Question

Find the analytic region of f(z) = (x - y)²+ 2i (x+y)​

Answer 1

Answer:

Sufficient condition for a function being analytic: The sufficient condition for a function f(z)=u+ivf(z)=u+iv to be analytic at all the points in region RR are

(a)∂u∂x=∂v∂y,∂u∂y=−∂v∂x(a)∂u∂x=∂v∂y,∂u∂y=−∂v∂x

(b)∂u∂x,∂v∂y,∂u∂y,∂v∂x(b)∂u∂x,∂v∂y,∂u∂y,∂v∂x are continuous functions of xx and yy in region RR .

Here, u=2xyu=2xy , 

Answer 2

Answer :

{z = x + iy : x - y = 1}

Solution :

• Given : f(z) = (x - y)² + 2i(x + y)
• To find : Analytic region

We have ,

f(z) = (x - y)² + 2i(x + y)

Here ,

u = (x - y)² and v = 2(x + y)

→ ∂u/∂x = 2(x - y) and ∂v/∂x = 2

∂u/∂y = -2(x - y) and ∂v/dy = 2

Clearly ,

All the first order derivatives ∂u/∂x , ∂u/∂y , ∂v/∂x and ∂v/∂y are continuous everywhere .

Now ,

The function f(z) will be analytic if the Cauchy Riemann equations are satisfied .

ie. ∂u/∂x = ∂v/∂y and ∂u/∂y = -∂v/∂x

→ 2(x - y) = 2 and -2(x - y) = -2

→ x - y = 1 and x - y = 1

→ x - y = 1

Hence ,

f(z) is analytic in the region {z = x + iy : x - y = 1}