The Complex Plane and Modulus: Practice

Writing the number in the form $a + bi$ gives $7 + (-3)i$, so $a = 7$ and $b = -3$.

$\text{imaginary part} = b = -3$

The imaginary part is the real number $-3$, not the term $-3i$. It is the vertical coordinate of the point, and coordinates are real numbers.

The point for $-2 + 5i$ is $(-2, 5)$: real part $-2$, imaginary part $5$.

$a = -2 < 0, \qquad b = 5 > 0$

A negative horizontal coordinate with a positive vertical coordinate places the point up and to the left, which is Quadrant II.

Conjugation sends $a + bi$ to $a - bi$, so the point $(a, b)$ moves to $(a, -b)$.

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

The horizontal coordinate is unchanged and the vertical coordinate changes sign, which is a mirror reflection in the real axis. It is not a single downward slide, because points at different heights move by different amounts (and points below the axis move up).

The modulus is the square root of the sum of the squares of the parts.

$|6 + 8i| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$

Adding the parts ($14$) or stopping at the sum of squares ($100$) skips a step of the Pythagorean computation.

Add the real parts and the imaginary parts separately.

$(2 + 1) + (3 + 1)i = 3 + 4i$

Geometrically, adding $1 + i$ slides the point $(2, 3)$ one unit right and one unit up to $(3, 4)$.

Adding $c + di$ moves every point $c$ units across and $d$ units up, with negative values meaning the opposite direction.

$w = 2 + (-1)i \;\Longrightarrow\; 2 \text{ right}, \; 1 \text{ down}$

The real part $2$ pushes the point right, and the imaginary part $-1$ pushes it down.

Multiply the number by its conjugate; the cross terms cancel.

$(3 + i)(3 - i) = 9 - i^2 = 9 + 1 = 10$

This matches the identity $z\,\overline{z} = |z|^2 = 3^2 + 1^2 = 10$. Note that $\sqrt{10}$ is $|z|$ itself, not $z\,\overline{z}$.

Square each part, add, then take the square root. The squaring erases the minus sign.

$|-12 + 5i| = \sqrt{(-12)^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$

Adding the parts ($17$), subtracting them ($7$), or subtracting the squares ($\sqrt{119}$) are not the Pythagorean computation.

Write the number as $0 + (-3)i$ and flip the sign of the imaginary part.

$\overline{-3i} = 0 - (-3)i = 3i$

The point $(0, -3)$ reflects across the real axis to $(0, 3)$. A number on the imaginary axis is never its own conjugate unless it is $0$.

As a complex number, $-9 = -9 + 0i$, so the modulus formula gives

$|-9| = \sqrt{(-9)^2 + 0^2} = \sqrt{81} = 9.$

On the real axis the modulus is the familiar absolute value: the distance from $-9$ to $0$ is $9$. A modulus is never negative.

Write $z = a + bi$ and set it equal to its conjugate.

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

So $z = a$ is real. Geometrically, conjugation is the reflection across the real axis, and the points it leaves fixed are exactly the points on the mirror, the real numbers. $z$ does not have to be zero: every real number works.

Apply the modulus formula with $a = 3$ and $b = -4$.

$|3 - 4i| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

The point $(3, -4)$ is the hypotenuse point of a $3$-$4$-$5$ right triangle, so its distance from $0$ is $5$. The value $25$ is $z\,\overline{z}$, the modulus squared.