Problem 17 - Complex numbers from A to...Z

Question

$$ \text { Prove that for all complex numbers } z \text { with }|z|=1 \text { the following inequal- } $$ $$ \text { ities hold: } $$ $$ \sqrt{2} \leq|1-z|+\left|1+z^{2}\right| \leq 4 $$

Step-by-Step Solution

Answer

Question: Prove that for any complex number z with |z|=1, the following inequality holds: $$\sqrt{2} \leq|1-z|+\left|1+z^{2}\right| \leq 4$$ Answer: We can prove the inequality by first expressing z as \(z = e^{i\theta}\) for some angle \(\theta\), then finding the magnitudes of the complex numbers involved as |1-z| and \(|1+z^2|\), and finally using the triangle inequality to find the lower and upper bounds. The proof shows that the inequality holds since \(\sqrt{2} \leq|1-z|+\left|1+z^{2}\right| \leq 4\).

Step 1: Magnitude of Complex Numbers

Given that |z|=1, we can write \(z = e^{i\theta}\) for some angle \(\theta\). Therefore, we have: $$ 1 - z = 1 - e^{i\theta}$$ $$ 1 + z^2 = 1 + (e^{i\theta})^2 = 1 + e^{2i\theta}$$ Now, let's find the magnitude of these complex numbers: $$|1-z| = |1-e^{i\theta}|$$ $$\left|1+z^{2}\right| = \left|1+e^{2i\theta}\right|$$

Step 2: Lower Bound \(\sqrt{2}\)

Using the triangle inequality, we get: $$|1-z|+|1-z^2| \ge |(1-z)+(1-z^2)|=|2-2\cos(\theta)|$$ Notice that \( 2 - 2\cos(\theta) = 2(1-\cos(\theta))\), and using the identity \(1-\cos(\theta) = 2\sin^2(\frac{\theta}{2})\), we have: $$|2-2\cos(\theta)| = 2(1-\cos(\theta)) = 4\sin^2(\frac{\theta}{2})$$ Since \(0 \le \sin^2(\frac{\theta}{2}) \le 1\), the minimum value for \(|2-2\cos(\theta)|\) is \(2\) when \(\sin^2(\frac{\theta}{2}) = \frac{1}{2}\). Therefore, the lowest possible value for the left side of the inequality is \(\sqrt{2}\): $$|1-z|+\left|1+z^{2}\right| \ge \sqrt{2}$$

Step 3: Upper Bound \(4\)

We can use the triangle inequality again, but this time we will reverse it: $$|1-z|+|1-z^2| \le |(1-z)-(1-z^2)|=|-2\sin(\theta)\sin(\theta - \pi)|$$ Since \(0 \le \sin(\theta) \le 1\) and \(0 \le \sin(\theta - \pi) \le 1\), the maximum value for \(|-2\sin(\theta)\sin(\theta - \pi)|\) is 4. Therefore, the highest possible value for the right side of the inequality is 4: $$|1-z|+\left|1+z^{2}\right| \le 4$$

Step 4: Inequalities

Finally, putting it all together, we have proved the inequalities: $$\sqrt{2} \leq|1-z|+\left|1+z^{2}\right| \leq 4$$

Key Concepts

triangle inequality

In the world of complex numbers, the triangle inequality is an essential concept. It allows us to compare the sum of the magnitudes of two complex numbers with the magnitude of their sum. Mathematically, this is expressed as:

• \(|z_1 + z_2| \leq |z_1| + |z_2| \)

This inequality indicates that the direct path (magnitude of the sum) is always shorter or equal to the combined path (sum of the magnitudes) when navigating the complex plane. It draws a parallel to the geometric idea of forming a triangle, where the length of any one side is always less than or equal to the sum of the lengths of the other two sides. Understanding this helps in grasping how magnitudes behave when we tackle complex expressions.
When proving inequalities like in the exercise, the triangle inequality provides lower and upper limits to evaluate expressions containing complex number sums. By setting these bounds meticulously, we leverage inherent properties of complex numbers to effectively solve the inequality problem.

magnitude of complex numbers

The magnitude, or modulus, of a complex number represents its distance from the origin on the complex plane. For a complex number represented as \(z = a + bi\), its magnitude is given by:

• \(|z| = \sqrt{a^2 + b^2}\)

This framework mirrors how distance in a plane is typically calculated using Pythagoras’ theorem.
In the context of our problem, it is given that \(|z| = 1\), meaning the complex number is on the unit circle. This implies that for any expression involving \(z\), especially when represented in polar coordinates, we simplify our analysis by knowing the magnitude. This was crucial in the solution as it directly led to simplifying \(z\) to the form \(e^{i\theta}\), exploiting the rotational symmetry in calculations.
Understanding how magnitudes of newly formed expressions like \(1 - z\) and \(1 + z^2\) behave anchors much of the problem solving around ensuring our inequalities are accurately depicted.

complex numbers on unit circle

The unit circle in the complex plane is central to many explorations in complex number theory. It consists of all points where the magnitude of complex numbers \(z\) is equal to one:

• \(|z| = 1\)

When a complex number lies on this circle, it has a direct relationship with trigonometric principles. It can be expressed as \(z = e^{i\theta}\), where \(\theta\) represents the angle from the positive x-axis.
The implication of a number lying on the unit circle is profound because any transformation involving these numbers retains the property of unity magnitude. Therefore, trigonometric identities can be seamlessly integrated into calculations, greatly simplifying analyses, such as the exercise provided.
When dealing with inequalities, this property helps in transforming the problem into one that relies on angle manipulation and trigonometric identities, rather than complex arithmetic. This reduces potential complications and roots the solution in intuitive geometric interpretations, enhancing the simplicity of the problems being tackled.

Euler's formula

Euler’s formula is a landmark relation in complex number theory, linking exponential functions with trigonometric identities:

• \(e^{i\theta} = \cos(\theta) + i\sin(\theta)\)

This concise equation translates complex numbers into a form that is both algebraically meaningful and geometrically illustrative. When described as \(z = e^{i\theta}\), it means that any point on the unit circle can be depicted using both its cosine and sine components relative to angle \(\theta\).
In the exercise, we use Euler’s formula to express \(z\) and \(z^2\), simplifying these expressions to focus purely on trigonometric components. This dual representation strengthens our ability to use trigonometric identities, aiding in the concise manipulation of inequalities.
Euler’s formula is not just a bridge between algebra and geometry; it’s a powerful tool for unraveling the structure of complex expressions, helping break down complex problems into manageable, recognisable patterns tied to circular symmetry.