[TI-M] Re: complexes, ti-89 & 92+, polar and cSolve
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[TI-M] Re: complexes, ti-89 & 92+, polar and cSolve
> e^(Pi * i * 2 / 9 ) != -1 ^ ( 2 / 9 )
If you add some parens, you'll see that this is correct:
e^(pi*i*2/9) = (e^(pi*i))^(2/9) = (-1)^(2/9)
Notice the (-1) must be in parens. That threw me for a while.
If you have Complex mode set to polar and angle to radians, it'll give you
that simplification. If not, it'll use this rule:
e^(i*theta) = cis(theta) = cos(theta)+i*sin(theta)
so e^(2/9*pi*i) = cos(2/9*pi)+i*sin(2/9*pi)
if your Complex mode is set to rectangular, you'll get the same value for
(-1)^(2/9). They're equivalent, and the calculator is correct.
I'm not sure if I understood what the original poster was asking (can't say
I've really gone into this stuff in any class), but he may want to be in
rectangular mode, not polar.
Here's an interesting AMS problem -- try entering:
(e^(2pi/9*i))^3 = e^(2pi/3*i)
The calculator returns _false_, which is the _wrong_ answer! If it can't
tell, it should at least feed my input back to me; of course, the correct
answer is true. It's not indeterminant, it's wrong.
Further investigation reveals the calculator evaluating to the following:
(e^(2pi/9*i))^3 = -.5+.86602540378444*i
e^(2pi/3*i) = -.5+.86602540378445*i
You can only find that last digit by evaluating it, then highlighting it in
the history and pressing enter (in the history, it'll display 12 digits max
in the mantissa; when you do this to paste it in the command line, you'll be
able to view the full 14 digits that the calc uses internally).
I'm going to send this one off to some people at TI. . . I'm not sure how
one would go about fixing it (ie, I don't know how you can guarantee this
won't happen when you've gotta round the mantissa at some point), but I'll
point it out.
-Scott