Problem 4 - Complex Analysis

Question

For \(z=x+\mathrm{i} y, w=u+\mathrm{i} v\), with \(x, y, u, v \in \mathbb{R}\), the standard scalar product in the \(\mathbb{R}\)-vector space \(\mathbb{C}=\mathbb{R} \times \mathbb{R}\) with respect to the basis \((1, \mathrm{i})\) is defined by $$ \langle z, w\rangle:=\operatorname{Re}(z \bar{w})=x u+y v $$ Verify by direct calculation that, for \(z, w \in \mathbb{C}\) $$ \langle z, w\rangle^{2}+\langle\mathrm{i} z, w\rangle^{2}=|z|^{2}|w|^{2} $$ and infer from this the CAUCHY-SCHWARZ Inequality in \(\mathbb{R}^{2}\) : $$ |\langle z, w\rangle|^{2}=|x u+y v|^{2} \leq|z|^{2}|w|^{2}=\left(x^{2}+y^{2}\right)\left(u^{2}+v^{2}\right) $$ In addition, show the following identities for \(z, w \in \mathbb{C}\) by direct calculation: In addition, show the following identities for \(z, w \in \mathbb{C}\) by direct calculation: \(\begin{aligned}|z+w|^{2} &=|z|^{2}+2\langle z, w\rangle+|w|^{2} & &(\text { cosine law }) \\\|z-w|^{2} &=|z|^{2}-2\langle z, w\rangle+|w|^{2}, & & \\\|z+w|^{2}+|z-w|^{2} &=2\left(|z|^{2}+|w|^{2}\right) & & \text { (parallelogram law) } \end{aligned}\) (parallelogram law). Further, show that for each pair \((z, w) \in \mathbb{C}^{\bullet} \times \mathbb{C}^{\bullet}\) there is a unique real number \(\omega:=\omega(z, w) \in]-\pi, \pi]\) with $$ \cos \omega=\cos \omega(z, w)=\frac{\langle z, w\rangle}{|z||w|} $$ and $$ \sin \omega=\sin \omega(z, w)=\frac{\langle\mathrm{i} z, w\rangle}{|z||w|} $$ \(\omega=\omega(z, w)\) is called the oriented angle between \(z\) and \(w\) and will often be denoted by \(\angle(z, w)\) Show: \(\quad \angle(1, \mathrm{i})=\pi / 2, \angle(\mathrm{i}, 1)=-\pi / 2=-\angle(1, \mathrm{i}) .\)

Step-by-Step Solution

Answer

Cauchy-Schwarz inequality is verified. For angles: \(\angle(1, i) = \pi/2\), \(\angle(i, 1) = -\pi/2\).

Step 1: Verify with Direct Calculation

We begin by calculating \(\langle z, w\rangle^2 + \langle i z, w\rangle^2 \). For this, note that \( z \bar{w} = (x + iy)(u - iv) = (xu + yv) + i(yu - xv) \).Then:- \(\operatorname{Re}(z \bar{w}) = xu + yv = \langle z, w\rangle \)- \(\operatorname{Re}(i z \bar{w}) = \operatorname{Re}(i(x + iy)(u - iv)) = \operatorname{Re}((ix - y)(u - iv)) = yu - xv + i(xu + yv)\)- Thus, \(\langle i z, w\rangle = yu - xv \)Now,\[ \langle z, w\rangle^2 + \langle i z, w\rangle^2 = (xu + yv)^2 + (yu - xv)^2 = x^2u^2 + 2xyuv + y^2v^2 + y^2u^2 - 2xyuv + x^2v^2\]which simplifies to\[x^2u^2 + y^2v^2 + y^2u^2 + x^2v^2 = (x^2 + y^2)(u^2 + v^2) = |z|^2 |w|^2.\]

Step 2: Infer Cauchy-Schwarz Inequality

Using the result from Step 1, we have shown that \(\langle z, w\rangle^2 + \langle i z, w\rangle^2 = |z|^2 |w|^2\). Since both \(\langle z, w\rangle^2\) and \(\langle i z, w\rangle^2\) are non-negative, it follows that\[ \langle z, w\rangle^2 \leq |z|^2 |w|^2. \]This is precisely the Cauchy-Schwarz inequality in \(\mathbb{R}^2\):\[|x u + y v|^2 \leq (x^2 + y^2)(u^2 + v^2) = |z|^2 |w|^2.\]

Step 3: Verify Cosine Law

Calculate \(|z+w|^2\):\[z + w = (x + u) + i(y + v)\]Thus,\[|z+w|^2 = (x+u)^2 + (y+v)^2 = x^2 + 2xu + u^2 + y^2 + 2yv + v^2 = |z|^2 + |w|^2 + 2(xu + yv)\]which matches the cosine law:\[|z+w|^2 = |z|^2 + 2\langle z, w\rangle + |w|^2.\]

Step 4: Verify Second Identity

Now compute \(|z-w|^2\):\[z - w = (x - u) + i(y - v)\]Thus,\[|z-w|^2 = (x-u)^2 + (y-v)^2 = x^2 - 2xu + u^2 + y^2 - 2yv + v^2 = |z|^2 + |w|^2 - 2(xu + yv)\]which validates the identity:\[|z-w|^2 = |z|^2 - 2\langle z, w\rangle + |w|^2.\]

Step 5: Verify Parallelogram Law

Add the results from Steps 3 and 4 for the parallelogram law:\[|z+w|^2 + |z-w|^2 = (|z|^2 + 2\langle z, w\rangle + |w|^2) + (|z|^2 - 2\langle z, w\rangle + |w|^2)\]This simplifies to:\[2(|z|^2 + |w|^2)\]confirming the parallelogram law.

Step 6: Calculate Oriented Angle

We need to find the oriented angle, \(\omega\), where:\[\cos \omega = \frac{\langle z, w\rangle}{|z||w|} \quad \text{and} \quad \sin \omega = \frac{\langle i z, w\rangle}{|z||w|}\]For \(z=1, w=i\), we have:- \(\langle 1, i \rangle = \operatorname{Re}(1 \cdot \bar{i}) = \operatorname{Re}(0 - i) = 0\)- \(\langle i, 1 \rangle = \operatorname{Re}(-i \cdot 1) = \operatorname{Re}(-i) = 0\)- \(\langle i \cdot 1, i \rangle = \operatorname{Re}((-1) \cdot \bar{i}) = \operatorname{Re}(i) = 1\)Using these, \(\cos \omega = 0\) and \(\sin \omega = 1\) imply that \(\omega = \frac{\pi}{2}\) for \(\angle(1, i)\). Similarly, \(\angle(i, 1) = -\frac{\pi}{2}\).

Key Concepts

Scalar Product

The scalar product, also known as the dot product, is a fundamental concept in mathematics, especially in vector spaces. In the context of complex numbers, the scalar product of two complex numbers \( z = x + iy \) and \( w = u + iv \) is defined using the real part of their product. With the basis \((1, i)\), the scalar product \( \langle z, w \rangle \) becomes:

• \( \langle z, w \rangle = \operatorname{Re}(z \bar{w}) = xu + yv \)

Here, the bar over \( w \) signifies the complex conjugate of \( w \), changing the sign of its imaginary part. This product combines linear relationships of the real and imaginary components, and it is particularly useful in various geometric and algebraic calculations. The result is always a real number, indicative of the extent of correlation between the two vectors represented by \( z \) and \( w \).
Explore further to understand how this ties into more complex relationships such as the Cauchy-Schwarz Inequality.

Cauchy-Schwarz Inequality

The Cauchy-Schwarz Inequality is a fundamental inequality that applies to any inner product space, including complex numbers. It is crucial for bounding correlations and maintaining orthogonality between vectors. This inequality states that:

• \( |\langle z, w \rangle|^2 \leq |z|^2 |w|^2 \)

Here, \( |z| \) and \( |w| \) are the magnitudes (or norms) of the complex numbers \( z \) and \( w \), respectively. The inequality demonstrates how the absolute value of the scalar product is a measure of similarity but cannot exceed the product of the magnitudes of the two vectors. This is crucial in many fields, as it provides a maximum limit for the correlation between two data points or vectors.
The proof's essence lies in demonstrating that the sum of the squared scalar products is equal to the product of the squared magnitudes, securing the inequality’s validity.

Parallelogram Law

The Parallelogram Law provides a geometric interpretation of vector addition and subtraction using complex numbers. It claims that the sum of the squares of the diagonals of a parallelogram equals the sum of the squares of its sides. In mathematical terms:

• \(|z+w|^2 + |z-w|^2 = 2(|z|^2 + |w|^2)\)

This identity is validated by expressing the sum \( z + w \) and difference \( z - w \) of two complex numbers, and then calculating the magnitudes squared. Such relations allow us to represent vector positions and transformations in a two-dimensional plane, offering insights into the geometrical properties and symmetrical nature of shapes formed in a complex plane. Understanding this law can lead to a better grasp of the complex interplay between algebraic operations and geometric interpretations in vector spaces.