Let z=a+bi, then

|z-1|\cdot |z+1|

=\sqrt{(1-a)^2+(-b)^2}\cdot \sqrt{(1+a)^2+b^2}

=\sqrt{(a^2+b^2)+(1-2a)}\cdot \sqrt{(a^2+b^2)+(1+2a)}

=\sqrt{(a^2+b^2)^2+2(a^2+b^2)+1-4a^2}

Since a^2+b^2 represents the modulus of the complex number, let r=|z|. The following equation is established.

\sqrt{r^2+2r+1-4a^2} =2

Square both sides and simply

r^2+2r-3=4a^2

Since a^2\geq 0,

r^2+2r-3\geq 0

(r+3)(r-1)\geq 0

Solving the inequality gives

r\geq 1 or r\leq -3 (cancel)

Therefore, the minimum value of |z| is equal to 1