QJEEM24Mad2ef: JEE Main 2024 Mathematics - Class 11 JEEM24Mad2ef Solution: If is a complex number such that , then the minimum value of is :

To find the minimum value of $\left|z+\frac{1}{2}(3+4i)\right|$ , where $z$ is a complex number with $|z| \leqslant 1$ , we can think of this geometrically as the distance from any point inside or on the boundary of the unit circle in the complex plane to the fixed point $\frac{1}{2}(3+4i)$ .

To make this more clear, first write the expression as follows: $\left|z+\frac{1}{2}(3+4i)\right| = \left|z+\frac{3}{2}+\frac{4}{2}i\right| = \left|z+1.5+2i\right|$

This means we are looking for the distance between any point on the complex plane represented by $z$ (where $z$ has a magnitude of up to 1) and the point $1.5+2i$ .

The triangle inequality gives us the following relationship for any two complex numbers $z_1$ and $z_2$ : $\left|z_1 + z_2\right| \geqslant \left| \left|z_1\right| - \left|z_2\right| \right|$

Applying this to our specific case ( $z_1 = z$ and $z_2 = -1.5 - 2i$ ): $\left|z + 1.5 + 2i\right| \geqslant \left| \left|z\right| - \left|-1.5 - 2i\right| \right|$

Since $z$ is within the unit circle, $|z| \leqslant 1$ and $|-1.5 - 2i| = \sqrt{1.5^2 + 2^2} = \sqrt{2.25 + 4} = \sqrt{6.25} = 2.5$ .

So now we have: $\left|z + 1.5 + 2i\right| \geqslant \left| 1 - 2.5 \right|$

Which simplifies to: $\left|z + 1.5 + 2i\right| \geqslant 1.5$

Therefore, the minimum value of $\left|z+\frac{1}{2}(3+4i)\right|$ given that $|z| \leqslant 1$ is $\frac{3}{2}$ .

So the correct answer is:

Option C: $\frac{3}{2}$