). Therefore, \( \left(\frac{1+\mathrm{i} \sqrt{3}}{2}\right)^n = e^{i\frac{n\pi}{3}} \). So, the real and imaginary parts are \( \cos\left(\frac{n\pi}{3}\right) \) and \( \sin\left(\frac{n\pi}{3}\right) \) respectively.

Step 6: Evaluating \( \sum_{\nu=0}^{7}\left(\frac{1-\mathrm{i}}{\sqrt{2}}\right)^{\nu} \)

\( \frac{1-\mathrm{i}}{\sqrt{2}} \) is in geometric series with the first term 1 and common ratio \( \frac{1-\mathrm{i}}{\sqrt{2}} = e^{-i \frac{\pi}{4}} \). The sum of the series \( \sum_{u=0}^{7} r^u = \frac{1-r^8}{1-r} \), where \( r = e^{-i \frac{\pi}{4}} \). Since \( e^{-i 2\pi} = 1 \), the sum simplifies to 0 both real and imaginary as \( r^8 = 1 \).

Step 7: Evaluating \( \frac{(1+\mathrm{i})^{4}}{(1-\mathrm{i})^{3}} + \frac{(1-\mathrm{i})^{4}}{(1+\mathrm{i})^{3}} \)

Calculate \((1+\mathrm{i})^4 = (1+i)^2(1+i)^2 = (1+2i-1)^2 = -4+4i \). Similarly, \((1-i)^3 = (1-3i)((1-i)^2) = -2+2i \). The conjugates give identical magnitudes with opposite imaginary terms adding to zero. Result: pure real component sum. Both values are \( 6 \) so total = 12 (6 + 6). Both real.

Key Concepts

Real and Imaginary Parts

When dealing with complex numbers, it's essential to differentiate between their real and imaginary components. A complex number is generally represented as \( a + bi \), where \( a \) and \( b \) are real numbers. The term \( a \) is known as the real part, while \( bi \) represents the imaginary part. This distinction is crucial because it allows us to handle operations involving complex numbers effectively.

In the exercise, when given a fraction like \( \frac{3+4i}{1-2i} \), it's essential to manipulate it into the standard form \( a + bi \). To do this, we first eliminate the complex denominator by multiplying both the numerator and the denominator by the conjugate of the denominator (\( 1+2i \) in this case). This step is necessary because it converts the denominator into a real number, simplifying the fraction.

After simplifying, we get \( -1 + 2i \), which clearly identifies \( -1 \) as the real part and \( 2i \) as the imaginary part.

Polar Form

Sometimes, it's easier to work with complex numbers using their polar form instead of the standard form. The polar form represents a complex number in terms of its magnitude and angle, often expressed as \( re^{i\theta} \). Here, \( r \) is the magnitude of the number, and \( \theta \) (theta) is the angle formed with the positive real axis.

For example, \( \frac{1+i}{\sqrt{2}} \) can be represented in polar form. The magnitude is 1, and the angle \( \theta \) is \( \frac{\pi}{4} \). This is often written as \( e^{i \frac{\pi}{4}} \). Polar form is particularly useful when raising complex numbers to powers or taking roots, as it simplifies multiplication and division processes.

By converting \( \frac{1+i}{\sqrt{2}} \) to polar form, calculations such as powers become straightforward because we can directly use exponential properties. In polar form, multiplying by \( n \) results in \( e^{i\frac{n\pi}{4}} \), easily showing how the real and imaginary parts vary with \( n \).

Complex Conjugates

The complex conjugate of a complex number \( a + bi \) is \( a - bi \). This concept is fundamental in simplifying expressions involving complex numbers, especially when dealing with division or finding the modulus of a complex number.

Let's consider the division of complex numbers as demonstrated in the original solution: multiplying the numerator and the denominator by the conjugate of the denominator. This approach ensures the denominator becomes a real number, thus simplifying the complex fraction considerably. The conjugate leads to valuable symmetry in calculations because multiplying a complex number by its conjugate produces a real number, specifically the sum of their squares \( a^2 + b^2 \). This operation is vital in simplifying \( \frac{(1+i)^4}{(1-i)^3} \), as shown in the steps.

Utilizing conjugates helps to balance both parts of a complex computation, allowing simplification into a form that's easier to understand and further manipulate.

Powers of Imaginary Unit

The imaginary unit \( i \) is a fundamental aspect of complex numbers, as it allows us to represent the square root of negative one. Powers of \( i \) have a cyclical nature, which simplifies calculations involving powers of complex numbers.

The powers of \( i \) follow a specific pattern:

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

This cycle repeats every four powers, which means for any integer \( n \), \( i^n \) can be determined by the remainder of \( n \) divided by 4. If \( n \mod 4 = 0 \), \( i^n = 1 \). Other remainders yield values \( i \), \(-1 \), or \(-i \), depending on the position within the cycle.

Understanding these cyclical properties allows for straightforward conclusions about the real and imaginary parts when complex numbers are raised to powers, simplifying the original exercise and showing relational symmetries between real and imaginary components.