Arithmetic with Complex Numbers: Free Response

Combine real parts with real parts and imaginary parts with imaginary parts. For the sum:

$(5 - 7i) + (-2 + 4i) = (5 - 2) + (-7 + 4)i = 3 - 3i.$

For the difference, distribute the minus sign across both parts of the second number first:

$(1 + 6i) - (8 - 3i) = (1 - 8) + \bigl(6 - (-3)\bigr)i = -7 + 9i.$

Multiply every term of the first bracket by every term of the second, as with two ordinary binomials, leaving $i^2$ standing for a moment:

$(3 - 4i)(2 + 5i) = 6 + 15i - 8i - 20i^2.$

Replace $i^2$ with $-1$, so $-20i^2$ becomes $+20$, and collect:

$6 + 15i - 8i - 20i^2 = 6 + 7i + 20 = 26 + 7i.$

Look at where a factor of $i$ can be created versus merely carried along.

In a sum, every term comes from adding, never multiplying. The imaginary terms share a factor of $i$, so the distributive law pulls it straight out:

$(a + bi) + (c + di) = (a + c) + (b + d)i.$

No two factors of $i$ meet, no $i^2$ is created, and the real part is built from real parts alone. That is why the two lanes stay separate.

In a product, every term is multiplied by every other, so the two imaginary terms DO meet:

$(a + bi)(c + di) = ac + adi + bci + bd\,i^2 = (ac - bd) + (ad + bc)i.$

The term $bd\,i^2$ is where the lanes cross: born of two imaginary quantities, $i^2 = -1$ converts it into the real number $-bd$, added to the real part. So the real part, $ac - bd$, depends on the imaginary parts, and the imaginary part, $ad + bc$, depends on the real ones: neither can be computed without the other.

The denominator's conjugate is $2 + 3i$; multiplying above and below by it changes the quotient's form, not its value, since $\dfrac{2 + 3i}{2 + 3i} = 1$. The denominator becomes real:

$(2 - 3i)(2 + 3i) = 2^2 + 3^2 = 13.$

Expand the numerator, replacing $i^2$ with $-1$ as it appears:

$(4 + 7i)(2 + 3i) = 8 + 12i + 14i + 21i^2 = 8 + 26i - 21 = -13 + 26i.$

Divide each part of the numerator by the real denominator:

$\frac{-13 + 26i}{13} = -1 + 2i.$

Check by multiplying back: $(-1 + 2i)(2 - 3i) = -2 + 3i + 4i - 6i^2 = 4 + 7i$, the original numerator.

In standard form the denominator is $0 + 5i$, so $a = 0$, $b = 5$, and its conjugate is $-5i$. Multiply above and below by it:

$\frac{3}{5i} = \frac{3 \cdot (-5i)}{(5i)(-5i)} = \frac{-15i}{-25i^2} = \frac{-15i}{25} = -\frac{3}{5}i.$

The denominator $(5i)(-5i) = -25i^2 = 25$ is real and positive, exactly as $a^2 + b^2 = 0^2 + 5^2 = 25$ predicts. The answer $0 - \frac{3}{5}i$ is pure imaginary, with real part $0$.

Run the radical version first. The conjugate of $2 - 3\sqrt{3}$ is $2 + 3\sqrt{3}$, and the denominator becomes a difference of squares:

$(2 - 3\sqrt{3})(2 + 3\sqrt{3}) = 2^2 - (3\sqrt{3})^2 = 4 - 9 \cdot 3 = -23.$

The numerator expands to $(4 + 7\sqrt{3})(2 + 3\sqrt{3}) = 8 + 12\sqrt{3} + 14\sqrt{3} + 21 \cdot 3 = 71 + 26\sqrt{3}$, so the quotient is $\dfrac{71 + 26\sqrt{3}}{-23}$, with no radical left downstairs: the point of the move.

Set the two denominators side by side:

$(2 - 3\sqrt{3})(2 + 3\sqrt{3}) = 4 - 27 = -23, \qquad (2 - 3i)(2 + 3i) = 4 + 9 = 13.$

Same. Both multiply above and below by the conjugate, a disguised $1$, so neither changes the value. In both, the cross terms are exact opposites and cancel, removing the offending quantity ($\sqrt{3}$ or $i$) from the denominator, which is left built only from squares.

Different, and why. Everything turns on the square of the offending quantity. Squaring $\sqrt{3}$ gives $+3$, so $(x + y)(x - y) = x^2 - y^2$ leaves a SUBTRACTED term, here the negative $-23$. Squaring $i$ gives $-1$, flipping that subtracted term into an added one, so the denominator is $2^2 + 3^2 = 13$, a sum of two squares: never negative, and zero only when the denominator itself was.

Work in letters, so the conclusion covers every complex number at once, not the handful anyone has tried. Expand the product term by term:

$(a + bi)(a - bi) = a^2 - abi + abi - b^2 i^2.$

The cross terms $-abi$ and $+abi$ are exact opposites and cancel, removing $i$ from the answer. What is left carries the only $i$ still in play, inside $i^2$, and $i^2 = -1$ converts it:

$a^2 - b^2 i^2 = a^2 - b^2(-1) = a^2 + b^2.$

Since $a$ and $b$ are real, so are $a^2$ and $b^2$, so $a^2 + b^2$ is real. That is the first claim, holding for every real $a$ and $b$ with no case-checking required.

For the second claim, a square of a real number is never negative, so $a^2 \ge 0$ and $b^2 \ge 0$, and their sum is never negative. It equals $0$ only if $a^2 = 0$ and $b^2 = 0$ together, that is, only if $a = 0$ and $b = 0$. So for any complex number other than $0$, the product with its conjugate is strictly positive.

To reach standard form you need a real denominator, and part A just manufactured one. Multiply above and below by the conjugate $a - bi$:

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

Splitting the fraction into its two parts puts it in standard form:

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

The condition is the one part A turned up: $a^2 + b^2$ is zero only when $a$ and $b$ are both zero, so the formula is valid for every complex number except $0$, precisely the number you were never allowed to divide by anyway.

Check it on $3 + i$, where $a = 3$, $b = 1$, so $a^2 + b^2 = 10$:

$\frac{1}{3 + i} = \frac{3}{10} - \frac{1}{10}i.$

Multiplying back confirms it: $\left(\frac{3}{10} - \frac{1}{10}i\right)(3 + i) = \frac{9}{10} + \frac{3}{10}i - \frac{3}{10}i - \frac{1}{10}i^2 = \frac{9}{10} + \frac{1}{10} = 1$.

A claim that something HAS TO be so is destroyed by one honest example where it is not. Produce a pair and evaluate the product.

Take $z = 1 + i$ and $w = 2 - 2i$. Neither has imaginary part $0$, so neither is real, as required. Their product is

$(1 + i)(2 - 2i) = 2 - 2i + 2i - 2i^2 = 2 + 2 = 4,$

which is real. But the conjugate of $z$ is $\overline{z} = 1 - i$, and $w = 2 - 2i$ is not that number. The hypothesis holds and the conclusion fails, so the claim is refuted.

Now say what survives, since refuting a claim is not the same as showing there is nothing there. The example is not a coincidence: $w = 2 - 2i = 2(1 - i) = 2\overline{z}$, a real multiple of the conjugate. That always works, since scaling by a real $k$ keeps a real product real:

$z \cdot k\overline{z} = k(a^2 + b^2),$

real by part A for every real $k$. The converse holds too: if $zw$ is real, say $zw = r$, dividing by $z$ (nonzero) and clearing with its conjugate, the move of part B, gives $w = \dfrac{r}{z} = \dfrac{r\overline{z}}{z\overline{z}} = \dfrac{r}{a^2 + b^2}\,\overline{z}$, a real multiple of $\overline{z}$. So the honest version of the claim is that $w$ must be a REAL MULTIPLE of $\overline{z}$, and the student's version is the special case $k = 1$: the conjugate is not the only partner that makes a product real, just the one that does it with no scaling.

Test the lines one at a time rather than judging by the final answer.

The first line is a plain expansion of two brackets, and all four products are correct. The trouble is in the second line, in one term only. The conversions $3\sqrt{-9} = 9i$ and $2\sqrt{-4} = 4i$ are right. But the last term merges two negative radicands under a single root:

$\sqrt{-4}\cdot\sqrt{-9} \ne \sqrt{(-4)(-9)} = \sqrt{36} = 6.$

The rule $\sqrt{p}\cdot\sqrt{q} = \sqrt{pq}$ does not apply when both radicands are negative. Convert each root FIRST, so the two factors of $i$ meet and produce the sign:

$\sqrt{-4}\cdot\sqrt{-9} = (2i)(3i) = 6i^2 = -6.$

So the term should have been $-(-6) = +6$, and the line should read $6 + 9i - 4i + 6$.

Convert both roots before multiplying: $\sqrt{-4} = 2i$ and $\sqrt{-9} = 3i$. The product becomes an ordinary complex multiplication:

$(3 - 2i)(2 + 3i) = 6 + 9i - 4i - 6i^2.$

Replace $i^2$ with $-1$, so $-6i^2$ becomes $+6$, and collect:

$6 + 9i - 4i + 6 = 12 + 5i.$

Worth seeing what the error cost: the student's imaginary part, $5i$, was right all along; the mistake sat entirely in the real part, $6 - 6 = 0$ instead of $6 + 6 = 12$.

The student applied $(x - y)(x + y) = x^2 - y^2$ to a product whose two brackets are identical. That pattern needs the signs to DIFFER: it is the difference in sign that makes the cross terms opposites and cancels them. In a square, both brackets are $4 - 5i$, both cross terms are $-20i$, and they add rather than cancel.

Square it with the binomial pattern instead:

$(4 - 5i)^2 = 4^2 - 2(4)(5i) + (5i)^2 = 16 - 40i + 25i^2.$

Replace $i^2$ with $-1$, so $25i^2 = -25$:

$16 - 40i - 25 = -9 - 40i.$

The student's $41$ is not a random number: it is the correct value of $(4 - 5i)(4 + 5i)$, the conjugate product. They computed a real answer to a question that did not have one.

Expand each and compare the middle terms, where the two differ.

The conjugate product has brackets whose signs disagree, so its cross terms are exact opposites:

$(a + bi)(a - bi) = a^2 - abi + abi - b^2i^2 = a^2 + b^2.$

They cancel, and every trace of $i$ leaves with them. What remains sits in $-b^2i^2$, which $i^2 = -1$ converts into the real $+b^2$. The result is real for every real $a$ and $b$, with nothing to check.

The square has two IDENTICAL brackets, so its cross terms are equal rather than opposite, and reinforce:

$(a - bi)^2 = a^2 - abi - abi + b^2i^2 = (a^2 - b^2) - 2abi.$

The imaginary part is $-2ab$, and does not go away on its own. So the square is real exactly when $2ab = 0$: only when $a = 0$ (pure imaginary) or $b = 0$ (real to begin with). For a number with both parts nonzero, such as $4 - 5i$, the square is never real, which is why the second student's $41$ could have been rejected before a single term was checked.

Substitution means computing each term of the left-hand side at $x = 3 + 2i$ and adding them. Square first, with the binomial pattern, keeping $i^2$ visible until replaced:

$(3 + 2i)^2 = 9 + 12i + 4i^2 = 9 + 12i - 4 = 5 + 12i.$

The linear term scales both parts:

$-6(3 + 2i) = -18 - 12i.$

Add the three terms, collecting real with real and imaginary with imaginary:

$(5 + 12i) + (-18 - 12i) + 13 = (5 - 18 + 13) + (12 - 12)i = 0 + 0i = 0.$

Both parts vanish, so the left-hand side is $0$ and $3 + 2i$ is a root.

Substitute the conjugate and watch every imaginary quantity arrive with its sign flipped. The square is

$(3 - 2i)^2 = 9 - 12i + 4i^2 = 5 - 12i,$

and the linear term is $-6(3 - 2i) = -18 + 12i$. Adding the three terms:

$(5 - 12i) + (-18 + 12i) + 13 = (5 - 18 + 13) + (-12 + 12)i = 0.$

So $3 - 2i$ is a root too, and the equation has two.

Now the independent check. For $x^2 + bx + c = 0$ the roots sum to $-b$ and multiply to $c$; here $b = -6$, $c = 13$, so the two roots must sum to $6$ and multiply to $13$. The pair delivers both. The sum is

$(3 + 2i) + (3 - 2i) = 6,$

since the imaginary parts are opposites, and the product is the conjugate product from question 3:

$(3 + 2i)(3 - 2i) = 3^2 + 2^2 = 13.$

Both match, strong evidence nothing has gone wrong, and it explains why the coefficients came out real: the two roots conspire to hide their imaginary parts, cancelling in the sum and converting in the product.

Suppose $p + qi$ is a root, and turn that hypothesis into information. Square first:

$(p + qi)^2 = p^2 + 2pqi + q^2i^2 = (p^2 - q^2) + 2pq\,i.$

Add the remaining terms, $b(p + qi) = bp + bq\,i$ and $c$, and collect into standard form, legitimate precisely because $b$ and $c$ are real and contribute nothing imaginary of their own:

$(p + qi)^2 + b(p + qi) + c = \bigl(p^2 - q^2 + bp + c\bigr) + \bigl(2pq + bq\bigr)i.$

The hypothesis says this equals $0$, that is, $0 + 0i$. Two complex numbers are equal only when both parts match, so the hypothesis is really TWO real statements at once:

$p^2 - q^2 + bp + c = 0 \qquad \text{and} \qquad 2pq + bq = 0.$

Now substitute the conjugate and do the same collection. Squaring $p - qi$ flips the sign of the middle term, and $b(p - qi) = bp - bq\,i$ flips the sign of its imaginary term:

$(p - qi)^2 + b(p - qi) + c = \bigl(p^2 - q^2 + bp + c\bigr) - \bigl(2pq + bq\bigr)i.$

This is the same pair of real quantities as before, with only the sign in front of the imaginary one reversed. Both brackets are $0$, by the two statements the hypothesis handed us, so the whole expression is $0 - 0i = 0$, and $p - qi$ is a root.