Answer:
The given function is differentiable at y = 1.
At y = 1, f'(z) = 0
Step-by-step explanation:
As per the given question,
Let z = x + i y
Suppose,
On computing the partial derivatives of u and v as:
And
According to the Cauchy-Riemann equations
Now,
Therefore,
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
At y = 1
f'(z) = 0
Hence, the required derivative at y = 1 , f'(z) = 0