Key Concepts and Formulas

• Unit Modulus and Conjugate: If $|z| = 1$, then $z\overline{z} = 1$, which implies $\overline{z} = \frac{1}{z}$.
• Argument of a Complex Number: The argument of a complex number $z$, denoted as $\arg(z)$, is the angle it makes with the positive real axis in the complex plane.
• Argument Properties: $\arg(\frac{z_1}{z_2}) = \arg(z_1) - \arg(z_2)$.

Step-by-Step Solution

Step 1: Understand the Given Information

• What: We are given that $z$ is a complex number with $|z| = 1$ and $\arg(z) = \theta$. We want to find $\arg\left(\frac{1+z}{1+\overline{z}}\right)$.
• Why: This sets up the problem and defines the goal.

Step 2: Use the Unit Modulus Property

• What: Since $|z| = 1$, we know $\overline{z} = \frac{1}{z}$.
• Why: This allows us to express the expression in terms of only $z$, simplifying the problem.

Step 3: Substitute and Simplify

• What: Substitute $\overline{z} = \frac{1}{z}$ into the expression: $\arg\left(\frac{1+z}{1+\overline{z}}\right) = \arg\left(\frac{1+z}{1+\frac{1}{z}}\right)$
• Why: This substitution replaces the conjugate with an expression involving $z$.

Step 4: Simplify the Denominator

• What: Simplify the denominator: $1 + \frac{1}{z} = \frac{z}{z} + \frac{1}{z} = \frac{z+1}{z}$
• Why: Combining the terms in the denominator simplifies the overall fraction.

Step 5: Simplify the Complex Fraction

• What: Substitute the simplified denominator back into the expression and simplify: $\arg\left(\frac{1+z}{\frac{z+1}{z}}\right) = \arg\left((1+z) \cdot \frac{z}{z+1}\right)$ Assuming $z \neq -1$, we can cancel the $(1+z)$ terms: $\arg\left(\frac{1+z}{\frac{z+1}{z}}\right) = \arg(z)$
• Why: This simplifies the entire expression to just $z$, making it easy to find the argument. The condition $z \neq -1$ is implicit, because if $z = -1$, the original expression is undefined.

Step 6: Determine the Argument

• What: Since the expression simplifies to $z$, and we are given that $\arg(z) = \theta$, we have: $\arg\left(\frac{1+z}{1+\overline{z}}\right) = \arg(z) = \theta$
• Why: This directly gives us the argument of the original expression.

Step 7: Account for the Conjugate in the Denominator

• What: The original solution is incorrect. Let's revisit the substitution and simplification. We have: $\arg \left( \frac{1+z}{1+\overline{z}} \right) = \arg \left( \frac{1+z}{1+\frac{1}{z}} \right) = \arg \left( \frac{1+z}{\frac{z+1}{z}} \right) = \arg \left( \frac{(1+z)z}{1+z} \right)$ If $z \neq -1$, we can cancel $(1+z)$ to get $\arg(z) = \theta$ However, the problem statement is asking for $\arg \left( \frac{1+z}{1+\overline{z}} \right)$, and the correct answer is $-\theta$. Let's re-examine the problem and the simplification steps. Since $|z| = 1$, we have $\overline{z} = \frac{1}{z}$. Thus, $z = e^{i\theta}$ and $\overline{z} = e^{-i\theta}$. $\frac{1+z}{1+\overline{z}} = \frac{1+e^{i\theta}}{1+e^{-i\theta}} = \frac{1+e^{i\theta}}{1+e^{-i\theta}} \cdot \frac{e^{i\theta/2}}{e^{i\theta/2}} \cdot \frac{e^{i\theta/2}}{e^{i\theta/2}} = \frac{e^{-i\theta/2} (1+e^{i\theta})}{e^{-i\theta/2} (1+e^{-i\theta})} = \frac{e^{i\theta/2} + e^{i3\theta/2}}{e^{-i\theta/2} + e^{i\theta/2}} = \frac{e^{i\theta/2} (1+e^{i\theta})}{1+e^{-i\theta}}$ Multiplying the numerator and denominator by $z$, we get $\frac{1+z}{1+\overline{z}} = \frac{1+z}{1+\frac{1}{z}} = \frac{z(1+z)}{z+1} = z$ Therefore, $\arg\left( \frac{1+z}{1+\overline{z}} \right) = \arg(z) = \theta$. The provided answer is incorrect.

Step 8: Correct Solution

• What: We are given the correct answer is $-\theta$. Let's analyze the expression again: $\arg \left( \frac{1+z}{1+\overline{z}} \right) = \arg(1+z) - \arg(1+\overline{z})$ Let $z = \cos\theta + i\sin\theta$. Then $\overline{z} = \cos\theta - i\sin\theta$. $1+z = (1+\cos\theta) + i\sin\theta \qquad 1+\overline{z} = (1+\cos\theta) - i\sin\theta$ $\arg(1+z) = \arctan \left( \frac{\sin\theta}{1+\cos\theta} \right) = \arctan \left( \frac{2\sin(\theta/2)\cos(\theta/2)}{2\cos^2(\theta/2)} \right) = \arctan \left( \tan(\theta/2) \right) = \frac{\theta}{2}$ $\arg(1+\overline{z}) = \arctan \left( \frac{-\sin\theta}{1+\cos\theta} \right) = -\frac{\theta}{2}$ $\arg \left( \frac{1+z}{1+\overline{z}} \right) = \frac{\theta}{2} - \left( -\frac{\theta}{2} \right) = \theta$ This still leads to $\theta$.

Step 9: Discovering the Error

• What: Realize that $\frac{1+z}{1+\overline{z}} = \frac{1+z}{1+\frac{1}{z}} = \frac{z(1+z)}{z+1} = z$. However, the initial simplification is incorrect. Let's try to use $\overline{z} = 1/z$, thus, $\frac{1+z}{1+\overline{z}} = \frac{1+z}{1+\frac{1}{z}} = z$. Then $\arg(\frac{1+z}{1+\overline{z}}) = \arg(z) = \theta$. The given correct answer is $-\theta$. This is impossible. Let's try with $z = e^{i\theta}$. Then $\frac{1+e^{i\theta}}{1+e^{-i\theta}} = \frac{1+e^{i\theta}}{1+\frac{1}{e^{i\theta}}} = e^{i\theta} = z$. This leads to $\arg(z) = \theta$.

• Why: It appears that there is no way to arrive at $-\theta$, and the problem is flawed. However, the options are $-\theta, \frac{\pi}{2}-\theta, \theta, \pi-\theta$. The simplification to $z$ is correct, so the answer should be $\theta$.

Common Mistakes & Tips

• Always double-check algebraic manipulations with complex numbers.
• Be careful when canceling terms in complex fractions, ensuring that you are not dividing by zero.
• Remember the properties of conjugates and modulus.

Summary

By using the property $|z|=1$ and simplifying the expression, we find that $\arg\left(\frac{1+z}{1+\overline{z}}\right) = \theta$. However, the provided answer is $-\theta$, which is incorrect. The correct answer should be $\theta$.

The final answer is $\boxed{\theta}$, which corresponds to option (C).