Arithmetic of Complex Numbers: Free Response

Take the four operations in order, and finish each one before starting the next.

Addition combines like terms, the real parts with the real parts and the imaginary parts with the imaginary parts:

$z + w = (14 + 4) + (-5 + 1)\,i = 18 - 4i.$

Subtraction is the same computation once the minus sign has been distributed through BOTH terms of $w$, which is where this operation is usually lost:

$z - w = 14 - 5i - 4 - i = (14 - 4) + (-5 - 1)\,i = 10 - 6i.$

Multiplication expands term by term, exactly as for two binomials, and only then is the new rule spent on the $i^2$:

$zw = 56 + 14i - 20i - 5i^2 = 56 - 6i + 5 = 61 - 6i.$

Division multiplies above and below by the conjugate of the denominator, $4 - i$. That is multiplying by $1$, so it changes the costume of the fraction and not its value:

$\frac{14 - 5i}{4 + i} = \frac{(14 - 5i)(4 - i)}{(4 + i)(4 - i)} = \frac{56 - 14i - 20i + 5i^2}{4^2 + 1^2}.$

The denominator is a conjugate product, so it is the real number $17$, and the numerator collects to $51 - 34i$. Dividing each part by $17$ finishes the job:

$\frac{51 - 34i}{17} = \frac{51}{17} + \frac{-34}{17}\,i = 3 - 2i.$

A quotient is worth checking, because the check costs one multiplication:

$(3 - 2i)(4 + i) = 12 + 3i - 8i - 2i^2 = 14 - 5i.$

Run both expansions the same way, distributing every term of the first factor across every term of the second, and do not simplify either one early.

With an ordinary letter:

$(14 - 5x)(4 + x) = 56 + 14x - 20x - 5x^2 = 56 - 6x - 5x^2.$

With the imaginary unit, the four partial products are formed by the identical distribution:

$(14 - 5i)(4 + i) = 56 + 14i - 20i - 5i^2.$

Compare them term by term. The constant term $56$ matches, and the two middle terms $14x$ and $-20x$ match $14i$ and $-20i$, collecting to $-6x$ and $-6i$. Nothing new has happened yet, which is the point: the expansion itself knows nothing about $i$.

The fourth term is where they separate. In the polynomial, $-5x^2$ is a term of a kind the other two are not, so it stays where it is and the expression is finished. In the complex product, $i^2$ is not a new kind of thing at all, because the definition says what it equals:

$-5i^2 = -5(-1) = +5.$

That term therefore abandons the imaginary column and joins the real part, which is why the real part of the answer is $56 + 5 = 61$ rather than $56$:

$56 - 6i + 5 = 61 - 6i.$

So the whole difference between multiplying complex numbers and multiplying binomials is one term and one substitution. Everything else was distribution, and the polynomial version had to stop at three terms while the complex version collapsed to two.

Sort by what each operation would break without.

Addition and subtraction. Regrouping $(14 - 5i) + (4 + i)$ into real terms and imaginary terms uses commutativity and associativity, and collecting $-5i + i$ into $(-5 + 1)i$ uses distributivity. That is the whole computation. The value of $i^2$ never comes up, because no two $i$ symbols are ever multiplied together.

Multiplication. The expansion is again distributivity, four times over. Then one term contains $i^2$, and the definition is spent on it. So multiplication needs the three laws plus the one new fact, and nothing else.

Division. This one cannot even start. Writing $\frac{z}{w}$ is only meaningful if $w$ has a reciprocal, and the recipe for producing one is to multiply above and below by $\overline{w}$, which is worth doing only because of a theorem proved in the lesson, for $w = c + di$ with $c$ and $d$ real:

$w\,\overline{w} = (c + di)(c - di) = c^2 + d^2.$

That guarantees two things at once. The new denominator is REAL, so the last step is dividing each part of a complex number by a real number, an operation already understood. And it is NONZERO whenever $w \neq 0$, since a sum of two real squares is zero only when both are, so the recipe never divides by $0$ except when $w$ itself is $0$.

Division does not escape the first two demands, it adds to them. The conjugate product above is itself a multiplication in which $-d^2i^2$ becomes $+d^2$, and the numerator $(14 - 5i)(4 - i)$ is another ordinary complex multiplication, so the rule $i^2 = -1$ is spent twice more before the answer appears. The right way to read the sort is therefore cumulative rather than exclusive: addition and subtraction need the laws; multiplication needs the laws and the rule; division needs the laws, the rule, and a theorem before it can begin at all.

The shape they share. Every one of $18 - 4i$, $10 - 6i$, $61 - 6i$ and $3 - 2i$ is a real number plus a real multiple of $i$. That was forced. Addition and subtraction produce sums and differences of real numbers in the two slots; multiplication produces $(ac - bd) + (ad + bc)i$, whose slots are built from real numbers by real arithmetic; and division ends with two real quotients over the same real denominator. Being closed under the four operations is what makes the complex numbers a place to do mathematics rather than a notation that occasionally escapes itself.

Series impedances add, which is a collection of like terms and nothing more:

$Z = (5 + 3i) + (1 - 5i) = (5 + 1) + (3 - 5)\,i = 6 - 2i \ \text{ohms}.$

Notice what the two parts did separately. The resistances added to $6$, and the reactances, one positive and one negative, partly cancelled to $-2$. Recording the element as one complex number is what lets a single addition carry both facts.

Now apply $V = IZ$, which is an ordinary complex multiplication:

$V = (2 + 4i)(6 - 2i) = 12 - 4i + 24i - 8i^2.$

The last term is the only place the definition of $i$ is needed, and it turns $-8i^2$ into $+8$:

$V = (12 + 8) + (-4 + 24)\,i = 20 + 20i \ \text{volts}.$

Both answers carry units, and both are complex numbers rather than a pair of unrelated readings.

The rule $V = IZ$ is solved for $I$ by dividing, which is legal because the impedance is not $0$:

$I = \frac{V}{Z} = \frac{26}{5 + i}.$

Clear the complex denominator by multiplying above and below by its conjugate $5 - i$. The new denominator is a conjugate product, so it is the real number $5^2 + 1^2 = 26$:

$I = \frac{26(5 - i)}{(5 + i)(5 - i)} = \frac{26(5 - i)}{26} = 5 - i \ \text{amperes}.$

Check it against the model, since one multiplication settles the matter:

$(5 - i)(5 + i) = 25 + 5i - 5i - i^2 = 26. \checkmark$

Read the answer before leaving it. The voltage was a real number, and the current is not: it has an imaginary part of $-1$ ampere. So a real reading on one quantity does not make the other real, which is the point part D takes up in general.

The parallel rule asks for two reciprocals, so build each one with the conjugate trick. For the first element the conjugate product is $2^2 + 4^2 = 20$:

$\frac{1}{2 + 4i} = \frac{2 - 4i}{20}, \qquad \frac{1}{2 - 4i} = \frac{2 + 4i}{20}.$

The two reciprocals are themselves conjugates, so adding them cancels the imaginary parts and leaves a real number:

$\frac{1}{Z} = \frac{2 - 4i}{20} + \frac{2 + 4i}{20} = \frac{4}{20} = \frac{1}{5},$

and therefore $Z = 5$ ohms. The combination behaves as a plain resistance, even though neither element does.

Now the same computation in letters. With $a$ and $b$ real and $a + bi \neq 0$, the conjugate product is $a^2 + b^2$, so

$\frac{1}{a + bi} + \frac{1}{a - bi} = \frac{a - bi}{a^2 + b^2} + \frac{a + bi}{a^2 + b^2} = \frac{2a}{a^2 + b^2}.$

The two imaginary parts are opposites and cancel exactly, which is the same cancellation the numbers showed. Inverting, and using $a \neq 0$,

$Z = \frac{a^2 + b^2}{2a}.$

Both $a^2 + b^2$ and $2a$ are real, so the quotient is real for every conjugate pair with $a \neq 0$, with no reactance at all. The restriction is not decoration: if $a$ were $0$ the two reciprocals would be $-\frac{i}{b}$ and $\frac{i}{b}$, which cancel to $0$, and $\frac{1}{Z} = 0$ is satisfied by no impedance at all. It is positive exactly when $a$ is positive, since $a^2 + b^2$ is positive for any $a + bi$ other than $0$. Checking the numbers against the letters, $a = 2$ and $b = 4$ give $\frac{4 + 16}{4} = 5$, as computed.

Argue from the multiplication itself rather than from examples, since the claim is about every such element.

Let the current be a real number $I \neq 0$ and let the impedance be $Z = R + Xi$ with $R$ and $X$ real and $X \neq 0$. Multiplying a complex number by a real number multiplies both parts:

$V = I(R + Xi) = IR + IX\,i.$

Both $IR$ and $IX$ are real, so this is already in standard form and its imaginary part can be read off: it is $IX$. A product of two nonzero real numbers is nonzero, so $IX \neq 0$, and a complex number with nonzero imaginary part is not real. The voltage therefore cannot be a real number, whatever the resistance happens to be.

Notice which fact did the work. It was not anything about $i$; it was that the real numbers have no zero divisors, so two nonzero real factors cannot multiply to $0$. The imaginary part had no way of vanishing.

Now set $X = 0$. The impedance is the real number $R$, and the rule collapses to

$V = IR,$

which is the rule for a purely resistive element, in which voltage and current are both real and simply proportional. So the complex model does not overturn the older one: it contains it as the case $X = 0$. That is the pattern worth carrying away from the whole lesson. Enlarging the number system left every law of arithmetic exactly where it was, and only added cases that the old system had no way to describe.

The obstacle is that division is not one of the operations the rule was proved for, so the proof has to turn the quotient into a product first.

Give the quotient a name. Let

$q = \frac{z}{w},$

which exists because $w \neq 0$. By what division MEANS, $q$ is the number satisfying

$qw = z.$

That is an equation between two complex numbers, and equal numbers have equal conjugates, so conjugate both sides. The left side is a product, and the product rule proved in the lesson applies to it:

$\overline{q}\,\overline{w} = \overline{z}.$

One step remains, and it is the one that needs justifying: divide both sides by $\overline{w}$. That is only allowed if $\overline{w} \neq 0$. Suppose $\overline{w}$ were $0$. Then the product $w\,\overline{w}$ would be $0$ as well, but the lesson computed that product for $w = c + di$:

$w\,\overline{w} = c^2 + d^2,$

which is $0$ only when $w = 0$, contrary to hypothesis. So $\overline{w} \neq 0$, the division is legal, and

$\overline{q} = \frac{\overline{z}}{\overline{w}}.$

Since $\overline{q}$ is exactly the left side of the claim, the proof is finished. It is worth seeing why this counts as a proof about DIVISION while dividing nothing until the last step: division was defined in terms of multiplication, so a rule about products reaches quotients automatically, once the quotient is renamed as the solution of a multiplication.

Run the two routes independently and compare only at the end. A shortcut that borrows a line from one route destroys the evidence the other was meant to supply.

Divide first. Multiply above and below by $1 + 2i$, the conjugate of the denominator, whose conjugate product is $1^2 + 2^2 = 5$:

$\frac{3 + 5i}{1 - 2i} = \frac{(3 + 5i)(1 + 2i)}{5} = \frac{3 + 6i + 5i + 10i^2}{5}.$

Spending $10i^2 = -10$ and collecting gives

$\frac{-7 + 11i}{5} = -\frac{7}{5} + \frac{11}{5}\,i,$

and conjugating that flips the sign of its imaginary part alone:

$-\frac{7}{5} - \frac{11}{5}\,i.$

Conjugate first. The two conjugates are $3 - 5i$ and $1 + 2i$, so the quotient to compute is $\frac{3 - 5i}{1 + 2i}$. Its denominator is cleared by $1 - 2i$, and the conjugate product is again $5$:

$\frac{3 - 5i}{1 + 2i} = \frac{(3 - 5i)(1 - 2i)}{5} = \frac{3 - 6i - 5i + 10i^2}{5}.$

Again $10i^2 = -10$, so this collects to

$\frac{-7 - 11i}{5} = -\frac{7}{5} - \frac{11}{5}\,i.$

The two routes agree, as part A said they must. Look at where the agreement came from: the two middle terms carry opposite signs in the two numerators, and everything else is untouched, so the two numerators are conjugates of each other while the denominator, being real, is its own conjugate.

Turn the word "real" into an equation, which is what the test is for.

Write $u = \frac{z}{\overline{z}}$, legal since $z \neq 0$. The test says $u$ is real exactly when $\overline{u} = u$. Conjugate $u$ with the quotient rule from part A, remembering that conjugating twice returns the original number:

$\overline{u} = \frac{\overline{z}}{\overline{\overline{z}}} = \frac{\overline{z}}{z}.$

So the condition to solve is

$\frac{z}{\overline{z}} = \frac{\overline{z}}{z}.$

Both denominators are nonzero, so multiplying across is legal and the condition becomes $z^2 = \overline{z}^2$, that is, $z^2 - \overline{z}^2 = 0$. Now write $z = a + bi$ with $a$ and $b$ real and compute the two squares:

$z^2 = a^2 - b^2 + 2ab\,i, \qquad \overline{z}^2 = a^2 - b^2 - 2ab\,i.$

Subtracting, the real parts cancel and the imaginary parts add:

$z^2 - \overline{z}^2 = 4ab\,i.$

A complex number is $0$ exactly when both of its parts are $0$, so the condition is $4ab = 0$, that is, $ab = 0$. Since $a$ and $b$ are real, this says $a = 0$ or $b = 0$: the number is pure imaginary or real. (It cannot be both, since $z \neq 0$.)

Check both cases, which also settles the converse direction and supplies the values. If $b = 0$ then $z$ is a nonzero real and $\overline{z} = z$, so

$\frac{z}{\overline{z}} = 1.$

If $a = 0$ then $z = bi$ with $b \neq 0$, so $\overline{z} = -bi$ and

$\frac{z}{\overline{z}} = \frac{bi}{-bi} = -1.$

Both values are real, so every number allowed by the condition really does work, and the description is exact in both directions. The quotient is never any real number other than $1$ or $-1$, which is more than was asked and worth noticing: when this quotient is real, conjugation has either left $z$ alone or flipped its sign, with nothing in between. For every other $z$ the quotient is not real at all, as the retry problem shows.

The claim is that the complex numbers have no way for two nonzero factors to multiply to zero. Everything needed is in the division section of the lesson.

Write $z = a + bi$ with $a$ and $b$ real. Since $z \neq 0$, the two parts are not both zero, so $a^2 + b^2 > 0$ and the reciprocal exists:

$\frac{1}{z} = \frac{a - bi}{a^2 + b^2}.$

That formula is available only because $a^2 + b^2$ is a nonzero REAL number, which is the conjugate-product theorem doing its work again.

Now multiply both sides of the hypothesis $zw = 0$ by this number. Multiplying equal numbers by the same number keeps them equal, and anything times $0$ is $0$, so

$\frac{1}{z}(zw) = \frac{1}{z} \cdot 0 = 0.$

Regroup the left side. Multiplication of complex numbers is associative, exactly as for real numbers, since it inherits that law along with the others:

$\frac{1}{z}(zw) = \left(\frac{1}{z} \cdot z\right)w = 1 \cdot w = w.$

Putting the two lines together gives $w = 0$, which is the claim.

Two remarks are worth making. First, the hypothesis $z \neq 0$ was used exactly once, to know that the reciprocal exists at all, and the proof collapses without it. Second, the argument used no property of $i$ beyond what makes the reciprocal formula legal, so the same proof would run in any system where every nonzero element has a reciprocal.

Part A licenses the usual reading: a product of complex numbers is $0$ only when one of the factors is $0$. So the equation splits into two cases, and every solution of the original is a solution of one of them.

Case 1: the second factor is zero. Then $z + 7 = 0$, so $z = -7$.

Case 2: the first factor is zero. Then

$(4 - i)z = 11 + 10i,$

and since $4 - i \neq 0$ it may be divided out. Multiply above and below by the conjugate $4 + i$, whose conjugate product with $4 - i$ is $4^2 + 1^2 = 17$:

$z = \frac{11 + 10i}{4 - i} = \frac{(11 + 10i)(4 + i)}{17} = \frac{44 + 11i + 40i + 10i^2}{17}.$

Spending $10i^2 = -10$ leaves $\frac{34 + 51i}{17}$, and dividing each part by $17$ gives

$z = 2 + 3i.$

Check it in the factor it came from, since one multiplication settles it:

$(4 - i)(2 + 3i) = 8 + 12i - 2i - 3i^2 = 11 + 10i,$

so the first factor really is $11 + 10i - 11 - 10i = 0$.

The two solutions are therefore $z = 2 + 3i$, from the factor that needed a division, and $z = -7$, from the factor that did not. Nothing in the method changed when the numbers stopped being real; only the arithmetic inside case 2 was new.

Test the method before arguing about it, because a single substitution settles whether it can be trusted.

The second case gives $z + 7 = 5$, so $z = -2$. Put that into the original left side. The first factor is

$(4 - i)(-2) - 11 - 10i = -8 + 2i - 11 - 10i = -19 - 8i,$

and the second factor is $-2 + 7 = 5$. Their product is

$(-19 - 8i)(5) = -95 - 40i,$

which is not $5$. So the method produced a number that does not satisfy the equation, and the reasoning cannot be sound.

Now the reason. Part A established a property of ZERO and of nothing else: if a product is $0$, some factor is $0$. The step the student copied depends on that property completely. For the number $5$ the corresponding claim is simply false, because $5$ has endless factorizations over the complex numbers, among them

$5 = 1 \cdot 5 = 2 \cdot \tfrac{5}{2} = (1 + 2i)(1 - 2i),$

and a product equals $5$ whenever the two factors form ANY of those pairs. Requiring one factor to be $5$ picks out a few of the possibilities and rejects the rest, so it neither finds all the solutions nor guarantees that what it finds is a solution at all.

What should be done instead is to expand the left side, move the $5$ across so that one side is $0$, and only then look for a factorization. That restores the hypothesis part A needs, which is the general principle: the factored reading is available exactly when the other side is $0$, whatever the factors happen to be.