JEE Main 2024 — 1 February, Shift 2 — Question 12

We are asked to find the minimum value of $|z + \frac{1}{2}(3+4i)|$ given that $z$ is a complex number such that $|z| \geq 1$.

Let the expression be $E = |z + \frac{1}{2}(3+4i)|$. First, simplify the constant complex number: $\frac{1}{2}(3+4i) = \frac{3}{2} + 2i$.

So, we want to find the minimum value of $E = |z + (\frac{3}{2} + 2i)|$. Let $Z_0 = -(\frac{3}{2} + 2i)$. The expression $E$ can be written as $E = |z - Z_0|$. This represents the distance between the complex number $z$ and the fixed complex number $Z_0$ in the complex plane.

Next, calculate the modulus of $Z_0$: $|Z_0| = |-(\frac{3}{2} + 2i)| = |\frac{3}{2} + 2i|$ $|Z_0| = \sqrt{\left(\frac{3}{2}\right)^2 + (2)^2} = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{9+16}{4}} = \sqrt{\frac{25}{4}} = \frac{5}{2}$.

The condition $|z| \geq 1$ means that $z$ lies on or outside the circle centered at the origin with a radius of 1. Since $|Z_0| = \frac{5}{2} = 2.5$, and $2.5 > 1$, the point $Z_0$ lies outside the circle $|z|=1$.

To find the minimum distance from a point $z$ in the region $|z| \geq 1$ to the point $Z_0$, we consider the line segment connecting the origin to $Z_0$. The closest point $z$ on the boundary $|z|=1$ to $Z_0$ will lie on this line segment.

The minimum distance will be the distance from the origin to $Z_0$ minus the radius of the circle: Minimum distance $= |Z_0| - \text{Radius of the circle}$ Minimum distance $= \frac{5}{2} - 1$ Minimum distance $= \frac{5}{2} - \frac{2}{2} = \frac{3}{2}$.

This can also be understood using the triangle inequality: $|z_1 + z_2| \geq ||z_1| - |z_2||$. Let $z_1 = z$ and $z_2 = \frac{3}{2} + 2i$. Then $|z + (\frac{3}{2} + 2i)| \geq ||z| - |\frac{3}{2} + 2i|| = ||z| - \frac{5}{2}|$. Let $k = |z|$. We are given $k \geq 1$. We need to minimize the expression $|k - \frac{5}{2}|$. Since $\frac{5}{2} = 2.5$, and $k \geq 1$, the expression $|k - 2.5|$ is minimized when $k$ is as close as possible to $2.5$. The minimum occurs at $k = 1$ when $2.5$ is outside the range, but here $2.5$ is in the range of possible $k$ values (since $k \ge 1$ allows $k=2.5$). However, the actual minimum value for the expression $|z+w|$ for $|z| \ge R$ is $|w|-R$ if $|w|>R$. This happens when $z$ is in the direction opposite to $w$. More precisely, for $|z+w|$ with $|z| \ge R$, the minimum occurs when $z$ is on the boundary $|z|=R$ and $z$ is in the direction of $-w$. So $z = -R \frac{w}{|w|}$. In our case, $w = \frac{3}{2} + 2i$, $R=1$. The minimum value is $|w| - R = \frac{5}{2} - 1 = \frac{3}{2}$.

The final answer is $\boxed{\frac{3}{2}}$.