Complex Numbers with some absolute values.

Put $x=a+ib$ , with $a,b$ real. Then substituting into the given equation, we have $\left(a^2-b^2+4 \right)^2 + 4a^2 b^2 = \left(a^2-b^2-2b \right)^2 + \left(2ab+2a \right)^2$ . Expanding, cancelling, and factorising, this becomes $(b-1) \left(a^2+(b+2)^2 \right)=0$ , leading to two cases to analyse.

First, $b=1$ : we have $x=a+i$ , and $|x+i|$ is minimised when $a=0$ , giving $|x+i|=\sqrt2$ .

Second, $a^2+(b+2)^2=0$ . Since $a,b$ are real, this only happens when $a=0$ and $b=-2$ , ie when $x=-2i$ . We now get $|x+i|=\boxed1$ , which is the least possible value it can take.