View Full Version : Free algebra 2 answers
vanityb
Dec 2, 2009, 08:28 PM
Okay I'm Performing Operations with Complex Numbers and its prety confusing to me I have this equation and its somthin to have to do with imaginary numbers my equation is (6-3i) + (5+4i)
ROLCAM
Dec 2, 2009, 08:36 PM
1) (6-3i) + (5+4i)
2) 6 - 3i + 5 + 4i
3) 4i -3i = -6 - 5
4) I = -11
Unknown008
Dec 3, 2009, 03:44 AM
Uh... do you mean: (6-3i) = (5+4i)?
Because an equation has to have an equal sign, which I don't see in your question. What you gave is called an expression. Anyway, if it's like you said, then that's what you have to do:
(6-3i) + (5+4i) = 6-3i + 5+4i = 11 + i
And that's all. 11 is the real part, and i is the imaginary part.
KISS
Dec 3, 2009, 09:08 AM
rolcam:
I = sqrt(-1); a mathematical concept. Electrical Engineers use j rather than I to represent the same concept.
5+3i is an example of the rectangular form of a complex number. 155 @ an angle of 90 deg is the polar form/
Usually the "@ an angle" is written as a symbol that looks like a < sign laying on it's bottom side.
Unknown008
Dec 3, 2009, 09:15 AM
Or we use the trig form like
r(cos\theta + isin\theta) or in Euler's form re^{i\theta}
if my memory is good. Arg, I have to revise!! :(