Complex Numbers
Modulus Inequalities
Complex Numbers_PYQ
Grade 11
Question:
If $z$ is a complex number such that $|z|\geq 2$, then the minimum value of $\left|z+\dfrac{1}{2}\right|$
is equal to $\dfrac{5}{2}$
lies in the interval $\left(1,2\right)$
is strictly greater than $\dfrac{5}{2}$
is strictly greater than $\dfrac{3}{2}$ but less than $\dfrac{5}{2}$
Step-by-Step Solution
Key Concept: By the reverse triangle inequality the minimum of $|z+1/2|$ subject to $|z|\geq2$ is $|z|_{\min}-1/2=3/2$, which lies in $(1,2)$.
**Step 1: Apply reverse triangle inequality**
$\left|z+\dfrac{1}{2}\right|\geq\left||z|-\dfrac{1}{2}\right|=|z|-\dfrac{1}{2}$ (since $|z|\geq2>\dfrac{1}{2}$).
**Step 2: Find the lower bound**
$|z|-\dfrac{1}{2}\geq 2-\dfrac{1}{2}=\dfrac{3}{2}$. Minimum $\dfrac{3}{2}$ is achieved at $z=-2$: $\left|-2+\dfrac{1}{2}\right|=\dfrac{3}{2}$.
**Step 3: Match to options**
The minimum value is $\dfrac{3}{2}$, which lies in the open interval $(1,2)$. Option (b) is correct.
Correct Answer: 2