June 28th, 2017, 08:08 PM  #1 
Newbie Joined: Jun 2017 From: Long Beach Posts: 20 Thanks: 0  (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! 
June 28th, 2017, 08:58 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,517 Thanks: 910 Math Focus: Elementary mathematics and beyond 
$(1)^{2/3}$ 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 TI89? 
June 28th, 2017, 09:09 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,857 Thanks: 2230 Math Focus: Mainly analysis and algebra 
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. Last edited by v8archie; June 28th, 2017 at 09:18 PM. 
June 28th, 2017, 09:22 PM  #4 
Newbie Joined: Jun 2017 From: Long Beach Posts: 20 Thanks: 0 
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! 
June 28th, 2017, 09:42 PM  #5 
Newbie Joined: Jun 2017 From: Long Beach Posts: 20 Thanks: 0 
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? 
June 29th, 2017, 03:36 AM  #6 
Math Team Joined: Jul 2011 From: Texas Posts: 2,579 Thanks: 1276 
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 TINspire does the same. See screenshot. 
June 29th, 2017, 09:23 AM  #7 
Newbie Joined: Jun 2017 From: Long Beach Posts: 20 Thanks: 0 
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! 
June 29th, 2017, 09:32 AM  #8 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,857 Thanks: 2230 Math Focus: Mainly analysis and algebra 
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.

June 29th, 2017, 10:49 AM  #9 
Newbie Joined: Jun 2017 From: Long Beach Posts: 20 Thanks: 0 
oh, that makes sense! I understand now! thank you v8archie 