Question

Show that the function u(x, y)= ex cos y is harmonic. Determine its harmonic
conjugate v(x, y) and the analytic function f(z) = u + iv.

Answer 1

Answer:

∂u/∂x =e^x cos(y) , ∂2u/∂x2=e^x cos(y)

∂u/∂y= -e^x sin(y) , ∂2u/∂y2= -e^x cos(y)

∂2u∂x2+∂2u∂y2=e^x cos(y)+ (-e^x cos(y))

∂2u∂x2+∂2u∂y2=0.

∴ U is the harmonic function

Ux=Vy

∴Vy=e^x cos(y) .

f(z)=u+iv=  e^x cos(y)+i(e^x cos(y))=e^z+c

f(z)=e(x+iy)=e^x∗eiy=e^x(cosy+icos(y))=e^xcosy+i e^xcos(y).

Step-by-step explanation:

Answer 2

Answer:

du/dx =e^x cos(y), d2u/dx2=e^x cos(y)

du/dy=-e^x sin(y), d2u/dy2= -e^x cos(y)

d2udx2+d2udy2=e^x cos(y)+ (-e^x cos(y))

d2udx2+d2udy2=0. .. U is the harmonic function

Ux=Vy

.:Vy=e^x cos(y).

f(z)=u+iv=e^x cos(y)+i(e^x cos(y))=e^z+c

f(z)=e(x+iy)=e^x*eiy=e^x(

cosy+icos(y))=e^xcosy+i

e^xcos(y).