Cauchy-Riemann equations (or condition)

Cauchy-Riemann equations

Proof of Cauchy-Riemann Equation

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

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

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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
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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...
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