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Euler's Identity

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Hi,
I understand how to get i*e^(i*x) from e^(i*x). But I don't understand what does e^(i*x) mean. How can you know that e^(i*x) has the Helix?
( I understand e^2, e^x, e^n but not e^j2, e^ jn,...)

Hi again,


Well sometimes you have to sit back and think about this for a while, and you have to look at what is happening and what 'e' to an imaginary power means like e^(2*i).

First, from the preceding we discovered that we can convert e^(i*x) into a circular format:
e^(i*x)=cos(x)+i*sin(x)

and we went on to prove this using a Taylor Series. So what we have now is a new expression we can use to show what e^(i*x) means, and that is:
cos(x)+i*sin(x)

But what does this new expression really represent. Well to start, it has a real part and imaginary part, so it is at once a complex number:
X=a+i*b

where a and b are functions of x so they are not really constant, but if x is constant then it is just an ordinary complex number A+i*B.

And because it is a complex number, it can be represented on the complex plane as a single point with coordinates (a,b). Note we lost the imaginary operator 'i' when we did this. So now for constant x we have a single point (a,b) or we can just say that it is an ordered pair [a,b] which can be viewed as a vector. So a graph of this would be a point in the complex plane, where a would be along the real axis and b would be along the (usually vertical in 2d space) imaginary axis.

But if x varies from say 0 to some number like 7, we keep getting new values for a and b because:
a=cos(x), and
b=sin(x)

and note again we dont need the imaginary operator when we represent it this way.

So to graph this, we let x vary from 0 to some number and plot the values of a and b. We can do that just in the plane, and we would see a circular path, but since x is varying we would better represent this in 3d space. So as x varies a and b vary, and we plot a along the real axis, b along the imaginary axis, and x along the typical time axis which is usually vertical. What we get is a helix because as the numbers a and b get generated the point they plot moves in a circle while x increases linearly. Lets look at a few quick points...

First, the new representation of e^(i*t) using the variable 't':
cos(t)+i*sin(t)

or in component form:
x=cos(t)
y=sin(t)

Now we do a few values of t...

With t=0 we have:
t=0
x=1
y=0

So now x plots along the real axis, and y plots along the imaginary axis, and t=0 so the first point is in the x,y plane and it is at point (t,x,y) which is (0,1,0).

Next, we'll let t=0.1 so we have approximately:
t=0.1
x=cos(t)=0.9950
y=sin(t)=0.0998

so our next point is at (0.9950,0.0998) but since t also changed we have to move up along the time axis by 0.1 seconds, so we get the point (0.1,0.9950,0.0998).

One more plot point, with t=0.2 we have:
t=0.2
x=cos(t)=0.9801
y=sin(t)=0.1987

so the new point is plotted at (0.2,0.9801,0.1987). And here we see we got a new point (x,y) but also we moved up along the time axis one more 0.1 increment so this new point not only follows the curve in the x,y plane but it also moves vertically by 0.1 unit of time.

If we did this with more and more values for time 't' and chose smaller increments we would see a continuous curve in three dimensional space and that would map out a helix.
 
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