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PrepShorts · Study sheet · Class 11 Mathematics · Chapter 4, Complex Numbers and Quadratic Equations

Chapter 4 · Complex Numbers and Quadratic Equations

Why the powers of the new symbol run round a cycle of four

Doing arithmetic in the enlarged system12 minAditya Kedia

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12 min.

Multiply i by itself forever and nothing grows — the whole infinite list holds four numbers, repeating: i, minus one, minus i, one. Step two is not the return; step four is.

The idea

The powers of i repeat every four steps, and that is not a curiosity to be noticed — it is forced. Squaring i twice gives (−1)², which is 1, and once a power of something equals 1 every later power must repeat what came before it, because multiplying by that power changes nothing. So an infinite list of powers collapses onto four values, selected not by the size of the exponent but by its remainder on division by four. And because i has a multiplicative inverse, the same four values run backwards through the negative exponents, which is why a power like i⁻³⁵ is answered by a division, not by thirty-five multiplications.

What you should be able to do

Compute i³, i⁴, i⁵ and i⁶ from the single relation i² = −1
Explain why the repetition begins at the fourth power and not at the second
State the four values taken by i raised to an integer power, indexed by the remainder on division by four
Evaluate i⁻¹ and explain why it equals −i, using the inverse from §4.3.3
Reduce a large positive exponent to a remainder and read off the value
Reduce a negative exponent and evaluate it without expanding
Combine several powers of i in one expression and simplify to the form a + ib

Words to know

Term	Definition in one line	First introduced
power	the result of multiplying a number by itself a stated number of times	printed in this chapter, §4.3.5 heading, p. 79
integer	a whole number, positive, negative or zero — the range of exponents covered	printed in this chapter, §4.3.5, p. 79
multiplicative inverse	the number that multiplies a given non-zero number to give 1, used here for negative exponents	printed in this chapter, §4.3.3, p. 78
remainder	what is left when the exponent is divided by four, and the only thing that decides the value	an added term here; §4.3.5 on this page writes exponents as 4k, 4k+1, 4k+2 and 4k+3 and does not use the word
cycle of four	the repeating block i, −1, −i, 1 through which every integer power of i falls	an added phrasing; the pattern is printed on p. 79 but is given no name
index law	the rule that lets a power be split into a product of smaller powers	the explanation's shorthand, not printed in this chapter

Where people slip up

"i to a big power must be a big number." Every integer power of i is one of exactly four values, and none of them is large. Growth is not what repeated multiplication by i does.
"The pattern repeats every two, because i² = −1." Returning to −1 is not returning to the start. The list repeats only once a power equals 1, and that first happens at the fourth.
"Even exponents give 1 and odd exponents give i." Parity is the wrong invariant; i² and i⁴ are both even powers and they differ in sign. Only the remainder on division by four decides the value.
"Negative exponents need a new rule." They need only the inverse of i, which §4.3.3 already supplies. The same four values recur in reverse order.
"i⁻³⁵ means −i³⁵." The sign belongs to the exponent, not to the base: one is a reciprocal, the other a negation. Take care which exponent you demonstrate this on. At 35 the two happen to agree — both come to i — because an odd power of i is i or −i, and each of those two is its own negative reciprocal. To expose the difference you need an even exponent: with 2 the reciprocal is −1 while the negation is +1, and with 4 the reciprocal is 1 while the negation is −1. The brief's own value for Example 6(ii) is what makes the coincidence at 35 visible.
"You have to divide the exponent and use the quotient somehow." The quotient is discarded entirely. Everything the exponent contributes is its remainder.

Worked answers: Exercise 4.1 · Miscellaneous Exercise · this video explains Exercise 4.1 Q2, Exercise 4.1 Q3, Miscellaneous Exercise Q1, Miscellaneous Exercise Q14

Transcript1,643 words

Multiply a number by itself often enough and you expect it to get bigger. Two, four, eight, sixteen — growing is what repeated multiplication does. So here is a question with a surprising answer. Take the symbol whose square is minus one, and multiply it by itself, again and again, for ever. Nothing grows. The whole infinite list of powers only ever holds four different numbers — not four to begin with and then more, but four, for ever.

And which of the four you land on is not decided by how big the exponent is. It is decided by something much smaller than the exponent. This video is about why that happens, why it could not have happened sooner than the fourth step, and what exactly it is about an exponent that the answer depends on. One relation, iterated. That is the whole method. The symbol times itself is minus one. That is the single thing we are given, and nothing else is assumed anywhere.

So keep going, one step at a time. The third power is the second times the symbol: minus one, times the symbol, is minus the symbol. The fourth power is the third times the symbol again: minus the symbol, times the symbol, is minus the square — and the square is minus one, so the fourth power is one. Stop there and look at what has happened. We started at one, because a product of no factors at all is one.

Then the symbol. Then minus one. Then minus the symbol. And now we are back at one. Four steps, and the list has returned to exactly where it began. Someone always says the pattern repeats every two, because the square is minus one and minus one feels like an arrival. It is an arrival. It is not the starting value. Coming back to minus one is not coming back to the start, and the difference matters, because the argument that makes everything else work needs a power that is exactly one.

So walk it again, and turn down each step honestly. The first power is the symbol, and the symbol is not one. The second is minus one, and minus one is not one. The third is minus the symbol, and that is not one either. The fourth is one. Four is not chosen because four is a tidy number. Four is the first step at which the walk comes back, with one, two and three each visited and each rejected.

Now the step that turns four values into an infinite list. Suppose you know that some power of a thing equals one. Then multiplying by that power changes nothing at all, because multiplying by one changes nothing. So the power at any exponent, and the power four further along, have to be the same number. Split the exponent into the part you had and the extra four, and the extra four contributes a factor of one.

That splitting is the index law, and it is doing real work here, so it is checked rather than assumed. Add the two powers together instead of multiplying them and the splitting fails at every single one of the hundred and sixty-nine pairs tried. Multiply them, and it never fails once. So the moment the walk returns, it is committed to returning for ever — and in both directions. So the value repeats every four.

But that is a claim about which exponents give the same answer, and a claim like that deserves rivals. Here are eleven of them: the remainder on dividing by two, by three, by four, and so on all the way up to twelve. Each rival says the same kind of thing in its own way — exponents leaving the same remainder ought to give the same power. Test all eleven in exactly the same way.

Sixty-one exponents, running well into the negatives, give one thousand eight hundred and thirty pairs to compare. Three of the eleven survive: four, eight and twelve. Every other one is caught out. Parity — the idea that even exponents behave alike — gets four hundred and sixty-five of those pairs wrong. And you can see why in one line: two and four are both even, and they give minus one and one.

Of the three that survive, four is the smallest. That is exactly what the word period means. Divide an exponent by four and two things come out: a quotient and a remainder. Only one of them matters, and it is worth seeing how completely the other one does not. Fix a remainder — say one — and let the quotient run over sixteen different values, from minus five up to ten.

Sixteen different exponents. One value, every time. Do the same at each of the four remainders and the story repeats: sixteen exponents, one value. And the four remainders give four different values, so nothing has collapsed further than it should. Here is the mistake worth naming. Keep the quotient, and let it flip the sign whenever it is odd, and you get a reading that still repeats — but every eight, not every four.

It gets two hundred and thirty-two pairs wrong, and it makes the cycle look twice as long as it really is. Now go the other way. What is the symbol to the power minus one? The temptation is to answer minus the symbol, because there is a minus sign in the exponent — and that is the right answer for entirely the wrong reason. A negative exponent is a reciprocal. It is not a negation.

So take one over the symbol, and multiply above and below by the symbol. The bottom becomes the square, which is minus one, so the whole thing is the symbol over minus one — which is minus the symbol. The reciprocal really is the negative, here. But watch how nearly that coincidence fools you. Over the first forty exponents, the reciprocal of a power equals the negative of that power at exactly twenty of them — every odd one.

At every even one they part company. At two, the reciprocal is minus one while the negative is one. At four, the reciprocal is one while the negative is minus one. With a reciprocal in hand, the negative exponents are not new work. Minus one gives minus the symbol. Minus two gives minus one. Minus three gives the symbol. Minus four gives one. The same four values, in the reverse order, running away to the left as far as you care to go.

And it is worth being exact about what was done there. No rule was stretched by analogy. The value at a negative exponent was built by taking the reciprocal first and then multiplying it by itself, which is the same walk as before. The cycle running backwards is a result of that walk, not an assumption underneath it. The zeroth power is one from whichever side you come at it, because a product of no factors asks nothing of anybody.

So any exponent at all, however large and whichever sign, is answered by one division with remainder. Take the symbol to the power minus thirty-five. Thirty-five is four eights and three left over. So the thirty-fifth power is the third power, which is minus the symbol. Now invert it. The reciprocal of minus the symbol is the symbol, because minus the symbol, times the symbol, is minus the square, which is one.

The answer is the symbol. Two more of the same kind. The ninth power plus the nineteenth: nine is four twos and one left over, so that is the symbol; nineteen is four fours and three left over, so that is minus the symbol; and the two cancel to nothing. And the power minus thirty-nine: thirty-nine is four nines and three left over, giving minus the symbol, whose reciprocal is the symbol again.

Two harder ones, both of them hard only until the cycle is available. First: take the eighteenth power, add the twenty-fifth power of one over the symbol, and cube what you get. Eighteen is four fours and two left over, so the eighteenth power is minus one. One over the symbol is minus the symbol, and its twenty-fifth power is minus the symbol. So inside the bracket sits minus one, minus the symbol.

Cube that, and it comes to two minus two times the symbol. Second: find the smallest positive whole number m for which one plus the symbol, over one minus the symbol, raised to the m, comes to one. Clear the bottom by multiplying above and below by one plus the symbol, and the whole fraction turns out to be the symbol itself. So the question was the cycle all along, and the answer is four.

One last question, and it is the one that says what all of this was resting on. Suppose the symbol had squared to something else. Try nine values, from minus four up to four. For seven of the nine the powers never come back at all — and that is not because a search went looking and found nothing. Every power of the symbol has one of its two slots empty, always.

So a power can only be one when a plain power of the chosen value is one; and a number bigger than one in size stays bigger than one when you multiply, so it can never come back down. At nought the powers do not grow at all — they simply vanish. Only two of the nine ever return. One squares to one, comes back after two steps, and holds nothing you did not already have.

The other squares to minus one, comes back after four, and holds four values. The cycle of four is what the minus one bought.

Where this fits

Taken from the notes each video was made from, not from the reading order — these are the ideas this one rests on and the ones that later rest on it.

Builds on

An equation with no real solution, and the symbol invented to solve itClass 11 · Ch 4, Complex Numbers and Quadratic Equations
Multiplying, then inverting: how division becomes possibleClass 11 · Ch 4, Complex Numbers and Quadratic Equations

Either side of this one

Square roots of negative numbers, and the surd rule that stops workingClass 11 · Ch 4, Complex Numbers and Quadratic Equations