# complex jw vector

Discussion in 'Math' started by TheSpArK505, Dec 12, 2014.

1. ### TheSpArK505 Thread Starter Member

Sep 25, 2013
92
0
Hello guys.
I have a complex vector that starts at a+(jw)b and terminates at c+(jw)d

do i calculate the length and angle as following or do Complex vectors have their own special formulas

length=sqrt[(c-a)^2+(d-b)^2]

angle= arctan[(d-b)/(c-a)]

2. ### Papabravo Expert

Feb 24, 2006
10,338
1,850
Not quite. You need to account for ω in your calculations.

$L = \sqrt{(c-a)^2+(d\omega-b\omega)^2}$

$\angle=\arctan(\frac{d\omega-b\omega}{c-a})$

3. ### TheSpArK505 Thread Starter Member

Sep 25, 2013
92
0
but w is not given

4. ### Papabravo Expert

Feb 24, 2006
10,338
1,850
Then you cannot calculate a specific length. You can only say that length is a function of ω, or give the length for several values of ω. If ω=1 then your original formulation is correct. In which case you might want to rethink your original question. Does that help you?

5. ### TheSpArK505 Thread Starter Member

Sep 25, 2013
92
0
OK here the problem attached

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Sep 25, 2013
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7. ### Papabravo Expert

Feb 24, 2006
10,338
1,850
Right, I have the same book. I see an ω as a label for the imaginary axis,but that ω is not used in the evaluation of points in the complex plane. A general point is expressed as

a + jb, or c+jd​

8. ### TheSpArK505 Thread Starter Member

Sep 25, 2013
92
0
OK but for the angle when s=-3+j4
I calculated the angle as theta =arctan[(4-0)/(-3-(-1))] i got -63.43 degree . how come???
@Papabravo

9. ### Papabravo Expert

Feb 24, 2006
10,338
1,850
The domain of the arctangent function is [-∞,+∞]
The range of the arctangent function is [-90°,+90°]
To get the postive angle you seek you need to add 180° to your result

-63.43° + 180° ≈ 116.6°​

Another way to look at it is that -63.43° is the angle from -3+j4 to the zero at -1+j0, while 116.6° is the angle from the zero at -1+j0 to the point a -3+j4
Does that clear things up?