# How do you graph abs(z-i) = 2 in the complex plane?

Feb 3, 2016

This is a circle with radius $2$ and centre $i$

#### Explanation:

To say $\left\mid z - i \right\mid = 2$ is to say that the (Euclidean) distance between $z$ and $i$ is $2$.

graph{(x^2+(y-1)^2-4)(x^2+(y-1)^2-0.011) = 0 [-5.457, 5.643, -1.84, 3.71]}

Alternatively, use the definition:

$\left\mid z \right\mid = \sqrt{z \overline{z}}$

Consider $z = x + y i$ where $x$ and $y$ are Real.

Then

$2 = \left\mid z - i \right\mid = \left\mid x + y i - i \right\mid = \left\mid x + \left(y - 1\right) i \right\mid$

$= \sqrt{\left(x + \left(y - 1\right) i\right) \left(x - \left(y - 1\right) i\right)} = \sqrt{{x}^{2} + {\left(y - 1\right)}^{2}}$

Square both ends and transpose to get:

${x}^{2} + {\left(y - 1\right)}^{2} = {2}^{2}$

which is the equation of a circle radius $2$ with centre $\left(0 , 1\right)$, i.e. $i$.