# rectangular to polar conversions

#### lemon

Joined Jan 28, 2010
125
Hi:
I am reading/working through the following page on ac Series-parallel R, L, and C circuits. On this page they make a conversion from rectangular to polar forms but I am having difficulty understanding how they came to the eventual polar angles. I am ok with doing basic conversions. Could someone explain that for me please, or direct me to the page that does. I read it somewhere on these pages earlier but can't recall it and can't seem to find it now. Thank you.

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#### lemon

Joined Jan 28, 2010
125
yeah I have studied that page before but am still confused. The thing that is confusing me is that in the conversion from rectangular form to polar form, we use arctan(b/a), which for the first example in the attached image, Zc1=0-j564.38, would be arctan(564.38/0), which is of course undefined. Yet their solution is 564.38<-90.

#### Papabravo

Joined Feb 24, 2006
20,600
The tan of 90 degrees is not undefined it is infinite; so the arctan of infinity is 90 degrees and the arctan of minus infinity is -90 degrees.

Undefined is something 0/0

Plot the points on the complex plane to see what is going on.

#### lemon

Joined Jan 28, 2010
125
The tan of 90 degrees is not undefined it is infinite; so the arctan of infinity is 90 degrees and the arctan of minus infinity is -90 degrees.

Undefined is something 0/0

Understood.
But - 564.38 on the complex plane puts it in the lower left quadrant. While that is minus, it isn't -90. Isn't is -155.62?

#### t06afre

Joined May 11, 2009
5,934
Take a look at this picture. The Y axis is equal to the j axis

#### Papabravo

Joined Feb 24, 2006
20,600
The tan of 90 degrees is not undefined it is infinite; so the arctan of infinity is 90 degrees and the arctan of minus infinity is -90 degrees.

Undefined is something 0/0

Understood.
But - 564.38 on the complex plane puts it in the lower left quadrant. While that is minus, it isn't -90. Isn't is -155.62?
No that is not correct. The complex number has two(2) components, just like a vector. 0 - j564.38 has a Real part of 0 so you stay on the vertical axis and you go 564.38 units in the negative j (aka y) direction and you arrive at the point (0, -564.38) in Cartesian coordinates which is equivalent to the the complex number 0 -j564.38 on the complex plane. Got it now?

#### lemon

Joined Jan 28, 2010
125
yep! I think so. I don't go round I go up and down and left and right

But that is only when there is a zero involved though, right?

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#### Papabravo

Joined Feb 24, 2006
20,600
Yes, Right