Linear-Algebra-Done-Right · Unit 10 · Video 1 · Interactive Practice

Conjugates, |z| and the Complex Triangle Inequality

IKey Formulas

Formula	Name	What it says
$\bar{z} = \operatorname{Re} z - (\operatorname{Im} z)\,i$ , $\;\|z\| = \sqrt{(\operatorname{Re} z)^2 + (\operatorname{Im} z)^2}$	Conjugate, absolute value	For $z = a + bi$ with $a, b \in \mathbf{R}$ : $\bar{z} = a - bi$ and $\|z\| = \sqrt{a^2 + b^2} \ge 0$ , the distance from $0$ to $(a, b)$
$z + \bar{z} = 2\operatorname{Re} z$ , $\;z - \bar{z} = 2(\operatorname{Im} z)\,i$ , $\;z\bar{z} = \|z\|^2$	Sum, difference, product rules	For $z \in \mathbf{C}$ ; $z = \bar{z}$ exactly when $\operatorname{Im} z = 0$ , i.e. when $z$ is real
$\overline{w + z} = \bar{w} + \bar{z}$ , $\;\overline{wz} = \bar{w}\,\bar{z}$ , $\;\|wz\| = \|w\|\,\|z\|$	Conjugation and products	For all $w, z \in \mathbf{C}$ : conjugation respects sums and products, and absolute value is multiplicative
$\|w + z\|^2 = \|w\|^2 + \|z\|^2 + 2\operatorname{Re}(w\bar{z})$	Triangle inequality	For all $w, z \in \mathbf{C}$ ; since $\operatorname{Re}(w\bar{z}) \le \|w\bar{z}\| = \|w\|\,\|z\|$ , this gives $\|w + z\| \le \|w\| + \|z\|$

Key Insight: the proof loses something only at $\operatorname{Re}(w\bar{z}) \le |w\bar{z}|$ , so $|w + z| = |w| + |z|$ exactly when $w\bar{z}$ is a nonnegative real number — when $w$ and $z$ point the same way, or one of them is $0$ .

IIVisualization 1 — Conjugate and Absolute Value in the Plane

With $z = a + bi$ placed at $(a, b)$ , $\bar{z}$ is its mirror image across the real axis and $|z|$ its distance from $0$ .

IIIVisualization 2 — The Triangle Inequality

The arrows $w$ , $z$ moved to the tip of $w$ , and $w + z$ form a triangle, so $|w + z| \le |w| + |z|$ .

💡 Challenge: keep $w = 3 + i$ and find a nonzero $z$ with $|w + z| = |w| + |z|$ .

IVVisualization 3 — The Proof, One Line at a Time

Expanding $|w + z|^2$ leaves one term, $2\operatorname{Re}(w\bar{z})$ , that can be negative — and it is at most $2|w|\,|z|$ .

w, z Step 1

Step 1 — product rule, then additivity of the conjugate

|w + z|^2 = (w + z)\,\overline{(w + z)} = (w + z)(\bar{w} + \bar{z})

Step 2 — distribute; $z\bar{w} = \overline{w\bar{z}}$

= |w|^2 + |z|^2 + w\bar{z} + \overline{w\bar{z}}

Step 3 — sum rule: a number plus its conjugate

= |w|^2 + |z|^2 + 2\operatorname{Re}(w\bar{z})

Step 4 — $\operatorname{Re}(w\bar{z}) \le |w\bar{z}| = |w|\,|\bar{z}| = |w|\,|z|$

\le |w|^2 + |z|^2 + 2|w|\,|z|

Step 5 — a perfect square; both sides are nonnegative, so take square roots

= (|w| + |z|)^2 \quad\Longrightarrow\quad |w + z| \le |w| + |z|

VQuiz Questions

Problem 1 · Conjugate and Absolute Value

Given: $w = 2 - 5i$ — find $\bar{w}$ and $|w|$ .

What is $\bar{w}$ ?

What is $|w|$ ?

✅ Correct!

\bar{w} = 2 + 5i

keeps the real part

2

and flips the sign of

\operatorname{Im} w = -5

;

|w| = \sqrt{2^2 + (-5)^2} = \sqrt{29}

❌ That is $-w$ . Conjugation keeps the real part:

\overline{a + bi} = a - bi

, so only the sign of the imaginary part changes.

❌ Wrong part flipped.

-2 - 5i

changes the real part and keeps the imaginary part; the conjugate does the opposite.

❌ The parts don't trade places.

\bar{w}

has the same real part as

w

, namely

2

, and imaginary part

-\operatorname{Im} w = 5

❌ That is $|w|^2$ .

w\bar{w} = 4 + 25 = 29 = |w|^2

; the absolute value is its square root.

❌ That adds the legs.

|2| + |-5| = 7

is the path along the axes;

|w|

is the straight distance from

0

(2, -5)

, the hypotenuse.

❌ Watch $i^2$ .

w\bar{w} = (2 - 5i)(2 + 5i) = 4 - (5i)^2 = 4 + 25

: since

i^2 = -1

the squares add, giving

29

, not

|4 - 25| = 21

❌ Not quite.

\overline{a + bi} = a - bi

with

a = 2

b = -5

❌ Not quite.

|a + bi| = \sqrt{a^2 + b^2}

with

a = 2

b = -5

Show solution

Here $\operatorname{Re} w = 2$ and $\operatorname{Im} w = -5$ (a real number, without the $i$ ).

\bar{w} = \operatorname{Re} w - (\operatorname{Im} w)\,i = 2 - (-5)i = 2 + 5i

|w| = \sqrt{2^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt{29} \approx 5.385

Check with the product rule: $w\bar{w} = (2 - 5i)(2 + 5i) = 4 + 10i - 10i - 25i^2 = 29 = |w|^2$ .

Problem 2 · Times the Conjugate, Not Itself

Given: $z = -1 + 4i$ — find $z\bar{z}$ .

✅ Correct!

z\bar{z} = (-1)^2 + 4^2 = 17 = |z|^2

— a nonnegative real number, as the product rule promises.

❌ That is $z^2$ .

(-1 + 4i)^2 = 1 - 8i + 16i^2 = -15 - 8i

multiplies

z

by itself; multiplying by

\bar{z} = -1 - 4i

cancels the imaginary parts.

❌ Sign slip in $(bi)^2$ .

z\bar{z} = a^2 - (bi)^2 = a^2 + b^2

; writing

a^2 - b^2 = 1 - 16

forgets that

i^2 = -1

❌ That is $|z|$ . The product rule gives

z\bar{z} = |z|^2

, the square of the absolute value, with no square root.

❌ Not quite. Use

z\bar{z} = |z|^2 = (\operatorname{Re} z)^2 + (\operatorname{Im} z)^2

Show solution

The conjugate is $\bar{z} = -1 - 4i$ . Multiplying out:

z\bar{z} = (-1 + 4i)(-1 - 4i) = 1 + 4i - 4i - 16i^2 = 1 + 16 = 17

This agrees with the product rule $z\bar{z} = |z|^2 = (-1)^2 + 4^2 = 17$ . By contrast $z^2 = -15 - 8i$ is not even real, so $z^2 \ne |z|^2$ .

Problem 3 · Expanding $|w + z|^2$

Given: $w = 2 - 5i$ and $z = -1 + 4i$ , and the identity $|w + z|^2 = |w|^2 + |z|^2 + 2\operatorname{Re}(w\bar{z})$ — find $\operatorname{Re}(w\bar{z})$ and $|w + z|^2$ .

What is $\operatorname{Re}(w\bar{z})$ ?

What is $|w + z|^2$ ?

✅ Correct!

29 + 17 + 2(-22) = 2

, matching

w + z = 1 - i

directly:

|1 - i|^2 = 1 + 1 = 2

❌ That is $\operatorname{Re}(wz)$ .

wz = (2 - 5i)(-1 + 4i) = 18 + 13i

; the identity needs

w\bar{z}

with

\bar{z} = -1 - 4i

❌ Check the signs.

w\bar{z} = (2 - 5i)(-1 - 4i) = -2 - 8i + 5i + 20i^2 = -22 - 3i

, so the real part is negative.

❌ That is the imaginary part.

w\bar{z} = -22 - 3i

; its real part is

-22

❌ The cross term keeps its sign.

90 = 46 + 2 \cdot 22

uses

|\operatorname{Re}(w\bar{z})|

; the identity has

2\operatorname{Re}(w\bar{z}) = -44

❌ Squares add.

w + z = 1 - i

and

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

, not

1^2 - 1^2

❌ The cross term is missing.

46 = |w|^2 + |z|^2

; in general

|w + z|^2 \ne |w|^2 + |z|^2

— add

2\operatorname{Re}(w\bar{z}) = -44

❌ Not quite. Compute

w\bar{z}

with

\bar{z} = -1 - 4i

, then take its real part.

❌ Not quite. Substitute

|w|^2 = 29

|z|^2 = 17

and

\operatorname{Re}(w\bar{z})

into the identity.

Show solution

Step 1: $\bar{z} = -1 - 4i$ , so

w\bar{z} = (2 - 5i)(-1 - 4i) = -2 - 8i + 5i + 20i^2 = -22 - 3i, \qquad \operatorname{Re}(w\bar{z}) = -22.

Step 2: $|w|^2 = 4 + 25 = 29$ and $|z|^2 = 1 + 16 = 17$ , so

|w + z|^2 = 29 + 17 + 2(-22) = 46 - 44 = 2.

Check: $w + z = 1 - i$ and $|1 - i|^2 = 2$ . The triangle inequality holds with plenty of room: $|w + z| = \sqrt{2} \approx 1.414$ while $|w| + |z| = \sqrt{29} + \sqrt{17} \approx 9.508$ — the negative cross term is what makes $w + z$ so short.

Problem 4 · When Is It an Equality?

Given: $w = 1 + i$ — find the $z$ for which $|w + z| = |w| + |z|$ .

✅ Correct!

z = 3w

points the same way as

w

w\bar{z} = 6

is a nonnegative real number, and

|4 + 4i| = 4\sqrt{2} = \sqrt{2} + 3\sqrt{2}

❌ Close, but not equal.

|3 + 2i| = \sqrt{13} \approx 3.606

while

\sqrt{2} + \sqrt{5} \approx 3.650

: here

w\bar{z} = 3 + i

is not real, so

\operatorname{Re}(w\bar{z}) = 3 < |w\bar{z}| = \sqrt{10}

❌ Same line, opposite direction.

|w + z| = |-1 - i| = \sqrt{2}

but

|w| + |z| = 3\sqrt{2}

; here

w\bar{z} = -4

is negative.

❌ Not equal.

w + z = 2

, so

|w + z| = 2 < \sqrt{2} + \sqrt{2} \approx 2.828

; here

w\bar{z} = 2i

has real part

0

❌ Not quite. Equality needs

\operatorname{Re}(w\bar{z}) = |w\bar{z}|

, i.e.

w\bar{z}

a nonnegative real number.

Show solution

From the proof, $|w + z|^2 = |w|^2 + |z|^2 + 2\operatorname{Re}(w\bar{z})$ and $(|w| + |z|)^2 = |w|^2 + |z|^2 + 2|w\bar{z}|$ , so equality holds exactly when $\operatorname{Re}(w\bar{z}) = |w\bar{z}|$ , that is, when $w\bar{z}$ is a nonnegative real number. Test each option:

\begin{aligned} z = 3 + 3i:&\quad w\bar{z} = (1 + i)(3 - 3i) = 6 \\ z = 2 + i:&\quad w\bar{z} = (1 + i)(2 - i) = 3 + i \\ z = -2 - 2i:&\quad w\bar{z} = (1 + i)(-2 + 2i) = -4 \\ z = 1 - i:&\quad w\bar{z} = (1 + i)(1 + i) = 2i \end{aligned}

Only $z = 3 + 3i$ gives a nonnegative real number. Directly: $|w + z| = |4 + 4i| = \sqrt{32} = 4\sqrt{2}$ and $|w| + |z| = \sqrt{2} + \sqrt{18} = \sqrt{2} + 3\sqrt{2} = 4\sqrt{2}$ .

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