Question

How do you graph $|z-i|=2$ in the complex plane?

Answer 1

Hint: We first assume the value of the complex number $z=x+iy$. We place the value in the equation $z-i$. Then we find the modulus value of $z-i$. We equate with 2 and then take the square of the equation. The equation becomes the form of a circle. We plot the equation in the graph.

Complete step-by-step solution:
We have to find the graph of $|z-i|=2$ in the complex plane.
Here $z$ works as a complex number. So, we assume the value as $z=x+iy$. Here $x$ and $y$ are real constants and $i$ works as the imaginary number.
The function of $\left|\right|$ is the representation of modulus value.
For general complex number $z=x+iy$, the modulus value will be $\left|z\right|=\sqrt{x^2+y^2}$.
Now we find the value of $z-i=(x+iy)-i=x+i(y-1)$.
Now we find the modulus value of $z-i$.
$|x+i(y-1)|=\sqrt{x^2+{(y-1)}^2}$.
We have been given the equation $|z-i|=2$.
We place the values and get $\sqrt{x^2+{(y-1)}^2}=2$.
We take the square on the both sides of the equation $\sqrt{x^2+{(y-1)}^2}=2$ and value of the equation becomes $x^2+{(y-1)}^2=2^2=4$.
The equation is an equation of a circle.
We equalise $x^2+{(y-1)}^2=4$ with the general equation of circle ${(x-a)}^2+{(y-b)}^2=r^2$.
For the general equation we have the centre as $(a,b)$ and the radius as $r$.
Now we find the centre and the radius for $x^2+{(y-1)}^2=2^2$.
We have the centre as $(0,1)$ and the radius as 2.
Now we plot the equation in the graph.

Note: We need to remember that in the complex plan the unit circle representation is always applicable for modulus values. The modulus value eliminates the imaginary part of the equation.