Well, it is just converting the numerator and denominator from complex numbers to magnitude-phase (phasor) notation, which is really short-hand of imaginary exponential.
Numerator
j6
is a vector with no real component (only has imaginary component). So in phasor terms, the magnitude is 6
, and the angle is 90°
(if it had been negative (-j6
), the angle would have been -90°
). The angle is measured positive in the anti-clockwise direction, with respect to the positive real axis.
$$ j6 = 6\angle{90°} $$
Denominator
-1-j
has real and imaginary components equal to -1
. So from the origin it looks like an arrow pointing to the bottom-left in the complex plane. So the angle is -135°
, and the magnitude is \$(1^2+1^2)^{1/2} = \sqrt{2} \$
$$ -1-j = \sqrt{2} \angle{-135°} $$
To answer the question in the comments of why \$-j=\frac{1}{j} \$:
When you divide phasors, the resulting magnitude is the quotient of the magnitudes, and the resulting angle is the difference between the angles.
$$ -j = 1\angle{-90} = \frac{1\angle{0}}{1\angle{90}} = \frac{1}{j} $$