The Need for a New Number: i and its Powers

Transcript

0:00Introduction

Hello everyone! Let me prove something shocking. One equals root one, which is root of minus one times minus one. Split it: root minus one, times root minus one, that is, root minus one whole square, which is minus one. So, one equals minus one? Something is wrong here, right? Today we meet this square root of minus one, and by the end, you will catch the broken step. Its powers come up often in JEE, so let's go.

0:32A number whose square is -1?

So, look: x square plus one equals zero, which gives x square equals minus one; that is what the double arrow means. Now tell me, which real number, squared, gives minus one? None, absolutely none! Square any real number and you get something greater than or equal to zero; that is the sign on screen. So no real solution, and for centuries, mathematicians simply said: no solution. Okay? But in the fifteen hundreds, the sixteenth century, Cardano and Bombelli, solving cubic equations, found square roots of negatives on the way to perfectly real roots! So these impossible numbers were useful. Euler later gave them the symbol i, for imaginary. Interesting, right?

1:17The imaginary unit: iota

So, the definition. The number i, read as iota, has iota square equal to minus one. So iota is the square root of minus one; that is the root sign on screen. With me? Next, for real a greater than zero, root of minus a equals iota root a. The minus sign comes out as iota, and the rest stays under the root. Quick check: what is root of minus sixteen? Four iota, yes, exactly! Iota times root sixteen, iota times four. And root of minus seven is iota root seven; seven is not a perfect square, so it stays under the root. Clear?

1:55Powers of i: a cycle of four

Now, keep multiplying by iota. Iota to the power one is iota; the small number up top is the power. Iota square is minus one. So, what is iota cube? Minus iota, correct! Iota cube is iota square times iota; the dot means multiply. And iota to the four is iota square, in brackets, whole square: minus one, whole square, which is one. Back to the start! Got it? So iota to the five is iota again, iota to the six is minus one, and so on; the three dots mean it continues. Iota, minus one, minus iota, one: a cycle of length four. Remember it, okay?

2:40Any power of i: use the remainder

Now, the whole cycle in one line, for every integer n: positive, zero or negative. Iota to the four n is one; four n plus one gives iota; four n plus two, minus one; and four n plus three, minus iota. Right? So the rule is really simple. To find iota to the power m, divide m by four and use only the remainder. The quotient does not matter! Now, one over iota. Is it iota? Is it minus iota? Think. Minus iota, yes! One over iota is iota to the minus one, and minus one is four times minus one, plus three: remainder three. And iota to the zero is one. With me?

3:21Adding four powers in a row

Now, any four powers of iota in a row. Take iota to the n common, and the bracket is one plus iota minus one minus iota. Everything cancels! So any four consecutive powers of iota add to zero. Such a time saver in long sums, okay?

3:39The trap: root a times root b is not always root ab

Now, back to our shocking proof, written with iota. One equals root one, equals root of minus one times minus one, equals root minus one times root minus one, equals iota square, equals minus one. So which step is wrong? Let's test one suspicious step, splitting one root into two roots, with numbers. Root of minus four is two iota, and root of minus nine is three iota. Multiply: six iota square, which is minus six. But multiply inside the root first: minus four times minus nine is thirty six, and root thirty six is plus six. Plus six, and minus six. Not equal! So root a times root b equals root of a times b breaks here. Here is the exception. That rule needs at least one of a and b to be greater than or equal to zero. Both negative? It fails. And look at our proof: it split root of minus one times minus one, both negative. That is the broken step. Caught it? So, the safe habit: always change root of minus a into iota root a first, and only then multiply. This mistake comes up often in exams. Don't fall for it!

4:52Example 1: i to the power 35, -50, 2026 and more

Now, example one: five parts, numbered in small Roman numerals. Iota to the thirty five, to the minus fifty, to the two thousand and twenty six; minus iota, in brackets, to the seven; and iota to the minus one plus iota to the minus three. Pause the video and try. Done? Okay, let's check. Thirty five divided by four: four eights are thirty two, remainder three. So iota to the thirty five is iota cube, which is minus iota. Next, a negative power. Fifty is four twelves plus two, remainder two, so iota to the fifty is minus one. And iota to the minus fifty is one over that: one over minus one, which is minus one. Now, iota to the two thousand and twenty six. Big power, small work. Four times five hundred and six is two thousand and twenty four: remainder two. So the answer is minus one. That's it! Now, minus iota, in brackets, to the power seven. Careful with the signs. Is the answer iota, or minus iota? Plus iota! The bracket means the minus sign also gets the power: minus one to the seven, times iota to the seven. That is minus one times iota cube, since seven leaves remainder three. So, minus of minus iota: plus iota. Don't stop at minus iota! Last part. Iota to the minus one is minus iota, as we saw. Iota to the minus three is one over iota cube: one over minus iota, which is plus iota. So the sum is zero. Very good!

6:31Example 2: a sum of powers of i (IIT-JEE 1998)

Now a harder one, from IITJEE, nineteen ninety eight. The big Greek letter sigma means add up, from n equals one, at the bottom, to thirteen, at the top. Each term is iota to the n, plus iota to the n plus one. Hint: take something common. Pause and try. Done? Okay, let's see. Take iota to the n common: each term is iota to the n, times one plus iota. One plus iota is common to every term, so it comes out of the sum. Right? Now we need iota, plus iota square, up to iota to the thirteen. What do the first twelve add up to? Zero, exactly! Twelve terms make three complete cycles of four, and each cycle adds to zero. Our four in a row result, used three times. So only the thirteenth term survives: iota to the thirteen. Thirteen is four threes plus one, so it is just iota. Finally, one plus iota, times iota: iota plus iota square, which is iota minus one. Answer: iota minus one. See how fast that was, once you know the cycle?

7:48Quick revision: i squared, remainder rule, root trap

So, quick revision. Iota square is minus one, and root of minus a is iota root a. For any power of iota, divide by four and keep the remainder; four powers in a row add to zero. And when both numbers under the roots are negative, change to iota first, because root a times root b equals root of a times b fails. Now go and try the practice problems on this page. All the best!