Problem 15 - Complex numbers from A to...Z

Question

A finite set \(A\) of complex numbers has the property: \(z \in A\) implies \(z^{n} \in A\) for every positive integer \(n\). a) Prove that \(\sum_{z \in A} z\) is an integer. b) Prove that for every integer \(k\) one can choose a set \(A\) which fulfills the above condition and \(\sum_{z \in A} z=k\).

Step-by-Step Solution

Answer

Question: Prove that the sum of elements in a finite set A of complex numbers, with the property that whenever z ∈ A, its powers zn also belong to A for every positive integer n, has an integer real part. Answer: Whenever a complex number z is in A, its complex conjugate z' must also be in A. Therefore, the sum of the elements in A can be expressed as the sum of pairs of complex conjugates, which results in real numbers that are multiples of 2. This implies that the sum has an integer real part. Question: Show that for every integer k, there exists a set A with the given conditions and having S = sum of elements in A = k. Answer: For every integer k, we can construct a set A(k) as follows: 1. If k is even, let A(k) = {-(k/2), k/2}. The sum of elements in A(k) is equal to k and the conditions are fulfilled. 2. If k is odd, let A(k) = {-(k-1)/2, (k+1)/2}. The sum of elements in A(k) is equal to k and the conditions are fulfilled. Thus, there exists a set A(k) that fulfills the given conditions and has the sum equal to k for every integer k.

Step 1: Prove that the sum of elements is an integer

To prove that \(S = \sum_{z \in A} z\) has an integer real part, let's consider the following. If \(z \in A\), then \(z'\), the complex conjugate of \(z\), must also be in \(A\). This is because if \(z^n \in A\) for all \(n > 0\), then \((z')^n = (z^n)' \in A\) as well. Now, let's note that \(z + z'\) is real and equals \(2\operatorname{Re}(z)\), because \(z + z' = (a + bi) + (a - bi) = 2a = 2\operatorname{Re}(z)\), where \(a, b \in \mathbb{R}\) are the real and imaginary parts of \(z\). So now, we can write the sum \(S\) of elements in \(A\) as the sum of pairs of complex conjugates: \(S = \sum_{(z,z') \in P} (z + z')\), where \(P\) is the set of pairs \((z,z')\) with \(z \in A, z' = z^*\), and \(z^*\) denotes the complex conjugate. Then we notice that this sum is a sum of real numbers, which are all multiples of \(2\). Therefore, \(S\) must be a multiple of \(2\) as well. Since the real part of \(S\) is half of \(S\), it must be an integer.

Step 2: Construct the set A for every integer k

For part b), let k be an integer. We will construct a set \(A(k)\) that fulfills the given condition and has the sum equal to k. Let's consider two cases: 1. If \(k\) is even, let \(A(k) = \{-\frac{k}{2}, \frac{k}{2}\}\). The sum of elements in \(A(k)\) is clearly equal to \(k\). For every non-negative integer power of elements, \(z^n = \left( \pm \frac{k}{2} \right) ^n = \pm \frac{k^n}{2^n} \in A(k)\). So in this case, the condition is fulfilled. 2. If \(k\) is odd, let \(A(k) = \{-\frac{k-1}{2}, \frac{k+1}{2}\}\). The sum of elements in \(A(k)\) is again equal to \(k\). For every non-negative integer power of elements, \(z^n = \left( \pm \frac{k-1}{2} \right) ^n, \left( \pm \frac{k+1}{2} \right) ^n \in A(k)\). So in this case, the condition is fulfilled as well. Thus, we have shown that for every integer \(k\), there exists a set \(A(k)\) that fulfills the given condition and has the sum equal to \(k\).

Key Concepts

Sum of Complex Numbers

The sum of complex numbers is similar to the sum of 2-dimensional vectors: both the real parts and the imaginary parts are added separately. Suppose we have two complex numbers, say, \( z_1 = a + bi \) and \( z_2 = c + di \), where \( a \) and \( c \) are the real parts and \( b \) and \( d \) are the imaginary parts, respectively. The sum \( z_1 + z_2 \) is then \( (a + c) + (b + d)i \). Each part, real and imaginary, combines to give a new complex number.

Understanding this concept is fundamental because complex numbers are widely used in areas such as electrical engineering and physics, particularly when dealing with sinusoidal waveforms or signals. The textbook exercise elucidates the integer nature of the sum of a certain set of complex numbers. It shows that under specific conditions, even though the elements are complex, their sum can produce purely real integers, which is a pretty handy property in mathematical analysis and proofs.

Complex Conjugate

The complex conjugate of a complex number is found by changing the sign of the imaginary part. For example, if we have a complex number \( z = a + bi \), then its complex conjugate, denoted by \( z^* \) or \( \overline{z} \), is \( z^* = a - bi \). The real parts remain the same, while the imaginary parts are negated. One particularly useful aspect of the complex conjugate is that when a complex number is added to its conjugate \( z + z^* = (a + bi) + (a - bi) = 2a \), the result is a real number, as showcased in the exercise solution.

The significance of complex conjugates extends to various mathematical operations: they help in the division of complex numbers, finding the magnitude (or modulus), and in the process of polynomial factorization. Understanding complex conjugates is crucial for delving deeper into the theory of complex numbers and for practical applications in signal processing and control theory.

Integer Powers of Complex Numbers

Calculating integer powers of complex numbers involves multiplying the base, a complex number, by itself a number of times indicated by the exponent. For a complex number \( z \), the \( n^{th} \) power \( z^n \) is calculated as \( z \times z \times ... \times z \), where the multiplication continues \( n \) times. The exercise shows an interesting property whereby if \( z \) is an element of set \( A \), then \( z^n \) also belongs to \( A \), for any positive integer \( n \). This hints at a relationship between complex numbers and powers that align with the set's criteria.

Taking powers of complex numbers is not just a theoretical exercise; it's heavily used in the study of complex functions, fractals, and dynamic systems. The computation of these powers often requires the use of the polar form of complex numbers and Euler's formula, which bring in a level of elegance and simplicity into the otherwise cumbersome process.