If $z$ is a complex number such that $|z|\ge 1$, then the minimum value of $\left|z+\frac{1}{2}(3+4i)\right|$ is:
[Note: In official NTA paper no option was correct.]
Step-by-Step Solution
Key Concept: The point $-\frac{1}{2}(3+4i)=(-\frac{3}{2},-2)$ has modulus $\frac{5}{2}>1$, so it lies outside the unit disk $|z|\ge 1$. The minimum of $|z-P|$ for $|z|\ge 1$ where $P$ is outside the unit circle is $|P|-1=\frac{5}{2}-1=\frac{3}{2}$. But the NTA originally had no correct option; MathonGo determined the actual minimum is 0.
The expression $|z+\frac{3}{2}+2i|$ represents the distance from $z$ to the fixed point $P=(-\frac{3}{2},-2)$. Since $|P|=\sqrt{\frac{9}{4}+4}=\frac{5}{2}>1$, $P$ lies outside the unit disk. For $|z|\ge1$, the minimum distance from $z$ to $P$ is $|P|-1=\frac{5}{2}-1=\frac{3}{2}$. MathonGo notes the actual minimum value of $|z+\frac{3}{2}+2i|$ is 0 (original NTA paper had no correct option).
Correct Answer: 4