Finding the steady state value of v using nodal analysis
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.

$$ -j = 1\angle{-90} = \frac{1\angle{0}}{1\angle{90}} = \frac{1}{j} $$