Evaluate ∮_C (f(z))dz where C is the unit circle centered at the origin

Evaluate ∮_C (f(z))dz where C is the unit circle centered at the origin?, without using poles and unused residue theorem, Cauchy's theorem using only
1)f(z)==1/(2z^2+1)
2)f(z)=sqrt{z-4}

1) Both singularities of f(z) are inside C.

One way to do this is as follows:
∫c dz/(2z^2 + 1)
= ∫c dz/((z√2 + i)(z√2 - i))
= ∫c₁ dz/((z√2 + i)(z√2 - i)) + ∫c₂ dz/((z√2 + i)(z√2 - i))
where C_i encloses i/√2 and -i/√2 only, respectively (and not the other singularity).

= (1/√2) [∫c₁ (1/(z√2 + i)) dz/(z - i/√2)) + ∫c₂ (1/(z√2 - i)) dz/(z + i/√2)]
= (1/√2) [2πi * (1/(z√2 + i)) {at z = i/√2} + 2πi * (1/(z√2 - i)) {at z = -i/√2}], by Cauchy
= (1/√2) [2πi * (1/(2i)) + 2πi * (1/(-2i))]
= 0.

2) Since z = 4 is outside C, this integral equals 0 by Cauchy's Theorem.

I hope this helps!