JEE Main 2014. Previous Year Question.
Problem
If $z$ is a complex number such that $|z|\geq2$, then the minimum value of $\left|z+\dfrac12\right|$:
Key insight. $\left|z+\tfrac12\right|$ is just the distance from $z$ to the fixed point $-\tfrac12$. Since $-\tfrac12$ sits well inside the disc $|z|<2$, the closest any point satisfying $|z|\geq2$ can get to it is along the straight line from the origin through $-\tfrac12$, out to where that line first meets the boundary circle.
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Approach
Write $z=r(\cos\theta+i\sin\theta)$ with $r\geq2$, so $|z+\tfrac12|^2$ becomes a function of $r$ and $\theta$. Minimise over $\theta$ first (which pins $\cos\theta=-1$), then over $r$ (which pins $r=2$, the smallest allowed value), since making either variable “more negative” or “smaller” only shrinks the distance.
Solution
Step 1 — Write z in polar form
$$z = r\cos\theta + ir\sin\theta, \qquad r\geq2$$
Step 2 — Express |z + 1/2|²
$$\left|z+\frac12\right|^2 = \left(r\cos\theta+\frac12\right)^2+r^2\sin^2\theta$$
Step 3 — Expand and simplify using sin²θ + cos²θ = 1
$$= r^2\cos^2\theta+r\cos\theta+\frac14+r^2\sin^2\theta = r^2+r\cos\theta+\frac14$$
Step 4 — Minimise over θ
Since $r>0$, the term $r\cos\theta$ is smallest when $\cos\theta=-1$:
$$\left|z+\frac12\right|^2_{\min\text{ over }\theta} = r^2-r+\frac14$$
Step 5 — Minimise over r ≥ 2
Increasing $r$ only increases $r^2-r+\tfrac14$ for $r\geq2$ (since $2r-1>0$ there), so the smallest value occurs at the boundary $r=2$:
$$4-2+\frac14 = 2+\frac14 = \frac{9}{4}$$
Step 6 — Take the square root
$$\left|z+\frac12\right|_{\min} = \sqrt{\frac94} = \frac32$$
Step 7 — Match against the options
The minimum value is exactly $\dfrac32$. Checking the options: it’s not equal to $\dfrac52$ (rules out a), it’s not strictly greater than $\dfrac32$ since it equals $\dfrac32$ (rules out d), and it’s not strictly greater than $\dfrac52$ (rules out c). Since $0<\dfrac32<2$, option (b) — “lies in the interval $(0,2)$” — is the one that correctly describes it.
Answer
Minimum value is $\dfrac32$, which lies in the interval $(0,2)$ — option (b).
Common mistakes
- Picking option (d) because $\dfrac32$ “sounds like” the boundary it mentions. Option (d) requires the value to be strictly greater than $\dfrac32$, but the true minimum equals $\dfrac32$ exactly — a boundary case option needs the inequality to be checked as strict versus non-strict, not just numerically close.
- Minimising only over $\theta$ and stopping there. Since $r$ is also free to range over $[2,\infty)$, both variables need to be pushed to their most favourable values — here, $r=2$ (the smallest allowed radius) turns out to give the overall minimum.
Practise next
- If $|z|\geq3$, find the minimum value of $\left|z+1\right|$, using the same polar-form minimisation over both $\theta$ and $r$.
Show answer
$2$. $|z+1|$ is the distance from $z$ to $-1$, and $z$ ranges over the region $|z|\ge3$ — everything outside the circle of radius $3$.
In polar form, $|z+1|^2=r^2+2r\cos\theta+1$. For fixed $r$ this is smallest at $\cos\theta=-1$, giving $(r-1)^2$; and $(r-1)^2$ increases with $r$ for $r\ge3$, so the smallest $r$ is best.
So the minimum is at $r=3$, $\theta=\pi$ — the point $z=-3$ — and equals $3-1=2$.
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