Splitting a number into two parts, and when two such numbers agree

0:00Every ordinary number you have met can be pinned down with one piece of information. Say where it sits on the line, and you have said all there is to say. Two. Minus a half. Root three. One position each, one number each. The new kind of number is not like that. It is written as a first number, plus the symbol i, times a second number. a plus i b.

0:27And a and b there are both perfectly ordinary numbers — the kind that do sit on the line. So one of these carries two ordinary quantities, not one. That is not decoration. It is the shape of the thing. You cannot pin one down with a single number, because there are two independent things to say. Change either one of them and you have a different number. Two slots. Two quantities.

0:55Every question you ask about one of these will have to reckon with both. The two quantities need names. The first one — the a — is called the real part of the number. The second one — the b — is called the imaginary part. If the number is called z, those are written Re z and Im z. Take z equal to two plus i five. Its real part is two.

1:22Its imaginary part is five. Not five i. Five. Read those two names carefully, because they are about position, not about kind. The real part is the quantity sitting in the first slot. The imaginary part is the quantity sitting in the second slot. Both of them are ordinary real numbers. Both of them live on the line. The names tell you which slot a quantity came out of, and nothing else at all.

1:55That last point is where almost everyone slips. The imaginary part of two plus i five is five. The symbol is part of how the number is written down. It is not part of the value being named. Look at the form once more. a, plus, i, times b. The symbol sits between the plus and the b. It marks the slot. The quantity in the slot is b. Why does this matter so much?

2:24Because five and five i are not the same number. Put the symbol in front of five and you have moved to a different number entirely. Report the imaginary part as five i and you have produced a part that is not a real number. And once your parts stop being real numbers, you cannot compare them and you cannot solve with them. Strip the symbol. Report the number. Five.

2:53Here are three of these written out, to fix what the two slots are allowed to hold. Two, plus i three. Minus one, plus i root three. Four, plus i times minus one eleventh. Read the first slot off each of them: two, minus one, four. Now the second slot: three, root three, and minus one eleventh. Look hard at what turned up in that second slot. One of the three is a whole number.

3:23One of them is not a fraction at all — root three is irrational, and it is perfectly welcome there. And one of them is negative, and a fraction as well. There is no restriction whatever on the imaginary part. It is any real number you like. Negative, fractional, irrational — all allowed. And not one of the three carries the symbol. Now the real question of this video. When should two of these count as the same number?

3:57For ordinary numbers that is barely a question. Two points on a line are the same when they are in the same place. But here you have two objects, and each of them is carrying two quantities. a plus i b on one side. c plus i d on the other. Four ordinary numbers in play, and one question: are these two the same? You could imagine several answers. Perhaps only the first slots have to match, and the second is decoration.

4:28Perhaps some combination of all four has to come out right. Perhaps a shortfall in one slot could be paid for by a surplus in the other. That last one deserves taking seriously — it is what would happen if the two slots were not independent. So which is it? The answer is the strict one. a plus i b is the same number as c plus i d exactly when a equals c, and, separately, b equals d.

5:01Both of them. Independently. No trading between the slots. One question about these new numbers has just become two questions about ordinary ones. Say it the other way round. If either slot disagrees — even in the second slot only, even slightly — they are different numbers. Two plus i three and two plus i four are not nearly the same. They are different, and that is that. So that is the condition, and you could stop here and memorise it.

5:35But a condition you have only memorised is a condition you will misremember under pressure. It is worth a couple of minutes to see why it could not be anything else. Suppose the two are the same number, and see what that forces. a plus i b equals c plus i d. Subtract, and gather the ordinary numbers on one side and the symbol on the other. a minus c equals i, times d minus b.

6:07On the left, a minus c is an ordinary real number. On the right, so is d minus b, with the symbol standing in front of it. Now square both sides. On the left you get a minus c, all squared. On the right, i squared times d minus b, all squared — and i squared is minus one. So the right-hand side is minus, d minus b, all squared.

6:35Look hard at those two. The left is the square of a real number, and a square is never negative. The right is minus the square of a real number, and that is never positive. One of them lives at nothing or above. The other at nothing or below. And they have been set equal. Each shortfall is negative, nothing, or positive — three cases each, so nine pairs of cases in all.

7:04Work through the nine, and exactly one lets the two sides agree. It is the pair where both shortfalls are nothing. So a equals c, and b equals d. Both slots. Which is the condition. That argument is worth being exact about — what it uses, and what it proves. What it uses is subtraction, and squaring. Neither has been given a meaning at the moment the condition is first stated. So this is not a proof that comes first — it borrows arithmetic that arrives later.

7:41And it leans on something else, easy to miss. It leans on i squared being negative. Suppose the symbol squared to plus two instead, and run the same steps. You would arrive at a minus c, all squared, equals two times d minus b, all squared. And that has escapes. Take a minus c to be root two, and d minus b to be one. Two equals two. Both shortfalls are nonzero, and the equation is satisfied.

8:14With a plus, a shortfall in one slot really can be paid for by a surplus in the other. It is the minus that shuts that door. So: granted the ordinary laws of arithmetic and i squared equal to minus one, matching both slots is the only reading the algebra will tolerate. Not a rule to swallow whole. The only one that survives. Now the payoff, and it is the most useful single move in this whole subject.

8:48One equation between these numbers is worth two equations between ordinary ones. Here is why that is worth so much. You are handed one statement. It looks like one condition, and there are two unknowns to find, so it looks hopeless. But the statement is between two-slot numbers. Match the first slots, and that is one ordinary equation. Match the second slots, and that is a second ordinary equation. Two equations, two unknowns, and every method you already have applies to them.

9:24The step is worth a name, so call it equating parts. Get both sides into the standard form. Set first slot against first slot. Set second against second. Then solve. A statement that looked impossible has turned into school algebra. Here is one, exactly as it comes. Four x, plus i times three x minus y, equals three plus i times minus six. x and y are ordinary real numbers, and you must find them.

10:00Do not solve yet. Match first. The first slot on the left is four x. The first slot on the right is three. So four x equals three, and x is three quarters. The second slot on the left is three x minus y. On the right it is minus six. So three x minus y equals minus six. Now solve, and now substitute — in that order. y is three x plus six. x is three quarters, so three x is nine quarters, and six is twenty four quarters.

10:36y is thirty three quarters. Both slots matched, and there is the answer. And watch what each condition manages on its own. Put forty five candidate values on a list for each unknown. The first condition on its own leaves all forty five values of y standing. The second on its own leaves thirteen pairs. Together they leave exactly one. One last trap, and it is a good one. Here is a number. Three, plus i times, two plus i.

11:11What is its real part? The obvious answer is three, because three is what stands in front of the plus. The obvious answer is wrong. Work out what is actually there. i times two plus i is two i, plus i squared. And i squared is minus one. So the number is three, minus one, plus two i. Which is two, plus two i. Its real part is two. Its imaginary part is two.

11:41Not three, and certainly not two plus i. The rule is simple and has no exceptions. Get the number into the standard form first, then read its parts. Read them off an expression that has not been simplified and you are reading the page, not the number. Here the symbol was hiding inside the second slot, which is exactly where it can do the most damage. This matching step is not a one-off.

12:12Whenever two of these numbers get multiplied together, the answer has two slots, and each slot is a separate fact about the product. The real part of a product, for instance, is not what you might guess. It is the real part of the first times the real part of the second, minus the imaginary part of the first times the imaginary part of the second. That minus is the one you have been watching all the way through — it is where i squared equals minus one walks in.

12:46And the general shape turns up constantly: one condition between these numbers, two ordinary unknowns wanted. Every time, the move is the same. Standard form. Match the first slots. Match the second. Solve. Two independent quantities, two names for them, and one equation that was always two.