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June 28th, 2017, 09:08 PM   #1
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(-1)^(2/3) with TI89

hi guys

I haven't done math for a long time and I've started reviewing calculus!
I was trying to graph a function with TI 89 and it wasn't giving me the portion of the graph in the second coordinate!
when I went through the equation with the calculator, I noticed that TI89 is treating something like (-1)^(2/3) as a complex number that has an imaginary part:
(-1)^(2/3) = -0.5+(3^0.5)/2 * i ; I would've thought the expression in TI89 will be treated like (-1)^(2/3) = ((-1)^2)^(1/3)=1!
Can you help me see how TI89 is treating this expression to give this discrepancy? Thanks!
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June 28th, 2017, 09:58 PM   #2
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$(-1)^{2/3}$ is a constant (and it's equal to 1) not a function. Its graph is the line y = 1.

Can you post exactly what you entered into your TI-89?
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June 28th, 2017, 10:09 PM   #3
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Mmm... in the reals, yes. But in the complex numbers $$(-1)^{\frac23} = (e^{i\pi})^{\frac23} = e^{\frac23 i\pi} = -\tfrac12(1 - i\sqrt3)$$
is an obvious value.

The full calculation I guess is
$$(-1)^{\frac23} = (e^{i(1+2n)\pi})^{\frac23} = e^{\frac23 i(1+2n)\pi} = \left\{\begin{aligned} e^{\frac23 i \pi} &= -\tfrac12(1 - i\sqrt3) \\ e^{2 i \pi} &= 1 \\ e^{-\frac23 i \pi} &= -\tfrac12(1 + i\sqrt3) \end{aligned}\right.$$
And from there it depends how you go about picking your principal value, although $e^{2i\pi}=e^0=1$ is natural.
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Last edited by v8archie; June 28th, 2017 at 10:18 PM.
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June 28th, 2017, 10:22 PM   #4
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when I enter (−1)^(2/3) in TI 89, it gives me -0.5+(3^0.5)/2 * i=e^(2*i*pi/3)! I like to understand why TI 89 is treating (−1)^(2/3) as a complex number so I can avoid my function being treated as a complex function when I meant it to be treated as a real function in TI89!

what I was trying to graph is
y=(5^(2/3)-x^(2/3))^3/2; eq1
when I enter it as eq1, I only get the graph in the first quadrant.
but when I enter it as eq2:
y=(5^(2/3)-(x^2)^(1/3))^(3/2); eq2
I get the graph in both 1st and 2nd quadrants!!
it looked like the issue is how x^(2/3) is being treated different from (x^2)^(1/3) when x is negative; that's why I tried a number example like (-1)^(2/3) which is treated as a complex number while ((-1)^2)^(1/3) is treated as 1! Any feedback is appreciated
thanks!
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June 28th, 2017, 10:42 PM   #5
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Thank you for showing me this calculation!
but why is the calculator treating -1 as e^(i*pi) here instead of just treating it as -1 raised to the power of 2 raised to the power of 1/3? when is TI89 doing the calculations in complex?
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June 29th, 2017, 04:36 AM   #6
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Go to the MODE menu and check the complex format ... it should set for real

If set to polar or rectangular, you'll get outputs in complex form. My TI-Nspire does the same. See screenshot.
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June 29th, 2017, 10:23 AM   #7
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Thank you Skeeter! You solved my problem to make sure my equation is treated as a real equation not complex, I'll adjust the mode for how the values are treated as real not complex!
so I guess in the rectangle or polar complex mode all calculations are carried out in complex. This discrepancy in treating a value as real or complex happens when a negative number is raised to some rational power!
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June 29th, 2017, 10:32 AM   #8
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That's because exponentiation is not generally defined in the real numbers for negative bases. Even where it is defined, it's not well behaved and usually discontinuous.
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June 29th, 2017, 11:49 AM   #9
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oh, that makes sense! I understand now! thank you v8archie
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