Question

(a) Find all points where the function f(z) = (x^2+y^2-2y)+i(2x-2xy) is differentiable, and compute the derivative at those points.

Answer 1

Answer:

f is not complex-differentiable

Step-by-step explanation:

Let the functions u, v defined as follow

$u(x,y)=x^2+y^2-2y$

$v(x,y)=2x-2xy$

then f can be rewritten as

$f(x,y)=u(x,y)+iv(x,y)$

In order for f to be complex-differentiable in a point z=x+iy, it must satisfy the Cauchy-Riemann equations

$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$

$\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$

But

$\frac{\partial u}{\partial x}=2x$

$\frac{\partial v}{\partial y}=-2x$

and

$\frac{\partial u}{\partial x}\neq \frac{\partial v}{\partial y}$

As f does not satisfy the Cauchy-Riemann equations, f is not complex-differentiable.