i and (-1) with Multiple Powers
Date: 08/09/2001 at 22:32:03 From: Kenneth Martin Subject: Imaginary numbers (the order in which i and (-1) should be raised by when using multiple powers) My question is about the order in which i and (-1) should be raised when using multiple powers. The problem that I have run into is that when I change the order of powers I can obtain many different values. This is just one of the many examples I have: (-1)^(3/2)^(1/3) equals .8660254038-.5i, but when it is simplified to (-1)^(1/2) it of course equals i. Stranger still, when I switch the order of the powers to (-1)^(1/3)^(3/2), the value equals (-i). I have many more examples of this, but they're generally the same. Why is it possible to obtain so many different values? I am quite aware that i's have a nature all their own that negates rules in math that are followed when dealing with real numbers. So what are the rules we MUST follow when dealing with i?
Date: 08/11/2001 at 23:15:50 From: Doctor Peterson Subject: Re: Imaginary numbers (the order in which i and (-1) should be raised by when using multiple powers) Hi, Kenneth. It sounds as if you are finding powers on a calculator without stopping to think about what they mean. Let's think about that a moment. We're working with complex numbers here. Recall that x^(1/2), the square root, has _two_ values, not just one; in real numbers we can choose one as the primary root, but we can't even do that in complex numbers. Further, there are _three_ cube roots, _four_ fourth roots, and so on. So it really shouldn't surprise you at all that you can get several different values for the same expression; they are all correct! What should be surprising is that the calculator would tell you that one of them is "the" answer. That's the real cause of the difficulty; whatever rule they use to decide which answer to give produces different results when you do it in a different order. It would be better if it told you somehow that this is only one of a certain number of values. A side issue to consider is the meaning of x^y^z. We usually take this to mean x^(y^z); you appear to mean (x^y)^z, which is the same as x^(yz). I'll add brackets to avoid confusion. So let's try doing the calculation as if we were the calculator, doing just what we're told. The expression [(-1)^(3/2)]^(1/3) tells the calculator to raise -1 to the 3/2 power, which it probably does by taking "the" square root, i, and cubing it, to get -i. (Of course, this is actually the other square root of -1.) It would be equally valid to take -i as the square root and cube it to get i; but let's forget that for now. Now the calculator will take "the" cube root of this, which might be done either by thinking of it as having angle 270 degrees and dividing by 3 to get 90 degrees (giving the cube root as i); or it might think of the angle as -90 degrees and take the cube root with angle -30 degrees. The latter agrees with the answer you give. Since there are two square roots and three cube roots, there will be six possible values for this expression. You've found ways to get three of them. See if you can do the same thing I just did to explain [(-1)^(1/3)]^(3/2) I'd get i, myself; but we don't know for sure what rules the calculator is using. You might want to read the manual to see if they tell you, or keep it a secret. You can also read these answers from the Dr. Math archives for some additional insights: Multiplying Radicals of Negative Numbers http://mathforum.org/dr.math/problems/amy.7.12.00.html Cube Roots of Numbers http://mathforum.org/dr.math/problems/martin11.5.97.html - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Date: 08/12/2001 at 17:54:11 From: Kenneth Martin Subject: Thank you I recently submitted a question to Dr. Math about a problem I was having with i. Within a couple of days I received a response that helped with the problem that I had. Dr. Math explained enough for me to understand the solution, yet left enough unexplained to make me rethink the problem. You have managed to answer more for me in two days then my teachers could in two years! I just wanted to thank you and I hope you will continue helping students like myself for a long time to come.
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum