If $z$ is a complex number of unit modulus and argument $\theta$, then $\arg\!\left(\dfrac{1+z}{1+\bar{z}}\right)$ is equal to
Step-by-Step Solution
Key Concept: For $|z|=1$, both $1+z$ and $1+\bar{z}$ have the same real modulus $2\cos(\theta/2)$, so their ratio is the pure phase $e^{i\theta}$.
**Step 1: Write z in polar form**
$z = e^{i\theta}$, so $\bar{z} = e^{-i\theta}$.
**Step 2: Compute the ratio**
$1+z = 2\cos\dfrac{\theta}{2}\cdot e^{i\theta/2}$ and $1+\bar{z} = 2\cos\dfrac{\theta}{2}\cdot e^{-i\theta/2}$. Therefore $\dfrac{1+z}{1+\bar{z}} = e^{i\theta}$.
**Step 3: Read the argument**
$\arg\!\left(\dfrac{1+z}{1+\bar{z}}\right) = \arg(e^{i\theta}) = \theta$.
Correct Answer: 3