If z is a complex number such that |z| ≤ 1 , then the minimum value...

Given: $\left|z + \dfrac{3 + 4i}{2}\right|; \left|z\right| \le 1$

Let, $z = x + iy$

And \left|z\right| = 1$gives${x}^{2} + {y}^{2} = 1$, which represents a circle having centre at \left(0, 0\right)$ and radius as $1$ unit.

Now, \left|z + \dfrac{3 + 4i}{2}\right|$gives the distance of$z$from \left(- \dfrac{3}{2}, - 2\right)$

So, we need to find the minimum distance of $AP$.

$⇒ AP = OP - OA$

$⇒ AP = \sqrt{{\left(- \dfrac{3}{2} - 0\right)}^{2} + {\left(0 + 2\right)}^{2}} - 1$

$⇒ AP = \sqrt{\dfrac{9}{4} + 4} - 1$

$⇒ AP = \dfrac{5}{2} - 1$

$⇒ AP = \dfrac{3}{2}$