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

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Answer 1 · 2016-02-03T23:30:38

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

To say $abs(z-i) = 2$ 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:

$abs(z) = sqrt(z bar(z))$

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

Then

$2 = abs(z-i) = abs(x+yi-i) = abs(x+(y-1)i)$

$= sqrt((x+(y-1)i)(x-(y-1)i)) = sqrt(x^2+(y-1)^2)$

Square both ends and transpose to get:

$x^2+(y-1)^2 = 2^2$

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

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