RE: [tuning] A correction....

Sarn wrote,

>>Please, PLEASE, P-L-E-A-S-E, ---if somebody, anybody, could calculate:

>a) (2i^(1/12))^n
>b) (2i^(1/12i))^n
>c) (2^(1/12i))^n
>d) (2i^(1/12))^in
>e) (2i^(1/12i))^in
>f) (2^(1/12i))^in

Hi Sarn,

You seem to be leaving out some parentheses above, so I will augment this
list with some alternate versions, and hopefully the expressions you
intended are in there somewhere:

a1) (2i^(1/12))^n
a2) ((2i)^(1/12))^n
b1) (2i^((1/12)i))^n
b2) ((2i)^((1/12)i))^n
b3) (2i^(1/(12i)))^n
b4) ((2i)^(1/(12i)))^n
c1) (2^((1/12)i))^n
c2) (2^(1/(12i)))^n
d1) (2i^(1/12))^(in)
d2) ((2i)^(1/12))^(in)
e1) (2i^((1/12)i))^(in)
e2) ((2i)^((1/12)i))^(in)
e3) (2i^(1/(12i)))^(in)
e4) ((2i)^(1/(12i)))^(in)
f1) (2^((1/12)i))^(in)
f2) (2^(1/(12i)))^(in)

I'm going to e-mail you the answers off-list, Sarn.

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