Answer :

Solution :

$\oint \left(\frac{6\mathrm{z}}{2{\mathrm{z}}^4-3{\mathrm{z}}^3+7{\mathrm{z}}^2-3\mathrm{z}+5}\right)\mathrm{d}\mathrm{z}$
pole z = i

Check for singularity at pole z = i

f(z) = 2z⁴ - 3z³ + 7z² - 3z + 5

f(i) = 2(i)⁴ - 3(i)³ + 7(i)² - 3i + 5

f(i) = 2 ×1 - 3(-i) - 7 - 3i + 5 = 0 since, f(i) = 0 ⇒ z = i is a singular point From Residue theorem: