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)]
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?
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 and your original answer is correct.
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
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?