Q:

Is $|z+i| = |z-1|$ a circle with radius $\sqrt{1^2+1^2=1}$ and origin $(1,-i)$?

Is $|z+i| = |z-1|$ a circle with radius $\sqrt{1^2+1^2}=1$ and origin $(1,-i)$?
Because I know $|z+i| = 3$ is is a circle with radius $3$ and origin $(0,i)$.

A:

It is not a circle. To see what it is let's consider two approaches. Geometrically $|z-a|$ is the distance in the complex plan between the point represented by $z$ and the point represented by $a$. So the locus we're looking for is those points that are equidistant from $1$ and $-i$ so it is the perpendicular bisector as in @Omnnomnomnom comment.
Let's check that analytically by considering $z=x+iy$. Write $|z+i|=|z-1|$ which translates to $x^2+(y+1)^2=(x-1)^2+y^2$. This simplifies to $y=-x$ which is indeed the perpendicular bisector of $1$ and $-i$.

A:

For finding out the locus of point $z$, let's assume, $z=x+iy$ then substitute it in the relation given as follows
 $$|z+i|=|z-1| $$$$ \implies |x+iy+i|=|x+iy-1| $$$$ \implies \sqrt{x^2+(y+1)^2}=\sqrt{(x-1)^2+y^2}$$ $$\implies x^2+(y+1)^2=(x-1)^2+y^2$$
$$\implies x^2+y^2+2y+1=x^2-2x+1+y^2$$ $$\implies \color{blue}{y=-x} \quad \text{or} \quad \color{blue}{x+y=0}$$
It is obvious that the locus of the point $z$: $\color{blue}{|z+i|=|z-1|}$ is a straight line: $\color{blue}{x+y=0}$ passing through the origin having negative slope i.e. making an angle $135^o$ with positive real axis in the $\color{blue}{\text{complex plane}}$. 
Hence, $\color{blue}{|z+i|=|z-1|}$ represents a $\color{blue}{\text{straight line}}$ $\color{red}{\text{not a circle}}$.     

