Math Help - TI-84+ complex numbers

I can calculate i to any power very easily without using the TI-84+. Simply divide the power by 4. If the remainder is 0, then the answer is 1. If the remainder is 1 then the answer is i. A remainder of 2 yields -1 and a remainder of 3 yields -i.

Using the TI-84+, and in a + bi mode, i^6 = -1.

i^7 = -3E-13-i and since -3E-13 is so close to zero, you just use -i

i^40 = 1

i^41 = -1E-13+i and since -3E-13 is so close to zero, you just use i.

Sometimes it;s a simple answer. Sometimes it;s in scientific notation. Just Curious. It seems unnessary to me to display it any way except 1, i, -1, or -i.

Originally Posted by masters

I can calculate i to any power very easily without using the TI-84+. Simply divide the power by 4. If the remainder is 0, then the answer is 1. If the remainder is 1 then the answer is i. A remainder of 2 yields -1 and a remainder of 3 yields -i.

Using the TI-84+, and in a + bi mode, i^6 = -1.

i^7 = -3E-13-i and since -3E-13 is so close to zero, you just use -i

i^40 = 1

i^41 = -1E-13+i and since -3E-13 is so close to zero, you just use i.

Sometimes it;s a simple answer. Sometimes it;s in scientific notation. Just Curious. It seems unnessary to me to display it any way except 1, i, -1, or -i.

You see this kind of thing happening in the older computers too. (For all I know it still happens.) The algorithm that your calculator is using sometimes churns out very small, but "writable" numbers. By writable I mean something that is greater than . That's what it is on my calculator anyway. The "problem" is simply that algorithms on your calculator aren't perfect.

Think of it this way, your calculator might likely turn the expression into one involving logarithms (that's the way calculators used to work, anyway), do the problem, then use inverse logarithms, and finally display the answer. There can be any number of rounding errors along the way.

-Dan