The Complex Plane and Modulus: Practice

Write $z = a + 4i$ and impose the modulus condition.

$a^2 + 4^2 = 25 \;\Longrightarrow\; a^2 = 9 \;\Longrightarrow\; a = 3 \ \text{ or } \ a = -3$

So exactly two numbers work, $3 + 4i$ and $-3 + 4i$. Geometrically, the horizontal line at height $4$ cuts the circle of radius $5$ in two symmetric points.

The equation says $z$ is equidistant from $3$ and from $-3$, so it lies on the perpendicular bisector of the segment joining those two points. Algebraically, with $z = a + bi$,

$(a - 3)^2 + b^2 = (a + 3)^2 + b^2 \;\Longrightarrow\; -6a = 6a \;\Longrightarrow\; a = 0$

The solutions are the numbers with real part $0$: the imaginary axis, which is indeed the perpendicular bisector of the segment from $-3$ to $3$.

The triangle inequality caps the modulus of a sum.

$|z + w| \le |z| + |w| = 3 + 4 = 7$

The cap is achieved when the two arrows point the same way, for example $z = 3$ and $w = 4$, giving $|7| = 7$. The value $12$ is the product of the moduli, which bounds $|zw|$, not $|z + w|$.

Moduli always multiply across products, and conjugation is a reflection so it preserves modulus; the triangle inequality holds for every pair. But equality in the triangle inequality needs the arrows aligned, and it fails whenever they are not.

$z = 1, \; w = i: \qquad |z + w| = |1 + i| = \sqrt{2}, \qquad |z| + |w| = 2$

Since $\sqrt{2} \ne 2$, the claim that $|z + w| = |z| + |w|$ always holds is false.

A real number is $z = a + 0i$, so compute its distance to the center $4i$.

$|a - 4i| = \sqrt{a^2 + 16} = 5 \;\Longrightarrow\; a^2 = 9 \;\Longrightarrow\; a = \pm 3$

The circle has center $(0, 4)$ and radius $5$, and it crosses the real axis at the two points $(3, 0)$ and $(-3, 0)$: a $3$-$4$-$5$ triangle on each side of the imaginary axis.

The condition $z = \overline{z}$ forces $z$ to be real: writing $z = a + bi$ gives $2bi = 0$, so $b = 0$.

$z = a, \qquad |a| = 4 \;\Longrightarrow\; a = 4 \ \text{ or } \ a = -4$

Geometrically these are the points where the mirror line of conjugation (the real axis) meets the circle of radius $4$. The numbers $\pm 4i$ have modulus $4$ but fail $z = \overline{z}$.

The fourth vertex is the sum, so $w$ is recovered by subtracting.

$w = (z + w) - z = (3 - 2) + (4 - 1)i = 1 + 3i$

Check: $(2 + i) + (1 + 3i) = 3 + 4i$. The arrow from $z$ to $z + w$ is exactly a translated copy of $w$, which is why the difference recovers it.

The equation says $z$ is equidistant from $1$ and from $0$, so it lies on the perpendicular bisector of the segment between them. With $z = a + bi$,

$(a - 1)^2 + b^2 = a^2 + b^2 \;\Longrightarrow\; -2a + 1 = 0 \;\Longrightarrow\; a = \tfrac{1}{2}$

Every imaginary part $b$ is allowed, so the solution set is the whole vertical line through $\tfrac{1}{2}$, not just the single real point on it.

The right side $|z|$ is always a non-negative real number, so the left side must be one too. Writing $z = a + bi$, the conjugate $a - bi$ is real only when $b = 0$, so $z = a$ is real. The equation then reads

$a = |a|$

which holds exactly when $a \ge 0$. So the solutions are the non-negative real numbers: $0$ and the positive half of the real axis.

Multiplicativity says $|zw| = |z|\,|w|$, so solve for the unknown factor.

$30 = 6\,|w| \;\Longrightarrow\; |w| = 5$

The moduli behave exactly like lengths under multiplication, so the unknown length is the quotient $30 \div 6$.

Equality in the triangle inequality happens exactly when the two arrows point the same way. The candidate $4 + 4i = 2z$ points along $z$, and indeed

$|z + w| = |6 + 6i| = 6\sqrt{2}, \qquad |z| + |w| = 2\sqrt{2} + 4\sqrt{2} = 6\sqrt{2}.$

The others fail: $-2 - 2i$ points opposite ($|z + w| = 0$ while the sum of moduli is $4\sqrt{2}$), and $2 - 2i$ and $-4 + 4i$ point in different directions, giving strict inequality.

Inside means distance to the center $1 + i$ strictly less than $2$. Test the candidate:

$|(2 + 2i) - (1 + i)| = |1 + i| = \sqrt{2} < 2$

So $2 + 2i$ is inside. The others are not strictly inside: $|(3 + i) - (1 + i)| = 2$ and $|(-1 + i) - (1 + i)| = 2$ put those two exactly on the circle, and $|(1 + 4i) - (1 + i)| = |3i| = 3 > 2$ puts the last one outside.