The Complex Plane and Modulus: Additional Practice (Set 2)

The modulus multiplies across products, so take each factor's modulus and multiply.

$|2 + i| = \sqrt{5}, \qquad |3 - 4i| = 5, \qquad |1 + 2i| = \sqrt{5}$

Multiplying, $\sqrt{5} \cdot 5 \cdot \sqrt{5} = 5 \cdot 5 = 25$. No complex multiplication was ever needed, which is the whole point of multiplicativity.

Track the parts. In Quadrant II, $z = a + bi$ has $a < 0$ and $b > 0$.

$-\overline{z} = -(a - bi) = -a + bi$

The new real part $-a$ is positive and the imaginary part $b$ is still positive, so the point lands in Quadrant I. In fact $-\overline{z}$ is always the reflection of $z$ across the imaginary axis.

The reverse triangle inequality bounds a sum from below by the gap between the lengths.

$|z + w| \ge |w| - |z| = 4 - 3 = 1$

The bound is achieved with the arrows opposed, for example $z = 3$ and $w = -4$, giving $|-1| = 1$. Zero is impossible: $z + w = 0$ would force $w = -z$ and hence $|w| = |z|$, but $4 \ne 3$. A modulus can never be negative, ruling out $-1$ on sight.

Use the conjugate identity $z\,\overline{z} = a^2 + b^2$.

$z\,\overline{z} = (\sqrt{3})^2 + 1^2 = 3 + 1 = 4$

The value $2$ is $|z|$, one square root too far, and $2 + 2\sqrt{3}\,i$ is $z^2$, which is a different product entirely.

Subtract to get the arrow between the points, then take the modulus.

$(2 - 3i) - (-4 + 5i) = 6 - 8i, \qquad |6 - 8i| = \sqrt{36 + 64} = \sqrt{100} = 10$

The horizontal gap is $6$ and the vertical gap is $8$, a scaled $3$-$4$-$5$ triangle.

Adding a fixed number slides the point by its parts: $2$ right and $5$ down.

$(-1 + 2i) + (2 - 5i) = (-1 + 2) + (2 - 5)i = 1 - 3i$

The point $(-1, 2)$ moves to $(1, -3)$, crossing from Quadrant II to Quadrant IV.

The quantity $|z - 5|$ is the distance from $z$ to the point $5$. The points with $|z| = 2$ form the circle of radius $2$ about $0$, and $5$ sits outside it at distance $5$ from the center. The reverse triangle inequality makes this precise:

$|z - 5| \ge |5| - |z| = 5 - 2 = 3$

The bound is achieved at $z = 2$, the point of the circle closest to $5$, where $|2 - 5| = 3$.

The triangle inequality and its reverse fence the sum in on both sides.

$|w| - |z| \le |z + w| \le |z| + |w| \;\Longrightarrow\; 3 \le |z + w| \le 7$

Both ends are achieved (arrows opposed for $3$, aligned for $7$), and swinging $w$ continuously from opposed to aligned makes $|z + w|$ pass through every value between. Values below $3$ are impossible because the two lengths differ by $3$.

Use multiplicativity of the modulus.

$|3 + 4i| = \sqrt{9 + 16} = 5, \qquad |8 - 6i| = \sqrt{64 + 36} = 10$

So the product has modulus $5 \times 10 = 50$, with no need to expand $(3 + 4i)(8 - 6i)$ at all.

The modulus is $|z| = \sqrt{a^2 + b^2}$, so squaring removes the root.

$|z|^2 = a^2 + b^2 = z\,\overline{z}$

The expression $a^2 - b^2$ is the real part of $z^2$, and $(a + b)^2$ adds an unwanted cross term $2ab$; neither measures distance.

Negation flips both parts: $-(a + bi) = -a - bi$, sending $(a, b)$ to $(-a, -b)$.

$(a, b) \;\longmapsto\; (-a, -b)$

That is the point reflection through the origin: $z$ and $-z$ sit on the same line through $0$, on opposite sides, with $|-z| = |z|$. Reflection across the real axis alone is the conjugate, and across the imaginary axis alone is $-\overline{z}$.

Conjugation reflects across the real axis, so its fixed points are exactly the mirror line: $z = \overline{z}$ forces $2bi = 0$, hence $b = 0$, and every real number satisfies it.

$z = \overline{z} \;\Longleftrightarrow\; z \text{ is real}$

The others describe different sets: $|z| = 0$ is only the origin, $z + \overline{z} = 0$ says the real part is $0$ (the imaginary axis), and $|z| = z\,\overline{z}$ says $|z| = |z|^2$, which is the circle $|z| = 1$ together with $0$.