How does the loci of the equation $|z-(i+1)| = |1 + i|$ look like?
I can't seem to visualise any points on the complex plane satisfying the above except the 2 obvious ones (2,2) and (0,0)... Is that the loci - just 2 points?
P.S. I am new to complex numbers...
On
Simplify the equation using $z=x+iy$ to get $$(x-1)^2+(y-1)^2=2$$ Try to find this equation yourself using the definitions. So, the locus is a circle centered at $(1,1)$ and with radius $\sqrt{2}$.
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Well, its simple to visualize the loci. First, think of $z$ as a point in the Argand plane, such that $z=x+iy$. Then fit it in the equation you have. For the given equation, $$|z-(i+1)|=|1+i|$$ $$|(x-1)+i(y-1)|=|1+i|$$ Now take the magnitudes and equate. You would get something like $$(x-1)^2+(y-1)^2=1^2+1^2=2$$ This loci would be a circle. Any complex point on this circle (in argand plane) would thus satisfy, two of which you've given, $z=0$ and $z=2+2i$. Some other points would be $z=1+\sqrt{2}$ and $z=1-\sqrt{2}$.
You can compute the RHS value to be $|1+i|=\sqrt{2}$. Your equation then has the form $$|z-m|=r,$$ with $z, m \in \mathbb C$, $r \in \mathbb R^{\ge 0}$. This corresponds to the circle of points centered at $m$ with a radius $r$.