If $$z$$ is a complex number of unit modulus and argument $$\theta$$, then $$\arg\left(\frac{1+z}{1+\bar{z}}\right)$$ can be equal to (given $$z \neq -1$$)

We are given that the complex number $$z$$ has unit modulus, that is $$|z| = 1$$, and its argument is $$\theta$$. A standard representation for such a number is Euler’s form

$$z = e^{i\theta}$$

because $$|e^{i\theta}| = 1$$ and $$\arg(e^{i\theta}) = \theta$$. The conjugate of $$z$$, denoted $$\bar z$$, therefore becomes

$$\bar z = e^{-i\theta}$$

Now we must examine the complex fraction

$$\frac{1 + z}{1 + \bar z}$$

Substituting the expressions $$z = e^{i\theta}$$ and $$\bar z = e^{-i\theta}$$, we obtain

$$\frac{1 + e^{i\theta}}{1 + e^{-i\theta}}$$

To uncover its argument, we will simplify this fraction completely. A well-known algebraic identity is the Euler cosine decomposition

$$1 + e^{i\theta} = e^{i\theta/2}\Bigl(e^{-i\theta/2} + e^{i\theta/2}\Bigr) = e^{i\theta/2}\cdot 2\cos\!\left(\frac{\theta}{2}\right).$$

Likewise,

$$1 + e^{-i\theta} = e^{-i\theta/2}\Bigl(e^{i\theta/2} + e^{-i\theta/2}\Bigr) = e^{-i\theta/2}\cdot 2\cos\!\left(\frac{\theta}{2}\right).$$

Placing these factorizations into the original fraction, we get

$$ \frac{1 + e^{i\theta}} {1 + e^{-i\theta}} \;=\; \frac{e^{\,i\theta/2}\,2\cos\!\left(\dfrac{\theta}{2}\right)} {e^{-i\theta/2}\,2\cos\!\left(\dfrac{\theta}{2}\right)}. $$

The common factor $$2\cos\!\left(\dfrac{\theta}{2}\right)$$ cancels immediately (recall the question explicitly states $$z \neq -1$$, which guarantees $$\cos\!\left(\dfrac{\theta}{2}\right)\neq 0$$ and therefore we are not dividing by zero). This leaves us with

$$ \frac{1 + e^{i\theta}}{1 + e^{-i\theta}} \;=\; \frac{e^{\,i\theta/2}}{e^{-i\theta/2}} \;=\; e^{\,i\theta}. $$

But the complex number $$e^{\,i\theta}$$ has modulus $$1$$ and, crucially, its argument is

$$\arg\bigl(e^{\,i\theta}\bigr) = \theta.$$

Therefore, after all simplification, the argument of the original fraction is the same as the argument of $$z$$ itself, namely $$\theta$$.

Hence, the correct answer is Option A.