Triangle Inequality in Complex Analysis

Triangle Inequality in Complex Analysis

What is the Triangle Inequality?

In Complex Analysis, it states that any two complex number z1 and z2, the equality holds:

|z1 + z2| ≤ |z1| + |z2|


Here, |z| denotes the modulus of the complex number, and the modulus of complex number z = x + yi, where x is the real number and yi is the imaginary number. So, in complex analysis is defined as,

|z| = √(x2 + y2)

What is the Triangle Inequality number theory?

In Number Theory, the Triangle Inequality is discussed in the context of the absolute value function over the integers or the real number. It states that for any real number a and b, the Inequality holds:

What is the triangle inequality modulus formula?

As we discuess above,

|z1 + z2| ≤ |z1| + |z2|

Proof of Triangle Inequality in Complex Analysis

If z1 and z2 are two complex numbers, the prove that
|z1 + z2| ≤ |z1| + |z2|
Statement. |z1 + z2|2 = (z1 + z2)

If \( z_1 \) and \( z_2 \) are two complex numbers, prove that

\( |z_1 + z_2| \leq |z_1| + |z_2| \)

Proof:

\( |z_1 + z_2|^2 = (z_1 + z_2)(\overline{z_1 + z_2}) = (z_1 + z_2)(\overline{z_1} + \overline{z_2}) \)

\( |z_1 + z_2|^2 = (z_1 + z_2)(\overline{z_1 + z_2}) = (z_1 + z_2)(\overline{z_1} + \overline{z_2}) \)

\( = z_1\overline{z_1} + z_2\overline{z_2} + z_1\overline{z_2} + \overline{z_1}z_2 \)

\( = |z_1|^2 + |z_2|^2 + z_1\overline{z_2} + \overline{(z_1\overline{z_2})} \)

[∵ \( \overline{(z_1\overline{z_2})} = \overline{z_1}(\overline{\overline{z_2}}) = \overline{z_1}z_2 \) ]

\( = |z_1|^2 + |z_2|^2 + 2\operatorname{Re}(z_1\overline{z_2}) \)

[∵ \( \operatorname{Re}(z) = \frac{z + \overline{z}}{2} \), i.e., \( 2\operatorname{Re}(z) = z + \overline{z} \) ]

\( \leq |z_1|^2 + |z_2|^2 + 2 |z_1||z_2| \)

[∵ \( x^2 \leq x^2 + y^2 \), ∴ \( |x|^2 \leq |z|^2 \) or \( |x| \leq |z| \), i.e., \( \operatorname{Re} z \leq |z| \) ]

\( = |z_1|^2 + |z_2|^2 + 2 |z_1||\overline{z_2}| = |z_1|^2 + |z_2|^2 + 2 |z_1||z_2| \)

[∵ \( |\overline{z}| = |z| \) ]

\( = (|z_1| + |z_2|)^2 \)

∴ \( |z_1 + z_2|^2 \leq (|z_1| + |z_2|)^2 \)

∴ \( |z_1 + z_2| \leq |z_1| + |z_2| \)

Comments

Post a Comment

Popular posts from this blog

Renu Spends 68% of her income. When her income increases by 40 and expenditure by 30%. Her saving is?

Image
In this post, we are going to solve a math question which is very important for CHSL, CGL, SSC GD, SBI PO, IBPS etc. Q. Renu Spends 68% of her income. When her income increases by 40 and expenditure by 30%. Her saving is? Sol.  Step-by-Step  Step1: Assume that Renu's original income = 100 Her original expenditure = 68 of 100 = 68 Her original saving = 100 - 68 = 32 Step2: After 40% increase in income New income = 140 Step3: After 30% increase expenditure New expenditure = 30 of 68 = 20.4 Step4: Calculate new saving: New saving = New income + New expenditure = 140-88.4 =51.6 Step5: Increase in savings: Increase in saving = New saving - original saving = 51.6 - 32 = 19.6 Step6: Percentage increase in saving: Percentage increase = Increase in saving/Original saving * 100 Percentage increase = 19.6/32 * 100 = 61.25 Therefore, the Renu saves 61.25%. Below is the Youtube video where you can easily unde...
Read more

The Union of two subgroups of a group G may not be a subgroup of G

Image
Theorem: The Union of two subgroups of a group G may not be a subgroup of G    Proof:  Let G = Z be the additive group of integers.    ∀ ∈  H 1 ∪H 2 Let H1 = {2n: n = Z} = {....-4, -2, 0, 2, 4, 6, ....} H2 = {3n: n = Z} = {....-6, -3, 0, 3, 6, ....} Since          0 =   2(0)  ∈  H 1 , so  H 1 is a non-empty subset of G Let a, b =  H 1 . Then a = 2n 1 and b = 2n 2 for some n 1 , n 2   ∈  Z.  Now, a-b = 2n 1 - 2n 2 = 2(n 1 -n 2 )  ∈  H 1   a-b  ∈  H 1 a, b  ∈  H1 and so H1 is a subgroup of G.  Similarly, H2 is a subgroup of G. Also,  H 1 ∪H 2  = {..... -6, -4, -3, -2, 0, 2, 3, 4, 6, ...}  Now, 2, 3  ∈   H 1 ∪H 2 but 2-3=-1  ∉   H 1 ∪H 2  Thus,  H 1 ∪H 2  is not a subgroup of G. Youtube Video on this Theorem
Read more

Solve x^2 (dy/dx)^2-2xy dy/dx+2y^2-x^2=0 | Equation Solvable for p

  Equation Solvable for p  MathJax Example Solve \(x^2 \left(\frac{dy}{dx}\right)^2 - 2xy \frac{dy}{dx} + 2y^2 - x^2 = 0\). Solution: \[ \begin{align*} & \text{The given equation is} \\ & x^2 \left(\frac{dy}{dx}\right)^2 - 2xy \frac{dy}{dx} + 2y^2 - x^2 = 0 \\ & \text{or} \\ & \left(\frac{dy}{dx}\right)^2 - 2xy \frac{dy}{dx} + 2y^2 - x^2 = 0 \\ & \text{or} \\ & \left(\frac{dy}{dx}\right) = \frac{2xy \pm \sqrt{4x^2 y^2 - 4x^2 (2y^2 - x^2)}}{2x^2} \\ & \text{or} \\ & \left(\frac{dy}{dx}\right) = \frac{y \pm \sqrt{x^2 - y^2}}{x} \\ \end{align*} \] Put \(y = vx\) so that \(\frac{dy}{dx} = v + x \frac{dv}{dx}\). Therefore, (2) becomes \(v + x \frac{dv}{dx} = \frac{vx \pm \sqrt{x^2 - v^2 x^2}}{x}\). Therefore, \(v + x \frac{dv}{dx} = v \pm \sqrt{1 - v^2}\). Therefore, \(x \frac{dv}{dx} = \pm \sqrt{1 - v^2}\). Separating the variables, \(\frac{1}{\sqrt{1 - v^2}} dv = \pm \frac{1}{x} dx\). Ma...
Read more