Converting to polar - with zero real values

Thread Starter

KevinEamon

Joined Apr 9, 2017
284
So I've got this problem.
Usually when doing division of complex numbers, I like to convert into polar form, then divide them.
So I'm trying to do a problem on the internet

it's
-j50 / 5 -j10

Now I've seen folks do that multiplying by the complex conjugate bit and that's fine, however I would very much like to convert this whole thing into polar and divide from there.

Now 5-j10 no problems... I hope
Sqrt(5^2 + 10^2)
and
tan^-1(-10/5)

= 11.18 <-63.43

Great

But what about the -j50 up on top, all on it's lil lonesome?

Now I've been taught that, this is equaled to 0+(-j50)

But I mean what does that even mean, when I say for eg.
sqrt(0^2 + (-50)^2)
Just returns a dirty big math error. Same with the tan-1 (-50/0)...?
 

WBahn

Joined Mar 31, 2012
29,976
So I've got this problem.
Usually when doing division of complex numbers, I like to convert into polar form, then divide them.
So I'm trying to do a problem on the internet

it's
-j50 / 5 -j10
The result is simply -j20

What you wrote is

(-j50 / 5) - j10

Stop being sloppy. Write what you intended to write so that people (including yourself) don't have to guess.

But what about the -j50 up on top, all on it's lil lonesome?

Now I've been taught that, this is equaled to 0+(-j50)

But I mean what does that even mean, when I say for eg.
sqrt(0^2 + (-50)^2)
Just returns a dirty big math error.
What dirty big math error? 0^2 is 0. (-50)^2 is 2500. 0+2500 is 2500. sqrt(2500) is 50. What error?

Same with the tan-1 (-50/0)...?
What is the tangent of -90°?

You either need to recognize when you have a tangent that is infinite (either positive or negative) and then not call the arctangent function but rather set the angle to +/-90° as appropriate, or use a four-quadrant arctangent function, usually called atan2() in most programming language math libraries. You will want to use this anyway since the normal arctangent function can't distinguish between a vector in the first quadrant and one in the fourth, or one in the second and one in the third.
 

Thread Starter

KevinEamon

Joined Apr 9, 2017
284
I'm just thinking, I should say that:- this is an capacitor of -j10 in parallel with a resistor of 5.

So it's the original equation was :

5 * j10
--------
5 + j10
 

Thread Starter

KevinEamon

Joined Apr 9, 2017
284
Haha Wbahn I do love ya man. I do however always feel that I've been somehow violated when you give me an answer. It's like I'm standing there with a smile on my face, holding my new answer, yet a horrible burny feeling of being butt whipped.
It's a pleasure pain thing, some folks are into that shizzle I guess.

Lol jking cheers. ;)

...I'll bite here just to see the answer.
When I plug (50 / 5) - j10 into the calculator I get = -20??? Sooooo.... ammm?

Now I'm guessing putting the -j inside the brackets does something funny with the answer and that's why i'm being disciplined?

What is the tan (-90)?
I have no idea, neither does my sucky calculator. I'm guessing something like zero...? Maybe...

We've just breached this subject Wbahn we have 1 question in the exam. I'm guessing it's just going to be an inductor and capacitor in series with a resistor. But I'm making sure I'm prepared in case it gets more complex. But as I say, a lot of this is completely new to me so forgive me. Also I am not allowed to use an advanced calculator - I just have a casio fx83gt plus
 

crutschow

Joined Mar 14, 2008
34,280
....Also I am not allowed to use an advanced calculator - I just have a casio fx83gt plus
Don't complain.
All I had to use in my exams was a slide rule (way back in the dark ages before hand-held calculators of course).
It was great fun (not) doing polar-rectangular conversions on that.
I would have given an eye tooth for any calculator that could have done just trig functions and polar/rectangular calculations. :rolleyes:
 

Thread Starter

KevinEamon

Joined Apr 9, 2017
284
Ahhh the heady days, when men were men and sheep were afraid ...
Lol crutzchow... sorry I was responding to this part

or use a four-quadrant arctangent function, usually called atan2() in most programming language math libraries.
 

WBahn

Joined Mar 31, 2012
29,976
Haha Wbahn I do love ya man. I do however always feel that I've been somehow violated when you give me an answer. It's like I'm standing there with a smile on my face, holding my new answer, yet a horrible burny feeling of being butt whipped.
It's a pleasure pain thing, some folks are into that shizzle I guess.

Lol jking cheers. ;)

...I'll bite here just to see the answer.
When I plug (50 / 5) - j10 into the calculator I get = -20??? Sooooo.... ammm?
(50 / 5) - j10 = 10 - j10

Going back to your original post, you are neglecting order of operations -- multiplication and division are performed before addition and subtraction. So when you write

-j50 / 5 - j10

the division is performed before the subtraction.

(-j50 / 5) - j10

-j50/5 = -j10

so you now have

-j10 - j10 = -j20

Now I'm guessing putting the -j inside the brackets does something funny with the answer and that's why i'm being disciplined?
What I'm assuming you meant -- since I'm having to guess -- is

-j50 / (5 - j10)

since that's the only thing that makes sense given what you wrote immediately thereafter.

What is the tan (-90)?
I have no idea, neither does my sucky calculator. I'm guessing something like zero...? Maybe...
You should not need a calculator to do this and the fact that you do is a testament to the disservice that the educational system has done you. I wish I could say that it shocked me, but I've seen trig textbooks that have failed to provided the most fundamental information about the very definition of the basic trig functions and, instead, have just shown pictures of which buttons to press on a calculator. It's a travesty. But it is now up to YOU to correct that failing of your schooling by taking it upon yourself to learn the things they failed to teach you. Fortunately, there is plenty of information out there to assist you. You might start with

https://en.wikipedia.org/wiki/Trigonometric_functions

We've just breached this subject Wbahn we have 1 question in the exam. I'm guessing it's just going to be an inductor and capacitor in series with a resistor. But I'm making sure I'm prepared in case it gets more complex. But as I say, a lot of this is completely new to me so forgive me. Also I am not allowed to use an advanced calculator - I just have a casio fx83gt plus
If you NEED a calculator at all, then you have not learned anything except to let a piece of hardware designed by someone else do your thinking for you. It's one thing to use a calculator to perform the grunge work, but that's only the case if you KNOW how to do it without the calculator. THAT's the level of competency you need to be aiming at.
 

MrAl

Joined Jun 17, 2014
11,389
So I've got this problem.
Usually when doing division of complex numbers, I like to convert into polar form, then divide them.
So I'm trying to do a problem on the internet

it's
-j50 / 5 -j10

Now I've seen folks do that multiplying by the complex conjugate bit and that's fine, however I would very much like to convert this whole thing into polar and divide from there.

Now 5-j10 no problems... I hope
Sqrt(5^2 + 10^2)
and
tan^-1(-10/5)

= 11.18 <-63.43

Great

But what about the -j50 up on top, all on it's lil lonesome?

Now I've been taught that, this is equaled to 0+(-j50)

But I mean what does that even mean, when I say for eg.
sqrt(0^2 + (-50)^2)
Just returns a dirty big math error. Same with the tan-1 (-50/0)...?
Hi,

When you run into zero in some of these problems you cant evaluate it directly you have to take the limit as the quantity approaches zero and then you might get a reasonable answer.

Since you cant compute atan(-50/0) directly for example you have to think about what happens to the function atan(y/x) when x gets near zero. When x gets near zero that means we have a value on a graph that has a y value that is on the y axis itself. The angle therefore must be plus or minus 90 degrees.

However, i believe using atan(y/x) is not a good idea unless you have to, while using atan2(y,x) is better.
atan2 is the two argument atan function while atan is just the single argument function which can not resolve which quadrant the answer lies in. You might get lucky if the numbers for x and y happen to work out to the same result for both atan and atan2, but of course we dont want to depend on luck.

To find out what quadrant the result lies in we just need to refer to the x and y values. The following chart will help:
x,y
pos,pos: first (0 to 90 degrees)
neg,pos: second (90 to 180 degrees)
neg,neg: third (180 to 270 degrees)
pos,neg: fourth (270 to 360 degrees)

The idea here is to calculate atan(|y|,|a|) and then use the above to figure out if we need to add anything.
For example, find the angle from atan(-5/-5).
Using atan(abs(-5)/abs(-5)) we get 45 degrees, but the point -5,-5 lies in the third quadrant so the angle must really be between 180 and 270 degrees. If we add 180 degrees to 45 degrees we get 225 degrees which is the right answer.

To contrast, if we use atan2(-5,-5) we get 225 degrees (or the equivalent -135 degrees).

Note ONE of the problems with atan() is that the parameter itself may resolve to a positive number (BEFORE we even start to calculate the actual inverse tangent) when they were both negative to start with, which results in the wrong answer until it is corrected. It is easy to see that atan(x) is not equal to atan(-x).

So you can use atan() instead of atan2() but then you have to check the result to make sure it is correct and if not correct it.

Note the same thing will happen if we try to do atan(5/-5) instead of atan2(5,-5). In fact:
atan2(5,-5)=atan(|5|/|-5|)+90 (in degrees), the correction here being +90 degrees.
 
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Thread Starter

KevinEamon

Joined Apr 9, 2017
284
Thanks guys, well the last shots have been fired for the year. They think it's all over... It is now!
The only question that has to be answered now is - Who's got my bottle of the finest single malt that money can buy....hmmm who's got it? Oh here it is :)
Haha thanks for the year guys its been most informative. Special thanks to Wbahn and MrAi.
This won't go to waste Mr Ai. I'll be reading it near to September. At the mo its the last thing I wanna do. I'm sure you understand that.
Think I've done well enough in case anyone was wondering, maybe a solid 70 percent.
Thank you again guys you are all awesome and I wouldn't have been able to do as well without your great advise!
 
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