Key Concepts and Formulas

• Triangle Inequality: For any complex numbers $z_1$ and $z_2$, $|z_1 + z_2| \leq |z_1| + |z_2|$. Equality holds if and only if $z_1$ and $z_2$ have the same argument.
• Argument of a Complex Number: The argument of a complex number $z$, denoted as arg($z$), is the angle that the vector representing $z$ makes with the positive real axis in the Argand plane.
• Modulus and Conjugate: For any complex number $z$, $|z|^2 = z\bar{z}$, where $\bar{z}$ is the complex conjugate of $z$.

Step-by-Step Solution

Step 1: Analyze the given condition and relate it to the Triangle Inequality.

The problem states that for two non-zero complex numbers $z_1$ and $z_2$, $|z_1 + z_2| = |z_1| + |z_2|$ This is the equality condition of the Triangle Inequality. This means that the vectors representing $z_1$ and $z_2$ in the Argand plane are collinear and point in the same direction. This is because the sum of the lengths of the individual vectors equals the length of their sum, which can only occur if they are aligned.

Step 2: Interpret the geometric implication of the equality condition.

Since $|z_1 + z_2| = |z_1| + |z_2|$, the complex numbers $z_1$ and $z_2$ lie on the same ray originating from the origin. This geometric interpretation is crucial because it directly relates to the arguments of the complex numbers. If they lie on the same ray, their arguments must be equal.

Step 3: Relate the geometric interpretation to the arguments of the complex numbers.

The argument of a complex number is the angle it makes with the positive real axis. If $z_1$ and $z_2$ lie on the same ray from the origin, then they make the same angle with the positive real axis. Therefore, $\text{arg}(z_1) = \text{arg}(z_2)$

Step 4: Calculate the difference in arguments.

Since $\text{arg}(z_1) = \text{arg}(z_2)$, their difference is: $\text{arg}(z_1) - \text{arg}(z_2) = 0$

Step 5: Provide an algebraic proof for enhanced rigor.

We are given $|z_1 + z_2| = |z_1| + |z_2|$. Squaring both sides, we have $|z_1 + z_2|^2 = (|z_1| + |z_2|)^2$ Using the property $|z|^2 = z\bar{z}$, we get $(z_1 + z_2)(\overline{z_1 + z_2}) = (z_1 + z_2)(\bar{z_1} + \bar{z_2}) = |z_1|^2 + z_1\bar{z_2} + z_2\bar{z_1} + |z_2|^2$ Also, $(|z_1| + |z_2|)^2 = |z_1|^2 + 2|z_1||z_2| + |z_2|^2$. Equating the two expressions, we have $|z_1|^2 + z_1\bar{z_2} + z_2\bar{z_1} + |z_2|^2 = |z_1|^2 + 2|z_1||z_2| + |z_2|^2$ $z_1\bar{z_2} + z_2\bar{z_1} = 2|z_1||z_2|$ Since $z_2\bar{z_1} = \overline{z_1\bar{z_2}}$, we can write $z_1\bar{z_2} + \overline{z_1\bar{z_2}} = 2|z_1||z_2|$ $2\text{Re}(z_1\bar{z_2}) = 2|z_1||z_2|$ $\text{Re}(z_1\bar{z_2}) = |z_1||z_2|$ Let $z_1 = r_1e^{i\theta_1}$ and $z_2 = r_2e^{i\theta_2}$. Then $\bar{z_2} = r_2e^{-i\theta_2}$. $z_1\bar{z_2} = r_1r_2e^{i(\theta_1 - \theta_2)} = r_1r_2[\cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2)]$ $\text{Re}(z_1\bar{z_2}) = r_1r_2\cos(\theta_1 - \theta_2)$ Therefore, $r_1r_2\cos(\theta_1 - \theta_2) = r_1r_2$. Since $r_1, r_2 \neq 0$, $\cos(\theta_1 - \theta_2) = 1$ $\theta_1 - \theta_2 = 0$ $\text{arg}(z_1) - \text{arg}(z_2) = 0$

Common Mistakes & Tips

• Mistake: Confusing the equality condition of the triangle inequality with $z_1$ and $z_2$ pointing in opposite directions. The given condition specifically implies they point in the same direction.
• Tip: Always visualize complex numbers as vectors in the Argand plane. This helps in understanding the geometric implications of various operations and properties.
• Tip: Remember that the equality in the triangle inequality holds only when the complex numbers are collinear and point in the same direction.

Summary

The given condition $|z_1 + z_2| = |z_1| + |z_2|$ implies that $z_1$ and $z_2$ have the same argument. Therefore, the difference between their arguments is zero.

The final answer is \boxed{0}, which corresponds to option (C).