Question

Prove that $$ \overline{z+z^{\prime}}=\bar{z}+\overline{z^{\prime}} \text { and } \overline{z z^{\prime}}=\bar{z} \cdot \overline{z^{\prime}} . $$ Hence show by induction that $$ \overline{z^{n}}=(\bar{z})^{n}, n=1,2, \ldots, \text { and } \sum_{k=1}^{n} a_{k} z^{k}=\sum_{k=1}^{n} \bar{a}_{k} \bar{z}^{k} . $$

Short answer

Conjugate of sum is sum of conjugates, conjugate of product is product of conjugates; proved by induction for powers and linear combinations.

Step-by-step solution

Showing linearity of conjugation

If we have two complex numbers, \( z = a + bi \) and \( z' = c + di \), their sum is \( z + z' = (a+c) + (b+d)i \). The conjugate of this sum is \( \overline{z+z'} = (a+c) - (b+d)i \). Separately, \( \bar{z} = a - bi \) and \( \overline{z'} = c - di \). Adding these, \( \bar{z} + \overline{z'} = (a+c) - (b+d)i \). Hence, we have demonstrated that \( \overline{z+z'} = \bar{z} + \overline{z'} \).

Proving product of conjugates

For the same \( z \) and \( z' \), their product is \( zz' = (a+bi)(c+di) = (ac-bd) + (ad+bc)i \). The conjugate is \( \overline{zz'} = (ac-bd) - (ad+bc)i \). Separately, compute \( (a-bi)(c-di) = (ac-bd) - (bc+ad)i = \bar{z} \cdot \overline{z'} \). So, \( \overline{zz'} = \bar{z} \cdot \overline{z'} \).

Induction base case

For n=1, \( z^1 = z \) and \( \overline{z^1} = \overline{z} \), which is \( (\bar{z})^1 \). The equality holds for \( n=1 \).

Induction hypothesis

Assume for some \( n = k \), \( \overline{z^k} = (\bar{z})^k \). We'll prove for \( n = k+1 \).

Induction step

Using the induction hypothesis, \( \overline{z^{k+1}} = \overline{z^k \cdot z} = \overline{z^k} \cdot \overline{z} \). By the induction hypothesis, this equals \( (\bar{z})^k \cdot \bar{z} = (\bar{z})^{k+1} \). Hence, the statement is true for \( n=k+1 \).

Showing conjugate sums over powers

We need to show that \( \sum_{k=1}^{n} a_k z^k = \sum_{k=1}^{n} \bar{a}_k \overline{z}^k \). Using the linearity of conjugation, \( \overline{ \left(\sum_{k=1}^{n} a_k z^k \right)} = \sum_{k=1}^{n} \overline{a_k z^k} = \sum_{k=1}^{n} \bar{a}_k \cdot \overline{z^k} \). By induction, \( \overline{z^k} = (\bar{z})^k \), thus, \( \sum_{k=1}^{n} \bar{a}_k (\bar{z})^k \) as required.

Key concepts

Complex Conjugate

A complex conjugate is a fundamental concept in complex number arithmetic. To find the complex conjugate of a complex number, you only need to change the sign of the imaginary part. For a complex number \( z = a + bi \), its complex conjugate is represented as \( \bar{z} = a - bi \). This operation has some very interesting properties.

• **Linearity of Conjugation:** The conjugate of the sum of two complex numbers is the same as the sum of their conjugates. Mathematically, if you have two complex numbers \( z \) and \( z' \), then \( \overline{z + z'} = \bar{z} + \bar{z'} \).
• **Product of Conjugates:** Similarly, the conjugate of the product of two complex numbers is the product of their conjugates. If \( z \) and \( z' \) are complex numbers, then \( \overline{zz'} = \bar{z} \cdot \bar{z'} \).

Changing the sign of the imaginary part may seem like a simple operation, but it has profound implications in complex number calculations, making it a key tool for solving many problems in mathematics.

Mathematical Induction

Mathematical induction is a powerful proof technique used to establish the validity of statements over a set of natural numbers. Think of it as a chain reaction -- once you prove it true for one case, you can prove it true for all subsequent cases. It involves two key steps:

• **Base Case:** First, you verify that the statement holds for an initial value, usually \( n = 1 \). For our problem, when \( n = 1 \), \( \overline{z^1} = \overline{z} = (\bar{z})^1 \).
• **Induction Step:** Next, you assume it's true for \( n = k \), and use this to show it's true for \( n = k+1 \). This "domino effect" ensures if the base case is true, all subsequent cases are true too. Here, we assumed \( \overline{z^k} = (\bar{z})^k \) true and proved \( \overline{z^{k+1}} = (\bar{z})^{k+1} \).

Mathematical induction helps establish the validity of statements concerning sums, sequences, series, and other iteratively defined constructs in mathematics, and it's a crucial tool for proving the properties of complex numbers.

Sum and Product of Complex Numbers

The sum and product of complex numbers are straightforward yet essential calculations. Understanding these operations is key to mastering complex number manipulations. When adding two complex numbers, \( z = a + bi \) and \( z' = c + di \), the sum is simply \( z + z' = (a+c) + (b+d)i \). Notice that both the real and imaginary parts are added separately.

• **Sum of Complex Numbers:** An important aspect of summing complex numbers is that their order of addition doesn't change the result. It's commutative, just like regular addition. The linearity of conjugation concerning sums, \( \overline{z + z'} = \bar{z} + \bar{z'} \), also plays a role here.
• **Product of Complex Numbers:** The product involves distributing terms, generally calculated as \( zz' = ac - bd + (ad + bc)i \). Again, the order doesn't matter due to commutativity, and properties like \( \overline{zz'} = \bar{z} \cdot \bar{z'} \) highlight the consistency of conjugative properties.

These operations extend to various algebraic methods, such as solving systems of equations or evaluating polynomial expressions involving complex numbers, making it essential to grasp how these sums and products are handled in practice.