The Imaginary Unit and Complex Numbers: Free Response

Convert first, read second. Nothing can be classified or compared until each number is genuinely in the form $a + bi$ with $a$ and $b$ real.

A square root of a negative becomes $i$ times a real square root, and that real radical then simplifies as it always did. Since $32 = 16 \cdot 2$:

$\sqrt{-32} = i\sqrt{32} = 4\sqrt{2}\,i.$

The second number is already split into two pieces, so only its radical needs work, with $80 = 16 \cdot 5$:

$6 - i\sqrt{80} = 6 - 4\sqrt{5}\,i.$

The third is a power of a pure imaginary number. Separate the real factor from the power of $i$, and use $i^3 = -i$:

$(5i)^3 = 5^3 i^3 = 125(-i) = -125i.$

The fourth is settled by a remainder, since $34 = 4 \cdot 8 + 2$:

$i^{34} = i^2 = -1.$

So the four standard forms are $0 + 4\sqrt{2}\,i$, $6 + (-4\sqrt{5})i$, $0 + (-125)i$ and $-1 + 0i$. Read the slots: the real parts are $0$, $6$, $0$ and $-1$, and the imaginary parts are $4\sqrt{2}$, $-4\sqrt{5}$, $-125$ and $0$. Every one of those imaginary parts is a real number carrying its own sign. The third is $-125$, not $-125i$, and an imaginary part is allowed to be irrational, as the first two are.

The three boxes are decided entirely by which slot holds a zero, so work from the standard forms and not from the expressions they came from.

A complex number is REAL exactly when $b = 0$, PURE IMAGINARY exactly when $a = 0$ and $b \neq 0$, and neither when both parts are nonzero.

The first and third numbers both have a zero real part and a nonzero imaginary part:

$4\sqrt{2}\,i = 0 + 4\sqrt{2}\,i, \qquad -125i = 0 + (-125)i.$

Both are pure imaginary. The second has $a = 6$ and $b = -4\sqrt{5}$, and neither is zero, so it sits in neither of the first two boxes: it is a complex number that is not real and not pure imaginary. The fourth has a zero imaginary part:

$i^{34} = -1 + 0i,$

so it is a real number. Notice it did not BECOME real when you simplified it. It was the number $-1$ all along; the $i$ was a feature of how it had been written.

The claim confuses a number with a piece of notation, so the first move is to recall what the definition actually says. A complex number written in standard form $a + bi$, with $a$ and $b$ real, is a real number exactly when its imaginary part $b$ is $0$. That is a statement about a value, and it says nothing whatever about which symbols were used to write the number down.

So the claim is refuted by any number written with an $i$ whose imaginary part turns out to be zero, and such numbers are not rare. Take any real $c$ and any real $d$, and use the defining property $i^2 = -1$:

$c + d\,i^2 = c + d(-1) = (c - d) + 0i.$

The imaginary part is $0$, so every number of this shape is real, no matter how prominently the $i$ appears on the page. Part A supplies a second family: an even power of $i$ is $1$ or $-1$, both real, and

$i^{34} = -1$

is one of them.

What survives of the classmate's instinct is a correct statement in the other direction: if a number's standard form has $b \neq 0$, then it is not real, and no rewriting can make it real. The test is always the same, and it is applied to the standard form: a complex number is real exactly when its imaginary part is $0$.

Everything in this question rests on one small computation, so do it first. Squaring $i^2$ gives $i^4$, and the defining property supplies the value twice over:

$i^4 = \left(i^2\right)^2 = (-1)^2 = 1.$

That single equation is what makes large exponents cheap: a factor of $i^4$ can be inserted or removed at will, because it is a factor of $1$. So break each exponent into as many fours as it will hold, and see what is left over.

For $614$, taking out as many fours as it will hold leaves $2$ behind, since $614 = 4 \cdot 153 + 2$:

$i^{614} = \left(i^4\right)^{153} \cdot i^2 = 1^{153} \cdot i^2 = -1.$

For $83$, the same stripping leaves $3$, since $83 = 4 \cdot 20 + 3$:

$i^{83} = \left(i^4\right)^{20} \cdot i^3 = 1^{20} \cdot i^3 = -i.$

Notice that the quotients $153$ and $20$ never appeared in either answer. They were absorbed into powers of $1$, which is exactly why nothing but the leftover matters, and why both values land among $i$, $-1$, $-i$, $1$.

Both descriptions come from the same fact: $i^n$ depends only on the remainder $r$ of $n$ on division by $4$, and the four possible remainders give the four values

$i^0 = 1, \quad i^1 = i, \quad i^2 = -1, \quad i^3 = -i.$

Of those four numbers, $1$ and $-1$ are real, while $i$ and $-i$ are not (a real number cannot have a negative square, and both of these square to $-1$). So $i^n$ is real exactly when $r$ is $0$ or $2$, and those are precisely the even exponents:

$n = 4q \ \text{ or } \ n = 4q + 2 \iff n \ \text{is even}.$

For the second description, $i^n = 1$ needs $r = 0$, since the other three remainders give $i$, $-1$ and $-i$. Remainder $0$ says exactly that $4$ divides $n$.

Neither description leaves anything out, and the reason is the division algorithm rather than good luck: every positive integer leaves one of the four remainders $0$, $1$, $2$, $3$, and each remainder has been accounted for. Note what the first description does NOT say. Even does not mean $1$: an even exponent gives $1$ or $-1$ according to which of the two even remainders it leaves, and only the multiples of $4$ give $1$.

An "exactly when" claim is two claims, and they need separate arguments.

Equal remainders give equal powers. Write $m = 4q + r$ and $n = 4p + r$ with the same remainder $r$. Using $i^4 = 1$:

$i^m = \left(i^4\right)^q i^r = i^r, \qquad i^n = \left(i^4\right)^p i^r = i^r.$

Both reduce to the same number $i^r$, so $i^m = i^n$.

Different remainders give different powers. Suppose $m$ and $n$ leave different remainders $r$ and $s$. By the same reduction, $i^m = i^r$ and $i^n = i^s$, so the claim comes down to showing that the four numbers $1$, $i$, $-1$, $-i$ are pairwise different. This is the step that is usually skipped, and it is the only place the definition of $i$ is really needed.

First, $1 \neq -1$, since these are real numbers already known to be different. Next, $i \neq -i$: if they were equal, adding $i$ to both sides would give $2i = 0$, hence $i = 0$, and then $i^2 = 0$, contradicting $i^2 = -1$. Finally, neither $i$ nor $-i$ is real, because each squares to $-1$:

$i^2 = -1, \qquad (-i)^2 = i^2 = -1,$

and no real number has a negative square. So neither of them can equal the real numbers $1$ or $-1$. All four are distinct, so $i^r \neq i^s$ and therefore $i^m \neq i^n$.

Both directions hold, so the biconditional stands. It is worth seeing what the second direction buys you: without it, you would know that the cycle repeats, but not that it never repeats sooner. It is the distinctness of the four values that makes $4$ the true period rather than merely a period.

Both sides are in standard form, and $x$ and $y$ are real, so the four numbers involved are real and the matching rule applies. Equate the real parts and then the imaginary parts:

$2x + 3y = -1, \qquad 5x - y = 23.$

One complex equation has become an ordinary pair of linear equations in two unknowns. Solve it by substitution: the second gives $y = 5x - 23$, and putting that into the first eliminates $y$,

$2x + 3(5x - 23) = -1 \quad\Rightarrow\quad 17x = 68 \quad\Rightarrow\quad x = 4,$

and then $y = 5(4) - 23 = -3$.

Check the pair in the original complex equation rather than in the derived system, so that both parts are tested at once:

$\bigl(2(4) + 3(-3)\bigr) + \bigl(5(4) - (-3)\bigr)i = -1 + 23i.$

Both slots agree, so the values are right.

Each of the two words is a condition on one slot, so write the conditions down before solving anything. Since $s$ is real, the two brackets are the real part and the imaginary part.

The number is REAL exactly when its imaginary part is zero:

$s - 7 = 0 \quad\Rightarrow\quad s = 7.$

The number is PURE IMAGINARY when its real part is zero AND its imaginary part is not. Take the first condition first and factor:

$s^2 - 9s + 14 = (s - 2)(s - 7) = 0 \quad\Rightarrow\quad s = 2 \ \text{ or } \ s = 7.$

Now apply the second condition, which is the one that is easy to forget. At $s = 2$ the imaginary part is $2 - 7 = -5$, which is not zero, so that value does give a pure imaginary number, namely $-5i$. At $s = 7$ the imaginary part is $0$ as well, so both slots are empty and the number is

$0 + 0i = 0,$

which is a real number, not a pure imaginary one. So $s = 7$ answers the first question and is disqualified from the second, and $s = 2$ answers the second alone.

A hypothesis is doing work exactly when the conclusion fails without it, so the job is to find a failure.

Take $a = 1$ and $b = i$ on the left, and $c = 0$, $d = 0$ on the right. The left side collapses, because the defining property of $i$ turns the product of the two $i$ symbols into $-1$:

$a + bi = 1 + i \cdot i = 1 + i^2 = 1 - 1 = 0.$

The right side is $0 + 0i = 0$ as well, so the two sides are the same number. But the matching conclusion is false here: $a = 1$ is not $c = 0$, and $b = i$ is not $d = 0$. The rule has failed, and the only thing that has changed is that $b$ was allowed to be non-real.

Now look at where the lesson's proof would have blocked this. That proof supposes $d \neq b$, rearranges to $a - c = (d - b)i$, divides by $d - b$, and then argues that

$i = \frac{a - c}{d - b}$

is a quotient of two REAL numbers and therefore real, which contradicts the fact that no real number has a negative square. In the example above $d - b = -i$, which is not a real number, so the quotient is not a quotient of reals and no contradiction arises. The proof does not merely become harder; its key sentence stops being true.

The moral is not that the matching rule is unreliable. It is that the rule is a theorem with a hypothesis, and every time you split a complex equation into two real ones you are quietly checking that hypothesis: the four numbers in the two standard forms have to be real.

Get $x^2$ by itself before any root is taken. Every term here carries a factor of $5$, so scaling the equation down by it costs nothing:

$5x^2 + 140 = 0 \quad\Rightarrow\quad x^2 + 28 = 0 \quad\Rightarrow\quad x^2 = -28.$

No real number squares to a negative, so there is nothing to find among the reals. Among the complex numbers there is: take the square root of both sides, keep both signs, and convert the root of the negative into $i$-form before simplifying it. With $28 = 4 \cdot 7$,

$x = \pm\sqrt{-28} = \pm i\sqrt{28} = \pm 2i\sqrt{7}.$

Check one of them by squaring, factor by factor, and let $i^2$ collapse:

$\left(2i\sqrt{7}\right)^2 = 4 i^2 \cdot 7 = -28,$

and then $5(-28) + 140 = 0$. The other solution squares to the same value, since the minus sign disappears under the square.

Report the answer with its number system attached: the solutions are $x = \pm 2i\sqrt{7}$ among the complex numbers. "No solution" is not the answer; "no real solution" was, and still is.

Put the quadratic into the form $ax^2 + bx + c$ first. There is no $x$ term, so the middle coefficient is not missing, it is zero: $a = 5$, $b = 0$, $c = 140$. Then

$\Delta = b^2 - 4ac = 0^2 - 4(5)(140) = -2800.$

The sign is negative, and the previous chapter's theorem applies: a negative discriminant means the quadratic has no real roots. Read that verdict exactly as it was proved. It is a statement about the real numbers, delivered by an argument that used the ordering and sign rules of the reals throughout, and it says nothing at all about numbers outside them.

So the discriminant settles one thing and leaves another open. It settles that no real number satisfies the equation, which is why the parabola never meets the axis. It leaves open what happens in a larger number system, and that question was not even askable until $i$ was admitted. Be careful about what is being said here. The discriminant is computed from the equation's own coefficients, so the NUMBER $-2800$ is certainly a fact about the equation, and it reports faithfully what it was proved to report. What was never a fact about the equation is the VERDICT that used to be attached to it, that there was nothing to find: that was a fact about which numbers were allowed to compete for the job of being a root.

The general machinery for a quadratic whose discriminant is negative and whose middle coefficient is NOT zero belongs to the last lesson of this chapter. Here the equation was simple enough to solve directly.

Both observations are true, so the resolution has to be that they answer different questions.

Start with the graph. A point of the $xy$-plane pairs a real input with a real output, and an $x$-intercept is a point $(r, 0)$ with $r$ REAL and $5r^2 + 140 = 0$. No such $r$ exists, and the reason is the first proof of this lesson: a real square is never negative, so

$5r^2 + 140 \ge 140 > 0$

for every real $r$. The parabola misses the axis, exactly as the discriminant said it would.

Now the solutions. They are not real numbers, so they are not candidates for a position on the horizontal axis: asking where $2i\sqrt{7}$ sits between two real numbers is asking the wrong picture for an answer. The graph is a complete record of what this quadratic does to real inputs, and a complete record of nothing else.

So the picture and the algebra never disagreed. The graph reports that there is no REAL root, which is the same news the discriminant delivered, and neither of them was ever in a position to comment on numbers off the real line. Complex numbers do get a picture of their own, one in which these two solutions have definite positions, and building it is the work of a later lesson in this chapter.

Test the steps in order, because only one of them is wrong and the rest are impeccable.

The move from $\sqrt{-75} \cdot \sqrt{-3}$ to $\sqrt{(-75)(-3)}$ is the first step, and it is where the work breaks. It applies the product rule $\sqrt{a}\,\sqrt{b} = \sqrt{ab}$, whose proof assumed throughout that each radicand was zero or positive, in a case where both of them are below zero. Everything after that is correct arithmetic on an expression that should never have been written: $(-75)(-3)$ really is $225$, and $\sqrt{225}$ really is $15$. A correct calculation performed on an unjustified line is still an unjustified answer.

The safe procedure is to convert every square root of a negative into $i$-form before multiplying anything. With $75 = 25 \cdot 3$:

$\sqrt{-75} = i\sqrt{75} = 5i\sqrt{3}, \qquad \sqrt{-3} = i\sqrt{3}.$

Now multiply, collecting the two factors of $i$ and letting $i^2$ collapse:

$\left(5i\sqrt{3}\right)\left(i\sqrt{3}\right) = 5i^2 \cdot 3 = -15.$

The correct value is $-15$. The minus sign is not a convention to memorize: it is manufactured by $i^2 = -1$ the moment the two factors of $i$ meet, and the merged radical $\sqrt{225}$ has no way of producing it: multiplying the two radicands cancels their minus signs against each other, so nothing on that route ever manufactures an $i$ at all.

The hypothesis, then, is the whole lesson: the product rule was established only for radicands that are zero or positive, and both of the student's fall outside that.

By the first rule, $i$ must be positive, negative, or zero, and it must be exactly one of the three. Rule out each in turn, and the claim collapses.

Suppose $i > 0$. Then $i$ is itself a positive number, so multiplying both sides of $i > 0$ by $i$ preserves the inequality:

$i \cdot i > 0 \cdot i \quad\Rightarrow\quad i^2 > 0 \quad\Rightarrow\quad -1 > 0.$

But the comparison was assumed to agree with the usual order on the reals, where $-1$ is less than $0$. Contradiction.

Suppose $i < 0$. Add $-i$ to both sides, which the second rule permits, to get $0 < -i$. So $-i$ is positive, and multiplying both sides of $-i > 0$ by the positive number $-i$ preserves the inequality:

$(-i)(-i) > 0 \quad\Rightarrow\quad i^2 > 0 \quad\Rightarrow\quad -1 > 0.$

The same contradiction, reached by the same route: whichever sign $i$ is given, squaring destroys it, and the square is $-1$.

Suppose $i = 0$. Then $i^2 = 0$, while the definition of $i$ says $i^2 = -1$, and $0 \neq -1$. Contradiction.

Every placement the first rule allows has been ruled out, so no such comparison can exist. Notice what has been proved: not that nobody has yet found an ordering of the complex numbers, but that no ordering compatible with these rules is possible at all. This is why a question like "which of two complex numbers is bigger" has no answer, and it is a permanent feature of the new number system rather than a gap waiting to be filled.

First solve. Isolating the square gives $x^2 = -30$, and $30$ has no perfect-square factor, so

$x = \pm\sqrt{-30} = \pm i\sqrt{30}.$

The two solutions are $i\sqrt{30}$ and $-i\sqrt{30}$, and neither is real.

Now the phrase. "The larger solution" presupposes that the two can be ranked, and part B showed that no comparison with $<$ obeying those three rules can exist on the complex numbers. So the phrase does not fail because the two happen to be difficult to compare, or because the answer is not yet known. It names nothing: there is no such relation available to make it a question. With real roots the phrase always worked, and it worked because the real numbers are ordered, which is the property that did not survive the enlargement.

Equality is a different matter entirely, and it did survive. Two complex numbers in standard form are equal exactly when their real parts agree and their imaginary parts agree, and that test needs no ordering, only the ability to tell whether two real numbers are the same. Apply it here. Both solutions have real part $0$, and their imaginary parts are

$\sqrt{30} \quad \text{and} \quad -\sqrt{30},$

which are different real numbers, since $\sqrt{30} > 0$. So the two solutions are unequal, and that is a complete and honest answer to a question about them.

That is the shape of the bargain. Of the two ways you were used to relating a pair of numbers, one comes through the enlargement intact and one does not, and the difference is not a matter of difficulty: equality is decided part by part, while size needs an ordering compatible with the arithmetic, and part B showed that no such ordering can exist.