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:
- \(\frac{5}{2}\)
- 2
- 3
- \(\frac{3}{2}\)
- \( \frac{3}{2} \)
The Correct Option is
Approach Solution - 1
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: \(\frac{3}{2}\).
Approach Solution -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 \( \frac{3}{2} \).
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