The Complex Plane and Modulus: Practice

Square the parts, add, and take the root.

$|2 - 2i| = \sqrt{2^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2}$

Simplify the radical: $\sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2}$, a little more than $2.8$.

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

$(-3 + 4i)(-3 - 4i) = (-3)^2 - (4i)^2 = 9 + 16 = 25$

A number times its conjugate is always a non-negative real number, the squared distance from $0$, so answers like $-7$ or $25i$ are impossible on sight.

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

$(3 + 2i) - (-1 + 2i) = 4, \qquad |4| = 4$

The two points share the imaginary part $2$, so they sit on the same horizontal line and the distance is purely the horizontal gap, $3 - (-1) = 4$.

Write $z = a + bi$ and translate both sides.

$a - bi = -a - bi \;\Longrightarrow\; 2a = 0 \;\Longrightarrow\; a = 0$

The imaginary part cancels from both sides, so the only condition is $a = 0$: the solutions are $0$ and the pure imaginary numbers, the whole imaginary axis. Geometrically, reflecting across the real axis equals reflecting through the origin only for points on the vertical axis.

Add the parts separately.

$z + w = (5 - 3) + (-2 + 6)i = 2 + 4i$

The sum has positive real part and positive imaginary part, so its point $(2, 4)$ lies up and to the right, in Quadrant I.

The modulus multiplies across products, so compute each factor's modulus.

$|2i| = 2, \qquad |3 + 4i| = \sqrt{9 + 16} = 5, \qquad |2i\,(3 + 4i)| = 2 \times 5 = 10$

As a check, expanding gives $2i(3 + 4i) = 6i + 8i^2 = -8 + 6i$, and $\sqrt{64 + 36} = \sqrt{100} = 10$.

The equation $|z| = 6$ says the distance from $z$ to $0$ is exactly $6$.

$|z| = 6 \;\Longleftrightarrow\; \sqrt{a^2 + b^2} = 6 \;\Longleftrightarrow\; a^2 + b^2 = 36$

That is the circle of radius $6$ centered at the origin. A disk would be $|z| \le 6$, and the two points $6$ and $-6$ are only where the circle crosses the real axis.

In Quadrant IV the real part is positive and the imaginary part is negative: $a > 0$, $b < 0$.

$\overline{a + bi} = a - bi, \qquad a > 0, \; -b > 0$

The reflection across the real axis keeps the point on the right side and lifts it above the axis, into Quadrant I.

Multiply by the conjugate, or use $z\,\overline{z} = a^2 + b^2$.

$(1 + i)(1 - i) = 1 - i^2 = 1 + 1 = 2$

Compare $z^2 = (1 + i)^2 = 2i$: squaring gives a complex number, while multiplying by the conjugate gives the real number $|z|^2$. The value $\sqrt{2}$ is $|z|$, not $z\,\overline{z}$.

The equation $|z - c| = r$ collects the points at distance $r$ from $c$.

$|z - (2 + i)| = 4 \;\Longrightarrow\; c = 2 + i, \quad r = 4$

The center is the number being subtracted from $z$, namely $2 + i$; the sign flip to $-2 - i$ is the classic misread.

Solve for $w$ by subtracting: $w = (1 + 3i) - (2 + i)$.

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

Check: $(2 + i) + (-1 + 2i) = 1 + 3i$. The arrow $w$ carries the point $(2, 1)$ one unit left and two units up to $(1, 3)$.

The marked legs say $z = 2 + 3i$: two units right, three units up. The modulus is the hypotenuse of that right triangle.

$|z| = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}$

The sum $2 + 3 = 5$ and the unrooted $13$ both skip part of the Pythagorean computation; $\sqrt{13} \approx 3.6$.