# 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? ## 1 Answer

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? – theGreenCabbage Jun 8 '14 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? – theGreenCabbage Jun 8 '14 at 21:26
• $-j=-j\cdot\frac{j}{j}=\frac{-j\cdot j}{j}=\frac{1}{j}$ – Vladimir Cravero Jun 8 '14 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°. – apalopohapa Jun 8 '14 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. – theGreenCabbage Jun 8 '14 at 21:37