b)
if you think of z as x+iy
im(z) = y
so f(z) =f (x,y) = y - 2i
now try to find two paths along which, the limit in the defn. of the derivative gives two values (show that the limiting value is not unique)
a)
Find the deriviative of the following function and specify the domain of this derviative
f(z)=[Log(4z-i)]/z^2
b)
Use the defintion of the derivative to prove that the function
f(z)=Im z -2i
is not differentiable at 1+2i
a)
Now I have attempted this question using the quotient Rule and ended up with something like this
f'(z)=[z^2(4/4z-i)-(2z)Log(4z-i)](z^2)^2
=[4z^2/(4z-i)-2zLog(4z-i)]/z^4
Now i am not sure on how to simplfy it further. The value thats getting to me is the most is the 4z^2/(4z-i)/z^4. Hope fully some one can show me how to do it and also help on what the domain is.
b)
For this part i know i am to use the formula
[f(z)-f(a)]/z-a
But how to do it is really confusing. I think as well that the Imz part is making it probably seem harder then it actually is.
Hopefully someone can point me in the right direction
Thanx
Bex
Ok I understand what you mean, but if i were to choose 2 sequences, say 1/n and -1/n, how can i do f(1/n) when in the equation f(z)=y-2i there is no z value. Even if the sequences I've chosen are wrong, I am going to end up at some point doing f(1+2i), and again there is no z value so will end up as 0. Not sure where to go from here. Help
Bex