The path of integration is the circle with centre (0, 0) & radius =2

By Cauchy’s Integral formula value of the integral at z=1 i.e. $f(1) = ∫ \frac{(4z^2+z+5)}{(z-1)} \ dz$

= $2 πi [4z^2+z+5 ]_{z=1} = 20 πi$

Since z=i also lies inside C, $f(i) = ∫ \frac{(4z^2+z+5)}{(z-i)} \ dz$

= $2 πi [4z^2+z+5 ]_{z=i} = 2 πi (i-1)$

$f(a) = 2 πi [4z^2+z+5 ]_{z=a} = 2 πi [4a^2 + a +5 ]$

f’(a)= 2 πi [8a + 1]

∴ f’(-1)= 2 πi [8(-1) + 1 ] = -14 πi

Also f’’ (a) = 16 πi

∴ f’’(-i)= 16 πi