# Finding the steady state value of v using nodal analysis

I am currently working on a problem of determing the steady-state value of v using nodal analysis. I understand all the steps up until the step of converting my j6 / -1-j to the phasor value that I circled within the red rectangle below.

Could someone give me some insight on what was achieved here and why? 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.

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

• Thanks @apalopohapa! What I still do not entirely understand is why -j = 1 / j. Would you mind explaining this as well? Jun 8, 2014 at 21:16
• So from the origin it looks like an arrow pointing to the bottom-left in the complex plane. So the angle is -135° How did you determine this precise angle? Jun 8, 2014 at 21:26
• $-j=-j\cdot\frac{j}{j}=\frac{-j\cdot j}{j}=\frac{1}{j}$ Jun 8, 2014 at 21:30
• @theGreenCabbage An arrow to the right is 0°. To the bottom is -90°, to the left is -180° (or 180°). So the angle between bottom and left is -135°, which is -90°-45°. Jun 8, 2014 at 21:31
• Thanks a lot. It appears I can input the values into my TI-Nspire CX, and it would convert the magnitude to a number I desire. However, I am unable to get the <angle; the value I get is something like e^01.439*i. Jun 8, 2014 at 21:37