Answer
$\displaystyle k'(z) = \frac{-1}{2z^{2}}$
$\displaystyle k'(-1) = -\frac{1}{2}$
$\displaystyle k'(1) = -\frac{1}{2}$
$\displaystyle k'(\sqrt 2) = -\frac{1}{4}$
Work Step by Step
$k(z) = \frac{1-z}{2z}$
Using the derivative definition:
$k'(z) = \lim\limits_{h \to 0}\frac{\frac{1-(z+h)}{2(z+h)} - \frac{1-z}{2z}}{h}$
$k'(z) = \lim\limits_{h \to 0}\frac{1}{h}\frac{-2h}{4z(z+h)}$
$k'(z) = \lim\limits_{h \to 0}\frac{-1}{2z(z+h)}$
$k'(z) = \frac{-1}{2z^{2}}$
Now,
$k'(-1) = -\frac{1}{2}$
$k'(1) = -\frac{1}{2}$
$k'(\sqrt 2) = -\frac{1}{4}$
