Question

4. What is the analytic function whose real part is :
e* (x cos y – sin y)
Definecmooth neth​

Answer 1

Answer:

Let [y+ (e^x)cos(y)] =u(x,y). Then u_x = (e^x)cos(y) = u_xx. And u_y = 1 - (e^x)sin(y) and u_yy = -(e^x)cos(y) ==> u_xx + u_yy = 0 for all real x and y. Hence the given function u(x,y) is harmonic.

Suppose that its conjugate harmonic is v(x,y). Then by Cauchy-Riemann conditions v_y = u_x = (e^x)cos(y)……(1), and v_x = -u_y = (e^x)sin(y) -1……..(2). Integrating (1) with respect to y gives v(x,y) = (e^x)sin(y) + g(x)…….(3), and integrating (2) with respect to x gives that v(x,y) = v(x,y) = (e^x)sin(y) -x +h(y)……..(4). Equating (3) and (4) we get g(x)=(-x) and h(y) =0. Hence v(x,y) = (e^x)sin(y)-x.

Therefore f(z)=f(x+i.y) = u(x,y)+i.v(x,y) = [y+(e^x)cos(y)] +i.[(e^x)sin(y) - x] =

(e^x)[cos(y) +i.sin(y)] + (y-i.x) = (e^x)e^(i.y) -i(x+i.y) = e^(x+i.y) - i.z = e^z - i.z.