fun with i's

[Cogman]

Ok, first off Id like to startout by saying THIS IS NOT HOMEWORK, I aready finished HS and colledge has not started yet. Anyways.

i = (-1)^(1/2) correct?

Well if you had an equasion (x+i)*(x-i)=0 that would equal x^2+1=0, Right? So heres whats confusing me. in the first statement x can eqaul i or -i for the statement to be true. But in the secont x = 1, -1. Why is that?

 

[madnod]

actually we are talking complex math, so u have real part and imaginary part or as other may call it a magnitude and a phase.
as a definition i*i = -1= 1 at an angle 180degrees. (the root sqare of -1 = 1 with at an angle of 90degrees)
so yes the first statement is correct since i= 1 at an angle of 90 degrees.
btw the value of i can be get as follows:
the complex numbers have the following format: x= Acos(B)+i*B*sin(B)
where B is an angle. and A and B are the magnitudes
if x=i then cos(B)=0 and sin(B) =1 and A=B=1.
so i=1(the magnitude) at an angle of 90 degrees!
regarding the other statement (x+i)*(x-i)=0.
here u gave 2 roots
x =-i; and x=i.
take x=i root.
then x=1 angle 90degrees. if u sqare it then x^2=-1
(so -1+1=0 true! root verified!)
take x=-i
then x= 1 angle (90+180)=1 angel 270.(when u multiply a number by -1 is like adding 180 degrees to its phase).
x^2=1 angle (270*2=540) u can remove 360 from 540 then x^2= 180 =-1!
then u also have -1 +1=0. the other root is verified!
so the 2 solutions are correct for x^2+1=0
in the other hand.
x^2+1=0.
x^2=-1
x^2=i^2
x=+i or -i.
we return to up proof.
actually if u try to solve it graphically it's better and easier.
enjoy!

 

[TuxDave]

Originally posted by: Cogman
Ok, first off Id like to startout by saying THIS IS NOT HOMEWORK, I aready finished HS and colledge has not started yet. Anyways.

i = (-1)^(1/2) correct?

Well if you had an equasion (x+i)*(x-i)=0 that would equal x^2+1=0, Right? So heres whats confusing me. in the first statement x can eqaul i or -i for the statement to be true. But in the secont x = 1, -1. Why is that?

Uh....

(1)^2+1 != 0
(-1)^2+1 != 0

So... uh... no

 

[Cogman]

TuxDave, Lol, now I feal really stupid 🙂. And thanx madnod for the interesting text on i. Our teacher really did not teach much about i when we went through it so I was curious why (-1)^(1/2) was so useful.

 

[TuxDave]

Originally posted by: Cogman
TuxDave, Lol, now I feal really stupid 🙂. And thanx madnod for the interesting text on i. Our teacher really did not teach much about i when we went through it so I was curious why (-1)^(1/2) was so useful.

No probs. I didn't see much use practical use of 'i' until I started studying electrical engineering.

 

[madnod]

Originally posted by: TuxDave

Originally posted by: Cogman
Ok, first off Id like to startout by saying THIS IS NOT HOMEWORK, I aready finished HS and colledge has not started yet. Anyways.

i = (-1)^(1/2) correct?

Well if you had an equasion (x+i)*(x-i)=0 that would equal x^2+1=0, Right? So heres whats confusing me. in the first statement x can eqaul i or -i for the statement to be true. But in the secont x = 1, -1. Why is that?

Uh....

(1)^2+1 != 0
(-1)^2+1 != 0

So... uh... no

cogman
on what basis u did -1^(1/2) there is not such thing in real math.to get 2 roots of x=1,and -1 then ur solving x^2-1=0 not x^2+1=0.
by definition x^2 is positive so x^2+1 can't be zero.(in normal math again!)
BTW i made a Typo in my first post the phase is C not B (am used to use beta and alpha there).
bye

 

[Psych]

Small distributive mistake, I think. (x+i)(x-i) = x^2-i+i-i^2.

 

[TuxDave]

Originally posted by: madnod

Originally posted by: TuxDave

Originally posted by: Cogman
Ok, first off Id like to startout by saying THIS IS NOT HOMEWORK, I aready finished HS and colledge has not started yet. Anyways.

i = (-1)^(1/2) correct?

Well if you had an equasion (x+i)*(x-i)=0 that would equal x^2+1=0, Right? So heres whats confusing me. in the first statement x can eqaul i or -i for the statement to be true. But in the secont x = 1, -1. Why is that?

Uh....

(1)^2+1 != 0
(-1)^2+1 != 0

So... uh... no

cogman
on what basis u did -1^(1/2) there is not such thing in real math.to get 2 roots of x=1,and -1 then ur solving x^2-1=0 not x^2+1=0.
by definition x^2 is positive so x^2+1 can't be zero.(in normal math again!)
BTW i made a Typo in my first post the phase is C not B (am used to use beta and alpha there).
bye

....... what you say?? 😕 lol! j/k j/k....

 

[Calin]

Simply put, a complex number is written as a + i*b (I saw in some places a + j*b - physics maybe???), or r * (cos(alpha) + j*sin(alpha) ), with r=(a^2 + b^2)^.5 and ALPHA as cotangent of b/a

Calin

 

[TuxDave]

Originally posted by: Calin
Simply put, a complex number is written as a + i*b (I saw in some places a + j*b - physics maybe???), or r * (cos(alpha) + j*sin(alpha) ), with r=(a^2 + b^2)^.5 and ALPHA as cotangent of b/a

Calin

J might be used in physics, but I know it's definitely J in Electrical Engineering. The letter I is already used for current.

 

[unsp0ken]

Originally posted by: TuxDave


No probs. I didn't see much use practical use of 'i' until I started studying electrical engineering.

yep, i's or j's as we engineers or engineering students call it, are a huge part of electrical engineering.

 

[chuckywang]

Originally posted by: unsp0ken

Originally posted by: TuxDave


No probs. I didn't see much use practical use of 'i' until I started studying electrical engineering.

yep, i's or j's as we engineers or engineering students call it, are a huge part of electrical engineering.

Word.