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:

The given function is differentiable at y = 1.

At y = 1, f'(z)  = 0

Step-by-step explanation:

As per the given question,

$f(z)\ = (x^{2}+y^{2}-2y)+i(2x - 2xy)$

Let z = x + i y

Suppose,

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

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

On computing the partial derivatives of u and v as:

$u'_{x} =2x$

$u'_{y}=2y -2$

And

$v'_{x} =2-2y$

$v'_{y}=-2x$

According to the Cauchy-Riemann equations

$u'_{x} =v'_{y} \ \ \ \ \ \ \ and\ \ \ \ \ \ u'_{y} = -v'_{x}$

Now,

$(u'_{x} =2x) \neq (v'_{y}=-2x)$

$(u'_{y}=2y -2) \ = \ (- v'_{x} =-(2-2y) =2y-2)$

Therefore,

$u'_{y}=- v'_{x}$ holds only.

This means,

2y - 2 = 0

⇒ y = 1

Therefore f(z) has a chance of being differentiable only at y =1.

Now we can compute the derivative

$f'(z)=\frac{1}{2}[(u'_{x}+iv'_{x})-i(u'_{y}+iv'_{y})]$

$f'(z) =\frac{1}{2}[(2x+i(2-2y))-i(2y-2+i(-2x))]$

$f'(z) = i(2-2y)$

At y = 1

f'(z) = 0

Hence, the required derivative at y = 1 ,  f'(z)  = 0