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Topic: Imaginary Plane Problem
Replies: 2   Last Post: Mar 14, 2007 9:59 AM

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R. Brown

Posts: 54
From: texas
Registered: 12/29/05
Imaginary Plane Problem
Posted: Mar 13, 2007 3:35 PM
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consider the following:


2 - 1.4142135i.........represents a complex number.

then, could not the graph point on the imaginary plane be stated as:

x= 2........(where: x= axis of reals)
i= -1.4142135..........(where: i=axis of imaginaries)

At this point, if you 'agree' that what I have just stated is reasonable and coincides with standard geometry, consider this:

Wouldn't it make 'more' mathematical sense to have stated the coordinates as:

x= 2..........(where: x= axis of reals)

z/i= -1.4142135..........(where: z/i=axis of imaginaries)


Now I will illustrate why:

z/i = -1.4142135.....eq.101

Let's multiply both sides of eq. 101 by i:
i * z/i = -1.4142135 * i
z= -1.4142135i .....

We have No problem transferring the (i) to the RHS of the equation!

Now let's look at the standard statement:

i= -1.4142135......eq.201

If we now multiply both sides of eq.201 by i, what happens?

i * i= -1.4142135 * i......eq.201-a
i2= -1.4142135i.. is this true?
this becomes:
-1 =/= -1.4142135i

NO! The equation above is obviously not true!

Therefore the idea of graphing 'imaginaries' on an (x & i axes system) makes LESS sense than graphing
them on an (x & z/i axes system)!

#Puzzles and Ideas compliments of Col. Rbtx, the Barnyard physicist of Texas



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