Convert the given complex number in polar form: $1-i$.

Let the complex number be $z = 1-i$.
We represent $z$ in polar form as $z = r(\cos \theta + i \sin \theta)$,where $r \cos \theta = 1$ and $r \sin \theta = -1$.
Squaring and adding these equations,we get:
$r^2(\cos^2 \theta + \sin^2 \theta) = 1^2 + (-1)^2$
$r^2 = 2$
$r = \sqrt{2}$ (since $r > 0$).
Now,$\cos \theta = \frac{1}{\sqrt{2}}$ and $\sin \theta = -\frac{1}{\sqrt{2}}$.
Since $\cos \theta > 0$ and $\sin \theta < 0$,the angle $\theta$ lies in the $IV$ quadrant.
Thus,$\theta = -\frac{\pi}{4}$.
Therefore,the polar form is $\sqrt{2} \left[ \cos \left( -\frac{\pi}{4} \right) + i \sin \left( -\frac{\pi}{4} \right) \right]$.