Let A = zC : 1|z-(1+i)|2 and B = z A : |z-(1-i)|=1.

Question

Let $A = \{z\in\mathbb{C} : 1\leq|z-(1+i)|\leq2\}$ and $B = \{z\in A : |z-(1-i)|=1\}$. Then $B$:

Accepted Answer

Let, $z = x + iy$

Given, $$1 \le \left| {z - (1 + i)} \right| \le 2$$

$$ \Rightarrow 1 \le \left| {x + iy - 1 - i} \right| \le 2$$

$$ \Rightarrow 1 \le \left| {(x - 1) + i(y - 1)} \right| \le 2$$

$$ \Rightarrow 1 \le \sqrt {{{(x - 1)}^2} + {{(y - 1)}^2}}  \le 2$$

It represent two concentric circle both have center at (1, 1) and radius 1 and 2.

Also given,

$$\left| {z - (1 - i)} \right| = 1$$

$$ \Rightarrow \left| {x + iy - 1 + i} \right| = 1$$

$$ \Rightarrow \left| {(x - 1) + i(y + 1)} \right| = 1$$

$$ \Rightarrow \sqrt {{{(x - 1)}^2} + {{(y + 1)}^2}}  = 1$$

This represent a circle with center at (1, -1) and radius = 1.

In the common region infinite values of B possible.

Solutions · 1