Show the functions f= f(z); z=x+iy is nowhere di�fferentiable
f(z)=cosy -i siny
how do i do this? heres my idea
separate them into u(x,y) and v(x,y) then differentiate them and since the first derivative of u(x,y) and v(x,y) does not equal i can deduce that they are nowhere di�fferentiable?
hint on this question will be helpful. :)
That is correct. They have to satisfy the Cauchy-Riemann equations.
i see thanks :)
Your idea is on the right track! To determine if a function is nowhere differentiable, we need to check if the Cauchy-Riemann equations are satisfied. For a complex function f(z) = u(x, y) + iv(x, y), the Cauchy-Riemann equations are:
∂u/∂x = ∂v/∂y,
∂u/∂y = -∂v/∂x.
Let's apply this to the given function f(z) = cos(y) - i sin(y):
Separating the function into its real and imaginary parts, we have:
u(x, y) = cos(y),
v(x, y) = -sin(y).
Now, let's calculate the partial derivatives:
∂u/∂x = 0, since cos(y) does not involve x,
∂u/∂y = -sin(y),
∂v/∂x = 0, since -sin(y) does not involve x,
∂v/∂y = -cos(y).
Compared with the Cauchy-Riemann equations, we see that ∂u/∂x is not equal to ∂v/∂y. This means that the function f(z) = cos(y) - i sin(y) does not satisfy the Cauchy-Riemann equations and is nowhere differentiable.
Therefore, your conclusion is correct: the function f(z) = cos(y) - i sin(y) is indeed nowhere differentiable.