1. Let $z=x+iy$.

   Then $z-1=(x-1)+iy, \qquad z+1=(x+1)+iy.$

2. Use the property of arguments: $\arg\left(\frac{z-1}{z+1}\right)=\arg(z-1)-\arg(z+1)=\frac{\pi}{4}.$

   Geometrically, this means the angle between the vectors from $-1$ and $1$ to the point $z$ is $\frac{\pi}{4}$.

3. To get the Cartesian equation, rationalize: $\frac{z-1}{z+1}=\frac{(x-1)+iy}{(x+1)+iy} \cdot \frac{(x+1)-iy}{(x+1)-iy}.$

   So, $\frac{z-1}{z+1}=\frac{((x-1)+iy)((x+1)-iy)}{(x+1)^2+y^2}.$

4. Expand the numerator: $((x-1)+iy)((x+1)-iy)=(x^2-1+y^2)+2iy.$

   Hence, $\frac{z-1}{z+1}=\frac{x^2+y^2-1}{(x+1)^2+y^2}+i\,\frac{2y}{(x+1)^2+y^2}.$

5. Since the argument is $\frac{\pi}{4}$, the complex number must satisfy $\tan\frac{\pi}{4}=1=\frac{\text{Im}}{\text{Re}}.$

   Therefore, $\frac{2y}{x^2+y^2-1}=1$ which gives $x^2+y^2-1=2y.$

6. Rearranging, $x^2+y^2-2y-1=0.$

   Complete the square in $y$: $x^2+(y-1)^2=2.$

7. This is a circle with $\text{centre }(0,1), \qquad \text{radius }\sqrt{2}.$

8. Check options:

   • A: centre $(0,-1)$, radius $\sqrt2$ — incorrect
   • B: centre $(0,1)$, radius $\sqrt2$ — correct
   • C: centre $(0,0)$, radius $\sqrt2$ — incorrect
   • D: centre $(0,1)$, radius $2$ — incorrect

Therefore, the correct option is B.