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When dividing in Polar form 31.2/5.74 becomes 5.44. The angle for 31. 2 is 0. So when it comes to solving imaginary number 12. 8 why is the - 12.8 degrees?

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    \$\begingroup\$ Because 0 - 12.8 = -12.8. \$\endgroup\$ – The Photon Apr 8 '19 at 17:31
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It sounds like you don't know that polar form is actually shorthand for exponential form of complex numbers.

Complex numbers can be expressed in three equivalent forms: Polar, exponential, trigonometric cartesian form, and simple cartesian form (from left to right).

$$ M \angle \theta = Me^{j\theta} = M(cos \theta +jsin \theta) = x + jy $$

Polar form is just a shorthand form of exponential so you can write things more cleanly (it doesn't have any real mathematical meaning on its own).

Trigonometric Cartesian form just the expanded form of simple Cartesian form.

If you want to know why exponential form is equal to cartesian form, you should look it up. In words, when you expand the exponential form using the Taylor expansion. Each term has a j to which either an odd or even exponent is applied. The even terms cause the j to disappear while the odd terms cause j to remain. What's left is a group of real terms and a group of imaginary terms. The real terms happen to be the Taylor expansion of cos and the imaginary terms happen to be the Taylor expansion of sin. It's magical. I recommend you just take a glance at the math. That's enough for you to know in the back of your mind where it comes from.

Anyways, so when you divide this term, you divide the magnitude M as normal. You also divide the exponential as normal. However, when you divide exponentials you subtract the power:

$$ \frac{A \angle \theta} {B \angle \delta} = \frac{A e^{j \theta}} {B e^{j \delta}} = \frac{A}{B} e^{j( \theta - \delta)} = \frac{A}{B} \angle{(\theta - \delta)} $$

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The Photon probably gave you a way too terse answer. But it's correct, just the same. I'll provide a way too long answer. (tl;dr)

Simplistic Geometric Approach

It's best to think of a magnitude (the length of the hypotenuse or radius vector) and its angle as being based upon the circle and the trigonometry definitions that mathematicians have established as "standard." Zero degrees has the vector pointing to the right from (0,0). 90 degrees has the vector pointing straight up from (0,0). Etc. So as the angle grows larger, the vector sweeps around counter-clockwise around the circle.

Now, if you place this value in the divisor of a rational, then what you've done is reverse the direction of rotation. Now as you increase the angle, the resulting vector magnitude sweeps clockwise, instead. So you have to negate the angle given in the divisor because the resulting expression has the angle given in the numerator (with a divisor of 1) and that angle, by convention, should have the effect of rotating counter-clockwise.

Prelude to a Mathy Approach

I guess you might as well get used to it. So let's discuss the miracle of Euler's formula:

$$e^{i\theta}=\operatorname{cos}\left(\theta\right)+i\cdot\operatorname{sin}\left(\theta\right)$$

This formula maps wonderfully to the standard trigonometry definitions used for the unit circle on the Cartesian graph. The only oddity being the use of \$i\$ (which maps to the plotting on the old \$y\$-axis.) It's really a lot more than that, as the use of \$i\$ actually deals with rotation and the (x,y) approach does not, inherently. A lot of study went into rotation in 3D, as well, with the result eventually being quaternions. There was a push by a few mathematicians, in fact, to make quaternions the "standard." But there was overwhelming opposition and it never took on.

I'd like to recommend that you view 3Blue1Brown's wonderful Euler's formula with introductory group theory video. You almost cannot avoid learning a great deal from it and he is almost a magician at presenting it so that you can almost feel it in your guts so that it never leaves you. Please take a moment and watch it. It will really help a lot.

Mathy Approach

Complex numbers have both a Cartesian notation as well as two equivalent polar notations. The Cartesian notation, \$a+b\:i\$, is very hard to use in electronics but it is very easy to plot. A common polar notation used is \$r\left[\operatorname{cos}\left(\theta\right) + i\:\operatorname{sin}\left(\theta\right)\right]\$.

In electronics, \$i\$ is replaced by \$j\$ (convention) and \$\theta\$ is replaced by \$\omega\$ (convention again.) I don't know the history of it, though. It just is. So any arbitrary vector on the plane can be expressed as: \$e^{\sigma+i\:\omega}=e^\sigma\:e^{i\:\omega}=e^\sigma\left[\operatorname{cos}\left(\omega\right) + i\:\operatorname{sin}\left(\omega\right)\right]\$. Here, the length or magnitude of the vector is \$r=e^\sigma\$ and its angle is \$\theta=\omega\$. We can now create a special variable, \$s\$, such that \$s=\sigma+j\:\omega\$, then the entire polar notation becomes \$e^s\$, which is actually very compactly written.

Note: Multiplication in the complex domain combines rotation and also stretching (called scaling by mathematicians) in a single action. In Cartesian notation, it is rather difficult to work out the original angles and the resulting angle of rotation. But in polar notation, the effect of multiplication on rotation is quite easy to understand. (You only need to sit down and try out rotation by angle \$\theta\$, but only using Cartesian notation, to apprehend what I mean.)

Now, just suppose that we've decided to use the power of Euler's in electronics. It's not a bad idea because electronics has magnitudes (voltages, currents, etc) and has frequency-dependent rotation taking place. Euler's is a silver platter, of sorts, for this kind of thing. So it was a natural. A DC voltage might be written as \$\sigma=\operatorname{ln}\left(V\right)\$ and \$\omega=0\$ (other constant values for \$\omega\$ could also be used for DC -- the important bit is that there's no variable \$f\$ for frequency present in it.) An AC voltage might be \$\sigma=\operatorname{ln}\left(V_0\right)\$ and \$\omega=2\pi\:f\$. (Though once time is introduced, the use of \$\sigma=\operatorname{ln}\left(V_0\right)\$ would lead to a time-dependent spiral, so it's common when introducing time to set \$\sigma=0\$ in electronics unless the spiraling behavior is also important to some question in mind.)

Suppose we want to divide a voltage by a current and say that the voltage is DC, with \$\omega=0\$. Let's say the current is also DC, but has some given angle relative to the voltage, \$\omega=12.8^\circ\$. (This doesn't actually happen in electronics because the two types of energy storage devices that might cause an angle difference are all frequency dependent devices. But we can choose to ignore that fact, for now.) Then you might say something like this: \$\frac{e^{\operatorname{ln}\left(V\right)}}{e^{\operatorname{ln}\left(I\right)+j\:12.8^\circ}}=e^{\operatorname{ln}\left(V\right)-\operatorname{ln}\left(I\right)-j\:12.8^\circ}\$.

Note the change in sign for the angle??

When you divide with exponents, you can simply negate the exponent and bring it up to the numerator. This is just another way of saying the same thing I started out saying, which is that the divisor rotates things in the opposite direction as does the numerator.

That pretty much tells you way in a mathy way, I suppose. There are dozens of other ways to explain it, I'm sure.

Final Note: In electronics, \$\omega\$ represents the rate of rotation and \$\sigma\$ represents the vector magnitude (keeping in mind \$r=e^\sigma\$, as \$\sigma=0\$ means a vector magnitude of \$r=1\$.) You can introduce time quite simply by using \$e^{s\: t}\$, where the rotation now is a smooth function of time and the vector magnitude (as a function of time) stays constant only when \$\sigma=0\$. (Otherwise, it spirals outward [not usually a good thing in electronics] or inward [damped.])

When analyzing the frequency response, the vector magnitude stretching aspect isn't used. So \$\sigma=0\$ and only the \$j\:\omega\$ (rotation rate) portion is kept. For that kind of analysis, all that's needed is \$s=j\:\omega\$ (as \$\sigma=0\$.)

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  • \$\begingroup\$ Thank you for the comprehensive, well explained answer!! \$\endgroup\$ – ThePandeo Apr 10 '19 at 6:13

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