The i operator

[Ben Abbott]

On Jan 25, 2009, at 10:39 PM, Steve wrote:

> Hi,
>
> I’d appreciate some advice with Octave and use of the i imaginary  
> operator.
>
> I can get this to work:
>
> octave-3.0.3.exe:27> (2 * (pi) * 10.7e6 * 1.62e-6i)
>
> ans = 0.00000 + 108.91273i
>
>
> And this works where I define the constant w as 2*pi*10.7e6:
>
> octave-3.0.3.exe:28> w = 2 * (pi) * 10.7e6
>
> w = 6.7230e+007
>
> octave-3.0.3.exe:29> (w * 1.62e-6i)
>
> ans = 0.00000 + 108.91273i
>
> I’d like to also define the constant L and use it in the calculation,
> but I get this error:
>
> octave-3.0.3.exe:30> w = 2 * (pi) * 10.7e6
>
> w = 6.7230e+007
>
> octave-3.0.3.exe:31> L = 1.62e-6i
>
> L = 0.0000e+000 + 1.6200e-006i
>
> octave-3.0.3.exe:32> w * Li
>
> error: `Li' undefined near line 32 column 5
>
> error: evaluating binary operator `*' near line 32, column 3
>
>
> How does one use the i operator with constants so that Octave  
> recognizes
> the i as the imaginary operator?
>
> Thanks.
>
> Steve

I assume  you are asking for help, and not reporting a bug?

For the first example, the following works

	(2 * (pi) * 10.7e6 * 1.62*10^(-6i))
	ans = 3.4432e+07 - 1.0333e+08i

For the last, it can be wise to use 1i in the event "i" has been used  
as a variable.

	w = 2 * (pi) * 10.7e6;
	L = 1.62*10^(-6i);
	w * L * 1i
	ans = 1.0333e+08 + 3.4431e+07i

Notice the answers for each of the above. The "i" is not part of the  
exponent.

Evaluate 1e1i in either Matlab or Octave gives the same result.

	1e1i
	ans = 0 + 10i

Thus, just like the answers above, 1e1i is interpreted as (1e1)*1i

Ben