Question

10
16. If|z - 2il<=2, then the maximum value of|3+i (z-1)|is:
(a) root2
(b) 2root2
(c) 2+root2
(d)3+2root2​

Answer 1

Let $z=x+yi,$ then,

$\longrightarrow|z-2i|\leq2$

$\longrightarrow|x+yi-2i|\leq2$

$\longrightarrow|x+(y-2)i|\leq2$

$\longrightarrow x^2+(y-2)^2\leq4$

The solution to this inequality is the inside portion of the circle of radius 2 units with center (0, 2), including the circle.

Now,

$\longrightarrow p=|3+i(z-1)|$

$\longrightarrow p=|3+i(x+yi-1)|$

$\longrightarrow p=|3+xi-y-i|$

$\longrightarrow p=|(3-y)+(x-1)i|$

$\longrightarrow p^2=(x-1)^2+(y-3)^2$

This equation represents the circle of radius $p$ units with center (1, 3).

As value of $p$ is determined by $z,$ these circle should be tangent, i.e., they should touch each other at a single point.

That's why, maximum value of p is equal to the length of the line segment drawn from center of second circle (1, 3) to the first circle by passing through its center. [See fig.]

So maximum value of $p,$

$\longrightarrow p_{\max}=2+\sqrt{(1-0)^2+(3-2)^2}$

$\longrightarrow\underline{\underline{p_{\max}=2+\sqrt{2}}}$

Hence (c) is the answer.