1. Since $z$ has unit modulus and argument $\theta$, we can write $z=\cos\theta+i\sin\theta=e^{i\theta}.$ Hence, $\overline z=e^{-i\theta}.$

2. Now evaluate $\frac{1+z}{1+\overline z}=\frac{1+e^{i\theta}}{1+e^{-i\theta}}.$

3. Use the standard factorization: $1+e^{i\theta}=e^{i\theta/2}\left(e^{-i\theta/2}+e^{i\theta/2}\right)=2\cos\frac{\theta}{2}\,e^{i\theta/2},$ $1+e^{-i\theta}=e^{-i\theta/2}\left(e^{i\theta/2}+e^{-i\theta/2}\right)=2\cos\frac{\theta}{2}\,e^{-i\theta/2}.$

4. Therefore, \frac{1+z}{1+\overline z}=rac{2\cos(\theta/2)e^{i\theta/2}}{2\cos(\theta/2)e^{-i\theta/2}}=e^{i\theta}, provided $1+\overline z\neq 0$ (i.e. $z\neq -1$).

5. Hence, $\arg\left(\frac{1+z}{1+\overline z}\right)=\arg(e^{i\theta})=\theta.$

So the correct option is: $\boxed{\text{C: }\theta}$

6. Verification with stored answer:

• Derived answer: C
• Stored correct answer: C
• They match.