Answer
$ k'(z) = \dfrac{-1}{2z^{2}} \\k'(-1) = -\dfrac{1}{2}\\ k'(1) = -\dfrac{1}{2} \\ k'(\sqrt 2) = -\dfrac{1}{4}$
Work Step by Step
Consider $k(z) = \dfrac{1-z}{2z}$
We will use the derivative definition, such that
$k'(z) = \lim\limits_{h \to 0}\dfrac{\dfrac{1-(z+h)}{2(z+h)} - \dfrac{1-z}{2z}}{h}= \lim\limits_{h \to 0}(\dfrac{1}{h})(\dfrac{-2h}{4z(z+h)})= \dfrac{-1}{2z^{2}}$
Now, we have
$k'(-1) =\dfrac{-1}{2(-1)^{2}}= -\dfrac{1}{2}\\ k'(1) =\dfrac{-1}{2(1)^{2}}= -\dfrac{1}{2} \\ k'(\sqrt 2) =\dfrac{-1}{2(\sqrt 2)^{2}} = -\dfrac{1}{4}$
