Respuesta :
Answer:
The given function is differentiable at y = 1.
At y = 1, f'(z) = 0
Step-by-step explanation:
As per the given question,
[tex]f(z)\ = (x^{2}+y^{2}-2y)+i(2x - 2xy)[/tex]
Let z = x + i y
Suppose,
[tex]u(x,y) = x^{2}+y^{2}-2y[/tex]
[tex]v(x,y) = 2x - 2xy[/tex]
On computing the partial derivatives of u and v as:
[tex]u'_{x} =2x[/tex]
[tex]u'_{y}=2y -2[/tex]
And
[tex]v'_{x} =2-2y[/tex]
[tex]v'_{y}=-2x[/tex]
According to the Cauchy-Riemann equations
[tex]u'_{x} =v'_{y} \ \ \ \ \ \ \ and\ \ \ \ \ \ u'_{y} = -v'_{x}[/tex]
Now,
[tex](u'_{x} =2x) \neq (v'_{y}=-2x)[/tex]
[tex](u'_{y}=2y -2) \ = \ (- v'_{x} =-(2-2y) =2y-2)[/tex]
Therefore,
[tex]u'_{y}=- v'_{x}[/tex] 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
[tex]f'(z)=\frac{1}{2}[(u'_{x}+iv'_{x})-i(u'_{y}+iv'_{y})][/tex]
[tex]f'(z) =\frac{1}{2}[(2x+i(2-2y))-i(2y-2+i(-2x))][/tex]
[tex]f'(z) = i(2-2y)[/tex]
At y = 1
f'(z) = 0
Hence, the required derivative at y = 1 , f'(z) = 0
