The Complex Plane and Modulus: Additional Practice (Set 1)

Multiplicativity turns the square outside into a square of the modulus.

$|(1 + i)^2| = |1 + i|^2 = \left(\sqrt{1 + 1}\right)^2 = 2$

As a check, $(1 + i)^2 = 1 + 2i + i^2 = 2i$, and $|2i| = 2$.

The distance between two numbers is the modulus of their difference.

$z - \overline{z} = (4 + 3i) - (4 - 3i) = 6i, \qquad |6i| = 6$

Geometrically, $z$ sits $3$ above the real axis and its conjugate sits $3$ below, so the mirror pair is $3 + 3 = 6$ apart, twice the imaginary part.

For $z = a + bi$, the number plus its conjugate is twice the real part.

$z + \overline{z} = (a + bi) + (a - bi) = 2a = 6 \;\Longrightarrow\; a = 3$

The imaginary part $b$ is unconstrained, so the solutions form the entire vertical line through $3$: all numbers $3 + bi$.

Apply the modulus formula and solve.

$k^2 + 6^2 = 100 \;\Longrightarrow\; k^2 = 64 \;\Longrightarrow\; k = 8 \ \text{ or } \ k = -8$

Both signs work because squaring erases the sign: the points $8 + 6i$ and $-8 + 6i$ both lie on the circle of radius $10$, mirror images across the imaginary axis.

Multiplicativity applies with the real factor $3$, whose modulus is $3$.

$|3z| = |3|\,|z| = 3 \times 4 = 12$

Scaling a number by $3$ stretches its arrow to three times the length, so the distance from $0$ triples.

A circle $|z - c| = r$ passes through $0$ exactly when the center is at distance $r$ from $0$, that is, when $|c| = r$. Check each center against its radius:

$|2| = 2 \ne 1, \qquad |1 + i| = \sqrt{2} \ne 2, \qquad |4| = 4 \ne 2, \qquad |3i| = 3 = 3$

Only $|z - 3i| = 3$ has $|c| = r$, so it is the one through the origin: setting $z = 0$ gives $|-3i| = 3$, which is true.

The quantity $|z - 1|$ is the distance from $z$ to the point $1$, and $z$ ranges over the circle of radius $3$ about $0$. The triangle inequality gives the cap:

$|z - 1| \le |z| + |{-1}| = 3 + 1 = 4$

It is achieved at $z = -3$, the point of the circle diametrically away from $1$, where $|-3 - 1| = 4$.

The equation says $z$ is equidistant from $2i$ and $-4i$, so it lies on the perpendicular bisector of the segment joining $(0, 2)$ and $(0, -4)$. With $z = a + bi$,

$a^2 + (b - 2)^2 = a^2 + (b + 4)^2 \;\Longrightarrow\; -4b + 4 = 8b + 16 \;\Longrightarrow\; b = -1$

Every real part is allowed, so the set is the horizontal line at height $-1$, which is indeed the midline between heights $2$ and $-4$.

The circle has center $3$ (on the real axis) and radius $2$, so its points sit between distance $3 - 2$ and $3 + 2$ from the origin. The reverse triangle inequality states it precisely:

$|z| = |(z - 3) + 3| \ge |3| - |z - 3| = 3 - 2 = 1$

The minimum is achieved at $z = 1$, the point of the circle closest to $0$ (check: $|1 - 3| = 2$, so $1$ is on the circle, and $|1| = 1$).

First compute the conjugate product, which is already a non-negative real number.

$z\,\overline{z} = 5^2 + 12^2 = 25 + 144 = 169$

The modulus of the real number $169$ is just $169$. Note the contrast: $|z| = 13$, while $|z\,\overline{z}| = |z|^2 = 169$.

Square the parts, add, and take the root; the sign of $-8$ vanishes when squared.

$|-8 + 15i| = \sqrt{(-8)^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17$

Adding the parts ($23$), subtracting them ($7$), or subtracting the squares ($\sqrt{161}$) are the usual wrong turns.

Write $z = a + bi$ and square both conditions: $a^2 + b^2 = 25$ and $(a - 8)^2 + b^2 = 25$. Subtracting eliminates $b$:

$a^2 - (a - 8)^2 = 0 \;\Longrightarrow\; 16a - 64 = 0 \;\Longrightarrow\; a = 4$

Then $b^2 = 25 - 16 = 9$, so $b = \pm 3$, giving the two solutions $4 + 3i$ and $4 - 3i$. Geometrically, two circles of radius $5$ whose centers are $8$ apart cross in exactly two points, since $8$ is less than $5 + 5$ and more than $5 - 5$.