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6 paź 2019 · If I solve the polynomial it would be $e^{(x^2-y^2+2ixy)}$ and if I use the function exponential of complex number like the previous example, it would be $e^{(x^2-y^2)}(\cos 2xy+i\sin 2xy)$ and when I derive it, they won't be the same ($U_x$ is not $V_y$, $U_y$ is not $-V_x$).
Our expert help has broken down your problem into an easy-to-learn solution you can count on. Here’s the best way to solve it. This AI-generated tip is based on Chegg's full solution. Sign up to see more! To determine whether is analytic using the Cauchy-Riemann equations, start by expressing as and then rewrite in terms of and . 2.
We give a simple proof of an important special case of the famous theorem of J ́osef Siciak on separate analyticity. 1. Introduction. The well-known theorem of Hartogs states that a function u(z1, z2) of two complex variables which is separately analytic must be analytic.
Theorem: Suppose f(z) = u(x,y)+iv(x,y) is differentiable at z = z0. Then u, v must satisfy the Cauchy-Riemann equations: ux = vy, uy = −vx Furthermore, f′(z) = ux +ivx = vy −iuy. Proof: in class Example 1: f(z) = z2 is differentiable and indeed, C-R are satisfied. Note f′(z) computed earlier
Topic 2 Notes Jeremy Orloff 2 Analytic functions 2.1 Introduction The main goal of this topic is to define and give some of the important properties of complex analytic functions. A function ( ) is analytic if it has a complex derivative ′ ( ). In general, the rules for computing derivatives will be familiar to you from single variable ...
course, since z is not analytic, we cannot expect mixed polynomials that contain powers of bothz and z to be analytic. Parts (b) and (c) of the theorem imply that f/gis differentiable at z if f and g are and if g(z) ̸= 0. They also imply the quotient rule % f g & ′ (z)= f′(z)g(z)−g′(z)f(z) g2(z). The chain rule also holds for the ...
(i) f(z) = z is analytic in the whole of C. Here u = x, v = y, and the Cauchy–Riemann equations are satisfied (1 = 1; 0 = 0). (ii) f(z) = zn (n a positive integer) is analytic in C. Here we write z = r(cosθ+isinθ) and by de Moivre’s theorem, z n= r (cosnθ + isinnθ). Hence u = r cosnθ and
