Question

The analytic function whose real part is e^x cos y is *

Answer 1

Answer:

the analystic function whose real part is e^x cos y is harmonic

Answer 2

• 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

                $v_y = u_x = (e^x)cos(y)……(1),$

                $v_x = -u_y = (e^x)sin(y) -1……..(2)$.

• 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$f(z)=f(x+i.y) = u(x,y)+i.v(x,y)$

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

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

                                             $= (e^x)e^(i.y) -i(x+i.y)$

                                             $= e^(x+i.y) - i.z$

                                             $= e^z - i.z.$