Question

If $z$ is a complex number such that $|z| \geq 1$, then the minimum value of \[\left| z + \frac{1}{2}(3 + 4i) \right|\]is:

• A. \(\frac{5}{2}\)
• B. 2
• C. 3
• D. \(\frac{3}{2}\)
• E. \( \frac{3}{2} \)

Answer 1

Answer: (E)

To determine the minimum value of \(\left| z + \frac{1}{2}(3 + 4i) \right|\), given that the complex number \(z\) satisfies \(|z| \geq 1\), we will proceed as follows:

The expression \(\left| z + \frac{1}{2}(3 + 4i) \right|\) can be rewritten using a substitution. Let the complex number \(z\) be represented as \(z = a + bi\), where \(a\) and \(b\) are real numbers.

The given condition \(|z| \geq 1\) translates to the inequality:

\(a^2 + b^2 \geq 1\)

Now, rewrite the target expression:

\(\left| z + \frac{1}{2}(3 + 4i) \right| = \left| (a + bi) + \left(\frac{3}{2} + 2i \right) \right|\)

This simplifies to:

\(\left| \left(a + \frac{3}{2}\right) + \left(b + 2\right)i \right|\)

According to the properties of complex numbers, the modulus is given by:

\(\sqrt{\left(a + \frac{3}{2}\right)^2 + \left(b + 2\right)^2}\)

We seek to minimize this expression under the constraint \(a^2 + b^2 \geq 1\).

Geometrical Interpretation:

The expression \(|z + \frac{1}{2}(3 + 4i)|\) represents the distance from the point \((-3/2, -2)\) to any point \((a, b)\) on or outside the circle centered at the origin with radius 1, represented by \(|z| \geq 1\).

The shortest distance from the point \((-1.5, -2)\) to the circle centered at the origin occurs along the line passing through the origin and \((-1.5, -2)\). We find this distance by first calculating the distance from the origin to the point \((-1.5, -2)\):

\(\sqrt{\left(-\frac{3}{2}\right)^2 + (-2)^2} = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{25}{4}} = \frac{5}{2}\)

The closest point on the circle to \((-1.5, -2)\) would achieve this minimum distance reduced by the circle's radius, which is 1:

\(\frac{5}{2} - 1 = \frac{3}{2}\)

Thus, the minimum value of \(\left| z + \frac{1}{2}(3 + 4i) \right|\) under the condition \(|z| \geq 1\) is \(\frac{3}{2}\).

Therefore, the correct answer is: \(\frac{3}{2}\).

Answer 2

Given:

\[ \text{Minimize } \left| z + \frac{1}{2}(3 + 4i) \right| \quad \text{subject to } |z| \geq 1 \]

Let:

\[ w = \frac{1}{2}(3 + 4i) \] \[ w = \left(\frac{3}{2}, 2\right) \]

We need to find the minimum value of:

\[ |z + w| = \left| z + \left(\frac{3}{2} + 2i\right) \right| \]

subject to \( |z| \geq 1 \).

The minimum distance from a point \( w \) to any point \( z \) on or outside the unit circle occurs when \( z \) lies on the boundary of the circle.

Thus, we find the minimum distance between \( w \) and the unit circle centered at the origin.

Distance Calculation
 

The distance of \( w = \left(\frac{3}{2}, 2\right) \) from the origin is given by:

\[ |w| = \sqrt{\left(\frac{3}{2}\right)^2 + 2^2} = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{25}{4}} = \frac{5}{2} \]

Since \( |z| \geq 1 \), the minimum value of \( |z + w| \) is obtained when \( z \) lies on the circle of radius 1, making the minimum value:

\[ |w| - 1 = \frac{5}{2} - 1 = \frac{3}{2} \]

Conclusion: The minimum value of \( \left| z + \frac{1}{2}(3 + 4i) \right| \) is \( \frac{3}{2} \).