The Complex Plane and Modulus: Free Response

Read every point in the same order: the real part is the horizontal coordinate and the imaginary part is the vertical one.

For $p = -5 + 2i$ the two parts are $-5$ and $2$, so $P$ is the pair $(-5, 2)$, five units left and two up, which is Quadrant II. For $q = 2 + 6i$ the parts are $2$ and $6$, so $Q$ is $(2, 6)$, right and up, in Quadrant I.

The third number has no visible $i$ at all, and that is the case worth writing out in full:

$r = 4 + 0i \quad \Longrightarrow \quad R = (4, 0).$

A second coordinate of $0$ puts the point neither above nor below the horizontal axis but exactly on it, so $R$ lies on the real axis and belongs to no quadrant.

That is also what marks $r$ as the real number of the three. The real axis is defined to be the set of complex numbers whose imaginary part is $0$, which is the old number line sitting unchanged inside the plane, so a point lands on it exactly when the number is real. The test is the position, not the notation: $p$ and $q$ sit off the axis because their imaginary parts $2$ and $6$ are not zero.

A slide moves every point by the same amount, so the amount is settled by what it does to one point. If adding $w$ carries $p$ to $q$, then $p + w = q$, and $w$ is the difference:

$w = q - p = (2 - (-5)) + (6 - 2)\,i = 7 + 4i.$

Read that as an instruction rather than a number: every point of the plane moves seven units right and four units up.

Apply the same instruction unchanged to the other two points. For $R$,

$r + w = 4 + (7 + 4i) = 11 + 4i,$

and for the origin, $0 + w = 7 + 4i$. That last one is worth keeping. The image of $0$ under a slide is the slide's own number, which is why an arrow drawn from $0$ to $w$ is a picture of the entire motion, and why the same $w$ can be described either as a number or as a pair of instructions.

Conjugation replaces $a + bi$ by $a - bi$, so in coordinates it sends $(a, b)$ to $(a, -b)$. The horizontal coordinate is untouched and the vertical one changes sign, which is precisely the reflection across the real axis: every point trades places with the point directly across the horizontal line from it, at the same distance on the other side.

Applying it to the three numbers:

$\overline{p} = -5 - 2i, \qquad \overline{q} = 2 - 6i, \qquad \overline{r} = 4.$

So $\overline{p}$ sits at $(-5, -2)$ in Quadrant III, directly below $P$, and $\overline{q}$ sits at $(2, -6)$ in Quadrant IV, directly below $Q$. The third point does not move at all, because its imaginary part was already $0$ and changing the sign of $0$ changes nothing.

That fixed point is what settles the second question. Suppose a slide added some $w \ne 0$. Then it would send $r$ to $r + w$, and $r + w = r$ would force $w = 0$, contrary to the assumption. So a nonzero slide moves EVERY point of the plane, including $R$, while conjugation holds $R$ still. The two moves therefore disagree on at least one point, and they cannot be the same move.

The argument does not depend on the particular number $4$. Conjugation fixes every real number and only the real numbers, since $a + bi = a - bi$ forces $2bi = 0$ and hence $b = 0$, so a whole line of points stays put. No slide but the trivial one leaves so much as a single point behind.

The formula wants both parts, so write each number as $a + bi$ before touching it, even when one part is $0$.

For the first, $a = 12$ and $b = 35$:

$|12 + 35i| = \sqrt{144 + 1225} = \sqrt{1369} = 37.$

For the second, $-8i = 0 + (-8)i$, so $a = 0$ and $b = -8$:

$|-8i| = \sqrt{0 + 64} = 8.$

The point sits eight units down the imaginary axis, and eight is exactly how far that is from $0$. The minus sign disappears under the squaring, as it must: a distance cannot be negative.

For the third, $-11 = -11 + 0i$, so $a = -11$ and $b = 0$:

$|-11| = \sqrt{121} = 11.$

This is the case that earns the notation. On the real axis the new formula returns the old absolute value, so the bars have not been recycled for a different idea; the same idea, distance from $0$, has been extended from a line to a plane.

The gap between two numbers is the length of the arrow joining them, and that arrow is their difference. Subtract first, measure second:

$(6 + 27i) - (-14 + 6i) = 20 + 21i,$

and then

$|20 + 21i| = \sqrt{400 + 441} = \sqrt{841} = 29.$

Subtracting in the other order gives $-20 - 21i$, whose modulus is the same $29$, so the direction of travel does not matter.

For the comparison, both moduli are square roots that do not come out whole, and there is no need to evaluate either. Squaring is what the formula does last, so stop one step earlier:

$|6 + 27i|^2 = 36 + 729 = 765, \qquad |-14 + 6i|^2 = 196 + 36 = 232.$

Both moduli are nonnegative real numbers, and for nonnegative numbers the larger square belongs to the larger number. Since $765 > 232$, the number $6 + 27i$ is the farther of the two from $0$. Its exact modulus is $3\sqrt{85}$ against $2\sqrt{58}$, but neither value was needed to rank them.

The symbol between the two numbers is the problem. The first lesson of this chapter settled that no ordering of the complex numbers can survive alongside their arithmetic, so $6 + 27i > -14 + 6i$ is not a false statement waiting to be corrected; it is not a statement at all.

What the modulus does is convert each complex number into a real number, and real numbers can always be ranked. So the honest sentence is about the two moduli, not the two numbers, and it says one thing only: the point for $6 + 27i$ lies farther from $0$ than the point for $-14 + 6i$. It reports nothing about which is larger, because there is no such relation to report on.

Now for what the conversion costs. Take

$|3 + 5i| = \sqrt{34} = |5 - 3i|.$

These are different numbers: their real parts $3$ and $5$ disagree, so the equality test fails at the first slot. Yet the modulus cannot tell them apart, and neither can any comparison built from moduli.

The geometry explains why. A modulus records one number, the distance from $0$, so every point of a whole circle about $0$ carries the same one, and the two numbers above are two of the infinitely many points on the circle of radius $\sqrt{34}$. Distance from the origin is real information, and it is the only information a modulus keeps.

One fact does all the work here: for any two complex numbers, $|z - c|$ is the distance between the points they occupy. So there is nothing to solve and nothing to expand; the geometry is already an equation once the right two numbers are put in the right slots.

The beacon is at $c = 7 - 6i$ and the range is $41$, so the points at distance exactly $41$ from it are the solutions of

$|z - (7 - 6i)| = 41.$

That is a circle of radius $41$ centred at the beacon, and it is the edge of the reach.

Coverage is the same measurement with a comparison instead of an equality. A point receives the signal when its distance is $41$ or less, and the boundary is included, so the inequality is not strict:

$|z - (7 - 6i)| \le 41.$

Two details decide whether this is right. The centre is SUBTRACTED, signs and all, so the expression contains $-(7 - 6i)$ and not $+(7 - 6i)$; writing $|z + 7 - 6i|$ would describe a circle about $-7 + 6i$, a different place on the chart entirely. And the radius appears on the right as the distance itself, not as its square: the modulus has already taken the square root.

Each vessel needs one subtraction and one modulus. Subtract the beacon's position, then measure.

For the first vessel, $(47 + 3i) - (7 - 6i) = 40 + 9i$, so

$\sqrt{1600 + 81} = \sqrt{1681} = 41.$

That is the range exactly. The signal reaches the boundary, so this vessel is covered, and it is worth saying which way the boundary case was decided rather than leaving it unremarked.

For the second, $(7 + 33i) - (7 - 6i) = 39i$, a purely vertical gap, so

$\sqrt{0 + 1521} = 39.$

The vessel is due north of the beacon at $39$ km, inside the range and covered.

For the third, $(35 + 39i) - (7 - 6i) = 28 + 45i$, so

$\sqrt{784 + 2025} = \sqrt{2809} = 53.$

Since $53 > 41$, this vessel is out of reach. Notice that it is not the vessel with the largest coordinates that is farthest away; only the distance from the beacon decides, and that is what the subtraction is for.

Being $16$ km east of the origin fixes the real part and leaves the north coordinate free, so write the unknown point as $z = 16 + yi$ with $y$ real. Subtract the beacon and simplify inside the bars:

$z - (7 - 6i) = 9 + (y + 6)\,i.$

The condition is that this has modulus $41$. Squaring both sides of a statement about nonnegative numbers loses nothing, so

$81 + (y + 6)^2 = 1681 \quad \Longrightarrow \quad (y + 6)^2 = 1600.$

A positive square has two square roots, and dropping one of them would drop a genuine position: $y + 6 = 40$ or $y + 6 = -40$, giving $y = 34$ or $y = -46$. The two points are $16 + 34i$ and $16 - 46i$.

The picture says why there are exactly two. The line runs north to south through every point with real part $16$, and the beacon has real part $7$, so the shortest distance from the beacon to that line is $|16 - 7| = 9$ km, measured due east. Since $9$ is less than the radius $41$, the line passes through the interior of the region and must cross the edge once going in and once coming out.

The count changes when that shortest distance changes. If the line were exactly $41$ km from the beacon, at real part $48$ or $-34$, it would touch the edge at one point only, the near point of the circle, and nowhere else. If it were farther than $41$ km, the line would stay wholly outside and meet the edge nowhere, which is the algebraic statement that the equation for $y$ would ask a square to be negative.

Write $z = a + bi$ with $a$ and $b$ real, so that $\overline{z} = a - bi$, and compute the two sides separately.

The left side is exactly the product quoted in the stem, so no expansion is needed:

$z\,\overline{z} = (a + bi)(a - bi) = a^2 + b^2.$

The right side comes from the modulus formula, squared. Since $|z| = \sqrt{a^2 + b^2}$ and the quantity under the root is never negative,

$|z|^2 = \left(\sqrt{a^2 + b^2}\right)^2 = a^2 + b^2.$

The two sides are the same real number for every real $a$ and $b$, so $z\,\overline{z} = |z|^2$ holds for every complex number.

Now read it. The left side multiplies a point by its reflection across the real axis, which is what conjugation does geometrically. The right side is the square of a distance. So multiplying any number by its mirror image lands you on the real axis, never to the left of $0$, and at a place with a meaning: the squared distance from the origin. It also explains the division method of the previous lesson. Multiplying a denominator $w$ by $\overline{w}$ produced a real number, and that mystery real number now has a name, $|w|^2$, which is zero only when $w$ itself is zero.

The check on $z = 6 - 7i$, whose conjugate is $6 + 7i$:

$z\,\overline{z} = 36 + 49 = 85, \qquad |z|^2 = \left(\sqrt{85}\right)^2 = 85.$

The length of a product is the product of the lengths, and the rule applies to a product of any number of factors, since it can be used repeatedly on two at a time.

Measure the three factors separately:

$|1 + 2i| = \sqrt{5}, \quad |3 + i| = \sqrt{10}, \quad |2 - 3i| = \sqrt{13}.$

Multiplying the three and simplifying the root, since $650 = 25 \cdot 26$,

$\sqrt{5}\,\sqrt{10}\,\sqrt{13} = \sqrt{650} = 5\sqrt{26}.$

No complex multiplication happened at any point, which is the whole saving: expanding three binomials and collecting the powers of $i$ would take several lines and offers several chances to lose a sign.

A power is the same rule with equal factors. Since $|1 - 3i| = \sqrt{10}$ and the fourth power is four factors,

$|(1 - 3i)^4| = \left(\sqrt{10}\right)^4 = 100.$

Both answers are nonnegative real numbers, as every modulus must be, and neither required knowing the product itself.

The quotient $z / w$ is defined as the number that multiplies back to $z$, so name it $u = z / w$ and record what that means:

$z = u\,w.$

Now take the modulus of both sides and apply multiplicativity to the right:

$|z| = |u|\,|w|.$

This is an equation between three nonnegative real numbers, so ordinary real algebra finishes it, provided the division is legal. It is: $w \ne 0$ by hypothesis, and a modulus is $0$ only for the number $0$, so $|w| \ne 0$. That is the single step the hypothesis protects, and it is easy to skip. Dividing,

$|u| = \frac{|z|}{|w|},$

which is the claim, since $u$ was the quotient.

Notice what the proof did not do. It never wrote the quotient in standard form, never multiplied by a conjugate, and never touched the parts of either number. The product rule alone carries the whole argument, and the same trick will work for any operation defined as an inverse.

The check needs only two moduli. Since $|1 - 3i| = \sqrt{10}$ and $|2 + i| = \sqrt{5}$,

$\left|\frac{1 - 3i}{2 + i}\right| = \frac{\sqrt{10}}{\sqrt{5}} = \sqrt{2}.$

Carrying the division out the long way agrees: the quotient is $\dfrac{-1 - 7i}{5}$, whose modulus is $\dfrac{\sqrt{50}}{5} = \sqrt{2}$.

Two bounds pin the sum, and both come from the triangle inequality. Above,

$|z + w| \le |z| + |w| = 9 + 5 = 14,$

since a detour through $z$ is never shorter than the straight route. Below, the reverse reading of the same inequality gives $|z + w| \ge |z| - |w| = 4$, and also $|z + w| \ge |w| - |z| = -4$, which is no restriction at all because a modulus is never negative. So the binding lower bound is $4$.

Bounds are only worth having if they are reached, so exhibit a pair for each. Both moduli are satisfied by real numbers, which are the easiest to add:

$z = 9, \ w = 5 \ \Rightarrow \ |z + w| = |14| = 14, \qquad z = 9, \ w = -5 \ \Rightarrow \ |z + w| = |4| = 4.$

Check the conditions on the second pair, since it is the one that looks suspicious: $|9| = 9$ and $|-5| = 5$, as required, because a modulus ignores the sign.

So $|z + w|$ is confined to the window from $4$ to $14$, and both ends are genuinely available.

One example is enough to sink a claim made about every pair, so choose the pair whose sum is easiest to measure exactly. Put one number on each axis: $z = 9i$ has modulus $9$ and $w = 5$ has modulus $5$, so both conditions hold. Their sum is $5 + 9i$, and

$|z + w| = |5 + 9i| = \sqrt{25 + 81} = \sqrt{106}.$

Here $|z| + |w| = 14$, so compare by squaring rather than estimating: $106$ is not $196$, so the two are different and the claim is false. The value sits strictly inside the window from part A, which is exactly what the inequality allows.

Where does the classmate's reasoning go wrong? Lengths add only along a straight path. Walking from $0$ to $z$ and then a further $|w|$ to $z + w$ covers $9 + 5 = 14$ units of travel, but the straight-line distance back to $0$ is shorter unless the second leg continues in the direction of the first. The proof of the inequality locates the exact moment slack enters: it bounds the real part of $z\,\overline{w}$ by its modulus, and that bound is an equality only when $z\,\overline{w}$ is a nonnegative real number.

That is the condition, and the two examples confirm it. For the pair $z = 9$, $w = 5$ from part A, $z\,\overline{w} = 45$, a nonnegative real, and the bound was attained. For $z = 9i$, $w = 5$, $z\,\overline{w} = 45i$, which is not real, and the inequality is strict. Geometrically the equality case is the degenerate triangle in which the detour is no detour at all.

The bound is the triangle inequality itself, applied to a pair chosen so that the sum comes out to be $z$. The pair is $z + w$ and $-w$, because

$(z + w) + (-w) = z.$

Applying the inequality to those two numbers, and using $|-w| = |w|$ since the point $-w$ is the same distance from $0$ as $w$,

$|z| \le |z + w| + |w|.$

Subtracting $|w|$ from both sides, which is ordinary arithmetic on real numbers, gives $|z + w| \ge |z| - |w|$. Nothing was assumed about $z$ or $w$, so the bound holds for every pair.

Now use it. With $|z| = 9$ and $|w| = 5$, every pair meeting the conditions satisfies

$|z + w| \ge 9 - 5 = 4.$

Suppose some pair had $z + w = 2 + 3i$. Then its modulus would be $\sqrt{4 + 9} = \sqrt{13}$, and comparing squares rather than estimating the root, $13 < 16$, so $\sqrt{13} < 4$. That contradicts the bound just proved, so no such pair exists.

Read the conclusion carefully. It is not that this particular sum is hard to arrange, or that the pairs tried so far have failed. The bound holds for every pair satisfying the two modulus conditions, so the point $2 + 3i$ is closer to $0$ than any sum those conditions permit, and it is unreachable.