We are given the function $f(z) = j \frac{1-z}{1+z}$, where $j = \sqrt{-1}$. We need to find the mapping of the real axis by its inverse function, $f^{-1}(z)$.

Finding the Inverse Function

Let $w = f(z)$. To find the inverse function, we solve for $z$ in terms of $w$.
$w = j \frac{1-z}{1+z}$
$w(1+z) = j(1-z)$
$w + wz = j - jz$
$wz + jz = j - w$
$z(w+j) = j - w$
$z = \frac{j-w}{w+j}$

Replacing $w$ with $z$, the inverse function is $f^{-1}(z) = \frac{j-z}{z+j}$.

Mapping the Real Axis

The real axis is defined by $z=x$, where $x$ is a real number ($x \in \mathbb{R}$). We substitute $z=x$ into the inverse function $f^{-1}(z)$:
$f^{-1}(x) = \frac{j-x}{x+j}$

To determine the path traced by $f^{-1}(x)$, we express it in the form $u + iv$, where $u$ is the real part and $v$ is the imaginary part. We multiply the numerator and denominator by the conjugate of the denominator ($j-x$):
$f^{-1}(x) = \frac{j-x}{j+x} \times \frac{j-x}{j-x}$
$f^{-1}(x) = \frac{(j-x)^2}{(j)^2 - (x)^2}$
$f^{-1}(x) = \frac{j^2 - 2jx + x^2}{-1 - x^2}$
Since $j^2 = -1$:
$f^{-1}(x) = \frac{-1 - 2jx + x^2}{-1 - x^2}$
$f^{-1}(x) = \frac{-(1 - x^2) - 2jx}{-(1 + x^2)}$
$f^{-1}(x) = \frac{1 - x^2}{1 + x^2} + j \frac{2x}{1 + x^2}$

Thus, we have $u = \frac{1 - x^2}{1 + x^2}$ and $v = \frac{2x}{1 + x^2}$.

Identifying the Mapped Shape

To find the shape formed by these coordinates $(u, v)$, we calculate $u^2 + v^2$:
$u^2 + v^2 = \left(\frac{1 - x^2}{1 + x^2}\right)^2 + \left(\frac{2x}{1 + x^2}\right)^2$
$u^2 + v^2 = \frac{(1 - x^2)^2 + (2x)^2}{(1 + x^2)^2}$
$u^2 + v^2 = \frac{1 - 2x^2 + x^4 + 4x^2}{(1 + x^2)^2}$
$u^2 + v^2 = \frac{1 + 2x^2 + x^4}{(1 + x^2)^2}$
$u^2 + v^2 = \frac{(1 + x^2)^2}{(1 + x^2)^2}$
$u^2 + v^2 = 1$

The equation $u^2 + v^2 = 1$ represents a circle centered at the origin $(0, 0)$ with a radius of 1. This is the unit circle.

Conclusion

The inverse function $f^{-1}(z)$ maps the real axis to the unit circle with centre at the origin.