What's of points with condition $|z + i - 1| < 2$ in the plane?

Question

What's of points with condition $$ |z + i - 1| < 2 $$ in the plane?

if i write $|x + i y +i -1| < 2$ this conclude

$$ (x -1)^2+(y+i)^2 < 2 $$ which is a circle, is it true?

Answer 1

If you write $\vert x + iy + i - 1\vert < 2$, let's first compute the modulus, you get $$\vert x + iy + i - 1\vert = \vert (x - 1)+ i(y+1) \vert$$ That is $$\vert x + iy + i - 1\vert = \sqrt{(x - 1)^2 + (y+1)^2}$$ So you get $$(x - 1)^2 + (y+1)^2 < 4$$. Isn't this a disk of center $(1,-1)$ and radius $2$ ?

Answer 2

$\mid z-z_0\mid\lt r$ is the open disc of radius $r$ centered at $z_0$.

So we have the open disc of radius $2$ centered at $1-i$.

Remember if $z=x+iy$, then $\mid z\mid=\sqrt{x^2+y^2}$.

There are a couple errors in your calculation: $(x-1)^2+(y+\color{red}i)^2\lt\color{red}2$.

You want: $(x-1)^2+(y+\color{blue}1)^2\lt\color{blue}{2^2}$

This is because $z-1+i=(x-1)+(y+1)i$...

Also, if you replace $\lt$ by $=$ you get a circle.