Question

Show that for any set of \(n\) complex numbers \(\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}\), we have $$ \left|\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}\right|^{2} \leq n\left(\left|\alpha_{1}\right|^{2}+\left|\alpha_{2}\right|^{2}+\cdots+\left|\alpha_{n}\right|^{2}\right) $$

Short answer

The given inequality is true for any set of \(n\) complex numbers

Step-by-step solution

Establish Base Case for Mathematical Induction

Let's consider the smallest valid set of \(n\), i.e \(n = 1\). Here, \(\left| \alpha_{1} \right|^2\) is equal to \(1*\left| \alpha_{1} \right|^2\). So, the inequality holds true for \(n = 1\).

Assume the Statement is True for \(n = k\)

Assume that the inequality holds true for \(n = k\). So, \(\left| \alpha_{1} + \alpha_{2} + \cdots + \alpha_{k} \right|^2 \leq k\left( \left| \alpha_{1} \right|^2 + \left| \alpha_{2} \right|^2 + \cdots + \left| \alpha_{k} \right|^2 \right)\)

Prove the Statement for \(n = k + 1\)

Now, consider \(n = k+1\). Calculate \(\left| \alpha_{1} + \alpha_{2} + \cdots + \alpha_{k} + \alpha_{k+1} \right|^2\) = \(\left| \alpha_{1} + \alpha_{2} + \cdots + \alpha_{k}\right|^2 + 2Real(\sum_{i=1}^{k} \alpha_{i})\alpha_{k+1} +\left| \alpha_{k+1} \right|^2\)

Use assumed true statement and properties of complex numbers

Since we've assumed the inequality holds for \(n = k\), also knowing \(2Real(\sum_{i=1}^{k} \alpha_{i})\alpha_{k+1}\) is less than or equal to \(2*\left| \alpha_{k+1} \right|*\left| \alpha_{1} + \alpha_{2} + \cdots + \alpha_{k} \right|\) by the triangle inequality, we can observe that the entire expression is less than or equal to \(k+1\)*\(\left( \left| \alpha_{1} \right|^2 + \left| \alpha_{2} \right|^2 + \cdots + \left| \alpha_{k} \right|^2 + \left| \alpha_{k+1} \right|^2 \right)\). Hence, we've proved that the statement is true for \(n = k+1\), given it was true for \(n = k\)

Key concepts

Mathematical Induction

When dealing with mathematical proofs, one powerful tool is mathematical induction. This technique is particularly useful for asserting the truth of a statement for all natural numbers (or a sequence of them). The method comes in two essential steps. First, we establish a base case—usually for the smallest value in the sequence—showing that the statement holds true. In the context of our inequality with complex numbers, the base case is when there is just one complex number, and the inequality simplifies to an identity.

After proving the base case, we make an inductive assumption: we suppose the statement is true for a particular but arbitrary natural number, typically denoted as 'k'. Then, the essence of mathematical induction lies in the inductive step. Here, the goal is to prove that if the statement holds true for 'k', it must also hold true for 'k+1'. This leap from k to k+1 is critical, as it effectively acts like a 'domino effect', ensuring the statement is true for all subsequent natural numbers following the base case.

Complex Number Properties

Complex numbers, a vital component of advanced mathematics, come with a set of unique properties that make them distinct from real numbers. One key property is the modulus of a complex number, denoted by \( |z| \), which represents its 'distance' from the origin in the complex plane. The modulus is always non-negative and is found by taking the square root of the product of the complex number and its complex conjugate.

Another distinctive feature of complex numbers is related to their algebraic operations. For instance, the sum of two complex numbers corresponds to vector addition in the complex plane. Additionally, since complex numbers encapsulate magnitude and direction (through the modulus and argument, respectively), their behaviors can lead to new interpretations of inequalities, as can be seen in the problem discussed here. Understanding these properties is essential when dealing with inequalities involving sums or other operations on complex numbers.

Triangle Inequality

The triangle inequality is a fundamental principle in mathematics, and it applies in various contexts, including the study of complex numbers. In its simplest form, the triangle inequality states that for any triangle, the length of any one side must be less than or equal to the sum of the lengths of the other two sides. This notion can be extended to the complex plane.

The complex version of the triangle inequality asserts that the modulus of the sum of two complex numbers is less than or equal to the sum of their moduli: \( |z_1 + z_2| \leq |z_1| + |z_2| \). This property is instrumental when we are attempting to prove inequalities involving sums of complex numbers, as it allows us to relate the magnitude of a sum to the magnitudes of its individual terms. In the exercise provided, the triangle inequality helps us transition from the magnitude of a sum to a statement involving the sum of magnitudes, thus serving as a critical step in the proof by mathematical induction.