Question
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. :)
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. :)
Answers
Steve
That is correct. They have to satisfy the Cauchy-Riemann equations.
lim
i see thanks :)