Answer:
The given f(z) is analytic function
i) δU/δx = eˣ (cos y) = δV/δy
ii) δU/δy = eˣ (-sin y) = - δV/δx
Hence the CR - equations are satisfied
Step-by-step explanation:
Explanation:-
Analytic function:-
A point at which an analytic function ceases to posses a derivative is
called a singular point of the function.Thus the necessary and sufficient condition for a complex function f(Z) = U+i V is analytic in a region R are
δU/δx = δV/δy and δU/δy = -δV/δx ( C R equations)
Given f(Z) = eˣ(cos y+ i sin y)
Let U = eˣ(cos y) ....(i)
and
V = eˣ(sin y) ....(ii)
Differentiating equation(i) partially with respective to 'x'
δU/δx = eˣ (cos y)
Differentiating partially equation(i) with respective to 'y'
δU/δy = eˣ (-sin y)
Differentiating equation(ii) partially with respective to 'x'
δV/δx = eˣ (sin y)
Differentiating equation(ii) partially with respective to 'y'
δV/δy = eˣ (cos y)
Now
δU/δx = eˣ (cos y) = δV/δy
i) δU/δx = δV/δy
δU/δy = eˣ (-sin y) = - δV/δx
ii) δU/δy = - δV/δx
Hence the CR - equations are satisfied
There fore the given f(z) is analytic function