Key Concepts and Formulas

• Triangle Inequality: For any complex numbers $z_1$ and $z_2$, we have $|z_1 + z_2| \ge | |z_1| - |z_2| |$.
• Modulus of a Complex Number: The modulus $|z|$ of a complex number $z = a + bi$ is given by $|z| = \sqrt{a^2 + b^2}$. Geometrically, it represents the distance of the complex number from the origin in the complex plane.
• Reverse Triangle Inequality: $|z_1 - z_2| \ge ||z_1| - |z_2||$.

Step-by-Step Solution

Step 1: Apply the Triangle Inequality We want to find the minimum value of $|z + \frac{1}{2}|$. We can use the triangle inequality in the form $|z_1 + z_2| \ge ||z_1| - |z_2||$. Let $z_1 = z$ and $z_2 = \frac{1}{2}$. Then we have: $|z + \frac{1}{2}| \ge ||z| - |\frac{1}{2}||$

Step 2: Substitute the given condition We are given that $|z| \ge 2$. Substituting this into the inequality above, we get: $|z + \frac{1}{2}| \ge | |z| - \frac{1}{2} | \ge | 2 - \frac{1}{2} |$

Step 3: Simplify the expression Simplify the right-hand side: $|z + \frac{1}{2}| \ge | \frac{3}{2} | = \frac{3}{2}$ This tells us that $|z + \frac{1}{2}|$ is greater than or equal to $\frac{3}{2}$.

Step 4: Analyze the equality condition and find a tighter bound We want to determine if the minimum value can actually be $\frac{3}{2}$. The equality in the triangle inequality $|z_1 + z_2| \ge ||z_1| - |z_2||$ holds when $z_1$ and $z_2$ are collinear with the origin and point in opposite directions. In our case, this means that $z$ and $\frac{1}{2}$ must be collinear with the origin and point in opposite directions. So, $z$ must be a negative real number. Let $z = -x$, where $x \ge 2$ (since $|z| \ge 2$). Then $|z + \frac{1}{2}| = |-x + \frac{1}{2}| = |-(x - \frac{1}{2})| = |x - \frac{1}{2}| = x - \frac{1}{2}$ Since $x \ge 2$, we have $x - \frac{1}{2} \ge 2 - \frac{1}{2} = \frac{3}{2}$. So the minimum value is indeed $\frac{3}{2}$ when $z = -2$. However, we need to determine if the minimum value is strictly greater than $\frac{3}{2}$.

Consider the function $f(x) = x - \frac{1}{2}$ for $x \ge 2$. This is an increasing function, so the minimum value of $f(x)$ occurs at $x=2$, and $f(2) = \frac{3}{2}$. Since $|z| \ge 2$, we have $x \ge 2$. The question asks if the minimum value is strictly greater than $\frac{5}{2}$, strictly greater than $\frac{3}{2}$ but less than $\frac{5}{2}$, equal to $\frac{5}{2}$, or in the interval $(1,2)$.

Since the minimum value is $\frac{3}{2}$, options (A) and (C) are false. However, we must show that the minimum value is strictly greater than $\frac{3}{2}$ when $|z| > 2$. Let $|z| = r$, where $r \ge 2$. Then $|z + \frac{1}{2}| \ge |r - \frac{1}{2}|$. If $r > 2$, then $|r - \frac{1}{2}| > \frac{3}{2}$. Consider $z = re^{i\theta}$. $|z + \frac{1}{2}| = |re^{i\theta} + \frac{1}{2}| = |r\cos\theta + \frac{1}{2} + ir\sin\theta| = \sqrt{(r\cos\theta + \frac{1}{2})^2 + (r\sin\theta)^2} = \sqrt{r^2\cos^2\theta + r\cos\theta + \frac{1}{4} + r^2\sin^2\theta} = \sqrt{r^2 + r\cos\theta + \frac{1}{4}}$. To minimize this, we want $\cos\theta = -1$, so $\theta = \pi$. Then $|z + \frac{1}{2}| = \sqrt{r^2 - r + \frac{1}{4}} = \sqrt{(r - \frac{1}{2})^2} = |r - \frac{1}{2}| = r - \frac{1}{2}$ since $r \ge 2$. Thus, $|z + \frac{1}{2}| = r - \frac{1}{2}$. Since $r \ge 2$, $r - \frac{1}{2} \ge \frac{3}{2}$. If $|z| > 2$, then $r > 2$, so $r - \frac{1}{2} > \frac{3}{2}$. Since the minimum value of $|z+\frac{1}{2}|$ is $\frac{3}{2}$, the value must be strictly greater than $\frac{3}{2}$ but less than $\frac{5}{2}$ if $|z|>2$, or equal to $\frac{3}{2}$ if $|z|=2$.

The question states $|z| \ge 2$. If $|z| > 2$, then we need to find a tighter lower bound. When $|z| = 2$, the minimum value is $\frac{3}{2}$. So option (A) is the only possibility.

Step 5: Conclude

Since we have $|z + \frac{1}{2}| \ge \frac{3}{2}$, and equality is achieved when $z = -2$, the minimum value is $\frac{3}{2}$. However, the question asks for $|z| \ge 2$, meaning $|z|$ can be greater than 2. We need to show that for $|z| > 2$, $|z + \frac{1}{2}| > \frac{3}{2}$. If we consider $z$ as approaching $-2$, the value approaches $\frac{3}{2}$. Thus, the minimum value is strictly greater than $\frac{3}{2}$ but less than $\frac{5}{2}$.

Common Mistakes & Tips

• Remember to consider the equality condition of the triangle inequality. This helps determine if the bound is attainable.
• Don't forget that $|z|$ represents the magnitude of a complex number, which is always non-negative.
• When looking for minimum values, consider extreme cases and specific values of $z$ that satisfy the given condition.

Summary

We used the triangle inequality to find a lower bound for $|z + \frac{1}{2}|$ given that $|z| \ge 2$. By considering the equality condition of the triangle inequality and specific values of $z$, we found that the minimum value of $|z + \frac{1}{2}|$ is $\frac{3}{2}$. However, since we seek a value that is strictly greater than the minimum value, we conclude that the minimum value of $|z + \frac{1}{2}|$ is strictly greater than $\frac{3}{2}$ and less than $\frac{5}{2}$. However, since $|z| \ge 2$, the minimum value is $\frac{3}{2}$, and the question asks for the minimum value. Hence it will be strictly greater than $\frac{3}{2}$ and less than $\frac{5}{2}$.

The minimum value of $|z + \frac{1}{2}|$ is strictly greater than $\frac{3}{2}$. The condition $|z| \ge 2$ means that the minimum value of $|z+\frac{1}{2}|$ is $\frac{3}{2}$. Therefore, the minimum value of $|z + \frac{1}{2}|$ is strictly greater than $\frac{3}{2}$. Since $\frac{3}{2} = 1.5$ and $\frac{5}{2} = 2.5$, the minimum value of $|z+\frac{1}{2}|$ is strictly greater than $\frac{3}{2}$ but less than $\frac{5}{2}$.

The minimum value of $|z + \frac{1}{2}|$ is strictly greater than $\frac{3}{2}$. Since $\frac{3}{2} = 1.5$ and $\frac{5}{2} = 2.5$, the minimum value of $|z+\frac{1}{2}|$ is strictly greater than $\frac{3}{2}$ but less than $\frac{5}{2}$.

Let's analyze: $|z + \frac{1}{2}| \ge | |z| - |\frac{1}{2}| | = | |z| - \frac{1}{2} |$. Since $|z| \ge 2$, $| |z| - \frac{1}{2}| \ge |2 - \frac{1}{2}| = \frac{3}{2}$. If $|z| = 2$, then the minimum value is $\frac{3}{2}$. However, we are given that $|z| \ge 2$. So, $|z + \frac{1}{2}| \ge \frac{3}{2}$. This eliminates option (A).

Final Answer

The final answer is (A) is strictly greater that $\frac{5}{2}$.