Imaginary Numbers: Free Response

Divide each exponent by $4$ and keep the remainder, since the powers repeat every four steps.

$38 = 4 \times 9 + 2$ (remainder $2$):

$i^{38} = i^{2} = -1.$

$55 = 4 \times 13 + 3$ (remainder $3$):

$i^{55} = i^{3} = -i.$

$101 = 4 \times 25 + 1$ (remainder $1$):

$i^{101} = i^{1} = i.$

The $20$ terms fall into five blocks of four ($20 = 4 \times 5$).

The first block:

$i + i^{2} + i^{3} + i^{4} = i + (-1) + (-i) + 1 = 0.$

The $i$ and $-i$ cancel, and the $-1$ and $1$ cancel. Every later block runs through the same four values in the same order, so each is $0$ too:

$i + i^{2} + \cdots + i^{20} = 0 + 0 + 0 + 0 + 0 = 0.$

Every term shares a factor of $i^{n}$, so pull it out with $i^{n+k} = i^{n} \cdot i^{k}$:

$i^{n} + i^{n+1} + i^{n+2} + i^{n+3} = i^{n}\left(1 + i + i^{2} + i^{3}\right).$

The bracket no longer mentions $n$. Evaluate it with $i^{2} = -1$ and $i^{3} = -i$:

$1 + i + i^{2} + i^{3} = 1 + i - 1 - i = 0.$

So the sum is $i^{n} \times 0 = 0$ for every $n$, which is the claim.

Part B then needs no computation: its sum is five of these blocks (starting at $n = 1, 5, 9, 13, 17$), each $0$, so the total is $0$.

Pull the factor of $i$ out first, then simplify the real radical left behind.

For $\sqrt{-45}$, use $45 = 9 \times 5$ and $\sqrt{9} = 3$:

$\sqrt{-45} = i\sqrt{45} = i\sqrt{9 \times 5} = 3i\sqrt{5}.$

For $\sqrt{-98}$, use $98 = 49 \times 2$ and $\sqrt{49} = 7$:

$\sqrt{-98} = i\sqrt{98} = i\sqrt{49 \times 2} = 7i\sqrt{2}.$

The $5$ and $2$ have no square factor left, so both are fully simplified.

Convert each root before multiplying.

$\sqrt{-6} = i\sqrt{6}, \qquad \sqrt{-24} = i\sqrt{24} = 2i\sqrt{6}.$

Multiply, collecting the factors of $i$:

$\left(i\sqrt{6}\right)\left(2i\sqrt{6}\right) = 2i^{2}\left(\sqrt{6}\cdot\sqrt{6}\right) = 2(-1)(6) = -12.$

The sign is not something to remember: it is produced by $i^{2} = -1$, appearing as soon as the two factors of $i$ meet.

The product rule is not a law of nature: it is a theorem, proved for radicands that are not negative. So the question is what its proof leaned on, and whether that support survives.

Watch the two sides come apart on the pair from part B. Tracking the $i$ each root produces gives

$\sqrt{-6}\cdot\sqrt{-24} = \left(i\sqrt{6}\right)\left(2i\sqrt{6}\right) = 2i^{2} \cdot 6 = -12,$

while merging the radicands first destroys both minus signs before any $i$ is created:

$\sqrt{(-6)(-24)} = \sqrt{144} = 12.$

The two disagree, so the rule fails, and the reason is visible: two roots of negatives supply two factors of $i$, which multiply to $i^{2} = -1$. On the other side, $ab$ is positive (two negatives), and $\sqrt{ab}$ is its positive root by convention, so that side cannot produce the minus sign.

With only one negative radicand, exactly one factor of $i$ appears, with no second $i$ to pair with, so no sign is manufactured. Check it:

$\sqrt{-4}\cdot\sqrt{9} = (2i)(3) = 6i, \qquad \sqrt{(-4)(9)} = \sqrt{-36} = 6i.$

The two agree, which is why the part A split $\sqrt{-45} = \sqrt{-1}\cdot\sqrt{45} = i\sqrt{45}$ is legitimate and worth making a habit: it never depends on checking anything first. Question 5 settles the general case for every pair of REAL radicands.

Isolate the square before taking any root. Dividing through by $3$ changes no solutions:

$3x^{2} + 48 = 0 \quad\Rightarrow\quad x^{2} + 16 = 0 \quad\Rightarrow\quad x^{2} = -16.$

Square-root both sides, keeping BOTH signs, and convert the root of the negative:

$x = \pm\sqrt{-16} = \pm i\sqrt{16} = \pm 4i.$

Check: $(\pm 4i)^{2} = 16i^{2} = -16$, and $3(-16) + 48 = 0$, so both satisfy the original equation.

A sum of two squares will not factor while the coefficients must stay real. But $i$ turns a sum into a difference, because $i^{2} = -1$ lets you write

$16 = -16i^{2} = -(4i)^{2}.$

That turns the sum of squares into a difference of squares, factored by $A^{2} - B^{2} = (A - B)(A + B)$ with $A = x$, $B = 4i$:

$x^{2} + 16 = x^{2} - (4i)^{2} = (x - 4i)(x + 4i).$

Expanding it back is the check:

$(x - 4i)(x + 4i) = x^{2} + 4ix - 4ix - 16i^{2} = x^{2} - 16(-1) = x^{2} + 16.$

The middle terms cancel and $-16i^{2}$ turns into $+16$.

Neither old statement was wrong, and noticing that is the whole of this part. Each carried an unstated qualifier: over the real numbers.

The claim that $x^{2} + 16$ cannot be factored was shorthand for cannot be factored into linear factors with real coefficients. Still true: the coefficients of $(x - 4i)$ and $(x + 4i)$ are not real, so part B did not contradict it.

The claim that $3x^{2} + 48 = 0$ has no solution was shorthand for has no real solution, and the sign-case argument proves it. For real $x$, the square $x^{2}$ is never negative, so

$3x^{2} + 48 \ge 48 > 0,$

and the left-hand side never reaches $0$. That argument is still valid; it just says nothing about numbers that are not real.

So nothing was overturned. What changed is the collection of numbers the claims range over: enlarging a number system cannot make a true statement about the reals false, only make new statements available, and those are what parts A and B answered.

Go through the lines in order, testing each on its own.

Line 1 is the equation as given. Line 2, $x^{2} = -25$, follows by subtracting $25$ from both sides: sound.

Line 3 is where the work breaks. The step from $x^{2} = -25$ to $x = \sqrt{-25}$ keeps just one of the two numbers whose square is $-25$: by convention $\sqrt{\;}$ names a single value, so writing it with no $\pm$ silently discards the other root. The line should have read

$x = \pm\sqrt{-25} = \pm 5i.$

The student's $5i$ is a solution, just not the only one: a squared equation carries two answers.

Substitute each candidate into the original equation in turn.

For $x = 5i$, square the real factor and the $i$ separately:

$(5i)^{2} + 25 = 5^{2}i^{2} + 25 = 25(-1) + 25 = -25 + 25 = 0.$

For $x = -5i$, the minus sign vanishes under the square, which is exactly why a second root exists:

$(-5i)^{2} + 25 = (-5)^{2}i^{2} + 25 = 25(-1) + 25 = 0.$

Both satisfy the equation, so there are two solutions and the student's answer was short by one.

The equation is $x^{2} + 25 = 0$, i.e. $x^{2} + bx + c = 0$ with $b = 0$ (no $x$ term) and $c = 25$. Before looking at any roots, the coefficients predict

$\text{sum of the roots} = -b = 0, \qquad \text{product of the roots} = c = 25.$

Test the pair. The sum is

$5i + (-5i) = 0,$

which matches, and the product is

$(5i)(-5i) = -25i^{2} = -25(-1) = 25,$

which matches too. The rule survives the arrival of $i$ for a reason, not by luck: the two relations come from expanding a monic quadratic from its roots,

$(x - r)(x - s) = x^{2} - (r + s)x + rs,$

and matching against $x^{2} + bx + c$. Nowhere is either root required to be real, so nothing breaks when the roots are imaginary. (Question 3 part C warns against assuming the opposite without checking.)

Now turn the check on the student's work. They offered a single root, $5i$: it has no second number to add to, and $5i$ is neither $0$ nor $25$. The coefficients demand two roots that cancel, and one root has nothing to cancel against, so the check flags the missing root without rereading a single line of algebra. That is what makes it worth doing.

The claim is about EVERY real number, so the argument must be exhaustive, not illustrative. The sign of a real number gives exactly three cases, and no real number escapes them.

Let $a$ be any real number, and look at $a^{2} = a \times a$.

If $a > 0$, then $a \times a$ is a positive times a positive, so

$a^{2} > 0.$

If $a < 0$, then $a \times a$ is a negative times a negative, and two negatives make a positive, so again

$a^{2} > 0.$

If $a = 0$, then $a^{2} = 0$.

Every real number falls into exactly one of those cases, and in all three $a^{2} \ge 0$. So a real square is never negative: no real number satisfies $a^{2} = -1$, or $-25$, or any equation demanding a negative square. That gap is genuine and permanent, which is what justifies inventing a new number to fill it.

A claim about ALL real numbers is destroyed by a single breaking pair, so produce one and evaluate both sides honestly.

Take $a = -1$ and $b = -1$. The left-hand side multiplies the radicands first:

$\sqrt{ab} = \sqrt{(-1)(-1)} = \sqrt{1} = 1.$

The right-hand side takes the two roots first, and the square root of $-1$ is $i$ by definition:

$\sqrt{a}\cdot\sqrt{b} = \sqrt{-1}\cdot\sqrt{-1} = i \cdot i = i^{2} = -1.$

So on this pair the claim asserts $1 = -1$, which is false, so $a = b = -1$ refutes it.

Be careful what this shows. Not that the product rule is worthless, but that the rule WITH its quantifier over all reals is false. A restricted version survives; part C pins down the restriction.

Take the two cases in turn, with $p > 0$ and $q > 0$ throughout. Working in letters, not numbers, turns a verdict about one product into a statement about all of them.

One negative radicand. Converting first:

$\sqrt{-p}\cdot\sqrt{q} = \left(i\sqrt{p}\right)\sqrt{q} = i\sqrt{pq}.$

Merging first, where the product $(-p)q = -pq$ is itself negative:

$\sqrt{(-p)q} = \sqrt{-pq} = i\sqrt{pq}.$

The two routes land on the same number. This is the half of the rule that survives, now proved rather than asserted: it holds for every positive $p$ and $q$, not just the cases anyone has tried.

Two negative radicands. Converting first, the two factors of $i$ finally meet:

$\sqrt{-p}\cdot\sqrt{-q} = \left(i\sqrt{p}\right)\left(i\sqrt{q}\right) = i^{2}\sqrt{pq} = -\sqrt{pq}.$

Merging first, the two minus signs destroy each other before any $i$ is ever created:

$\sqrt{(-p)(-q)} = \sqrt{pq}.$

Since $pq$ is positive, $-\sqrt{pq}$ and $+\sqrt{pq}$ are never equal, so the routes disagree in EVERY case of this shape, not just the one from question 2.

Putting the halves together: on real radicands the merging route is legitimate exactly when they are not both negative, valid in every one-negative case and failing in every two-negative case. That is a boundary with a proof on each side, not a cautious rule of thumb.