Complex conjugate — properties, applications in division: Worked examples

Everything here rests on three facts you already built in the parent note:

$\overset{z}{ˉ} = a - bi$ (flip the sign of the imaginary part),
$z \overset{z}{ˉ} = a^{2} + b^{2} = ∣ z ∣^{2}$ (always a non-negative real number),
$\frac{z _{1}}{z _{2}} = \frac{z _{1} z ˉ _{2}}{∣ z _{2} ∣ ^{2}}$ (the division formula).

The scenario matrix

Every conjugate/division problem falls into one of these case classes. The examples below are each tagged with the cell they cover.

#	Case class	What's tricky	Example
A	Denominator in Quadrant I ( $c > 0, d > 0$ )	the "friendly" baseline	Ex 1
B	Denominator in Quadrant II/III/IV (mixed signs)	signs must not get lost	Ex 2
C	Purely imaginary denominator ( $c = 0$ )	dividing "by $i$ "	Ex 3
D	Purely real denominator ( $d = 0$ )	conjugate does nothing	Ex 4
E	Reciprocal $1/ z$ and geometry of $\bar z /	z	^2$
F	Zero / degenerate inputs ( $z = 0$ )	division is undefined	Ex 6
G	Real-world word problem (AC circuit impedance)	translate words → $z$	Ex 7
H	Exam-style twist (solve for $z$ , powers of $i$ )	conjugate inside an equation	Ex 8
I	Property check across a product/quotient	verify the algebra laws	Ex 9

Example 1 — Cell A: friendly Quadrant I denominator

Example 2 — Cell B: mixed-sign denominator (all quadrants)

Worked example Compute

$\frac{1 + 2 i}{- 3 + i}$ . Forecast: the denominator $- 3 + i$ sits in Quadrant II. Do NOT guess the sign of the result yet — watch how the minus rides through the algebra.

Step 1. Conjugate of $- 3 + i$ is $- 3 - i$ (flip only the imaginary part; the real part $- 3$ stays negative). Why this step? The single most common slip is "flipping" the real sign too. The mirror is the real axis — horizontal position never changes. $\frac{( 1 + 2 i ) ( - 3 - i )}{( - 3 + i ) ( - 3 - i )}$ Step 2. Denominator $= (- 3)^{2} + (1)^{2} = 9 + 1 = 10$ . Why? $z \overset{z}{ˉ} = a^{2} + b^{2}$ with $a = - 3, b = 1$ . Squaring kills every sign — it is always positive. Step 3. Numerator $= (1) (- 3) + (1) (- i) + (2 i) (- 3) + (2 i) (- i) = - 3 - i - 6 i - 2 i^{2} = - 3 - 7 i + 2 = - 1 - 7 i$ . Why? $- 2 i^{2} = + 2$ . Answer: $\frac{- 1 - 7 i}{10} = - \frac{1}{10} - \frac{7}{10} i$ — Quadrant III.

Verify: $(- 3 + i) (- \frac{1}{10} - \frac{7}{10} i)$ must equal $1 + 2 i$ . See =VERIFY=.

Example 3 — Cell C: purely imaginary denominator

Worked example Compute

$\frac{5 - 2 i}{4 i}$ . (Here $c = 0$ : the denominator is pure $i$ .) Forecast: "dividing by $i$ " scares people. Guess: does $i$ move things by rotating them?

Step 1. The conjugate of $4 i$ is $\overline{0 + 4 i} = 0 - 4 i = - 4 i$ . Why this step? Same rule — flip the imaginary part. A pure imaginary number's conjugate is its negative. $\frac{( 5 - 2 i ) ( - 4 i )}{( 4 i ) ( - 4 i )}$ Step 2. Denominator $= (4 i) (- 4 i) = - 16 i^{2} = 16$ . Why? $z \overset{z}{ˉ} = ∣ z ∣^{2}$ ; here $∣4 i ∣^{2} = 0^{2} + 4^{2} = 16$ . Confirms the shortcut. Step 3. Numerator $= (5) (- 4 i) + (- 2 i) (- 4 i) = - 20 i + 8 i^{2} = - 20 i - 8 = - 8 - 20 i$ . Why? $8 i^{2} = - 8$ . Answer: $\frac{- 8 - 20 i}{16} = - \frac{1}{2} - \frac{5}{4} i$ .

Shortcut worth knowing: dividing by $i$ is the same as multiplying by $- i$ (since $1/ i = - i$ ). Multiplying by $- i$ rotates a point $9 0^{\circ}$ clockwise on the Argand diagram.

Verify: $(4 i) (- \frac{1}{2} - \frac{5}{4} i)$ must equal $5 - 2 i$ . See =VERIFY=.

Example 4 — Cell D: real denominator (the degenerate "conjugate does nothing")

Example 5 — Cell E: reciprocal and its geometry

Worked example Find

$\frac{1}{z}$ for $z = 1 + 3 i$ , and locate it on the plane. Forecast: $z$ has modulus $2$ . Guess: will $1/ z$ be inside the unit circle or outside?

Step 1. $∣ z ∣^{2} = 1^{2} + (3)^{2} = 1 + 3 = 4$ . Why this step? $\frac{1}{z} = \frac{z ˉ}{∣ z ∣ ^{2}}$ needs $∣ z ∣^{2}$ in the denominator. Step 2. $\overset{z}{ˉ} = 1 - 3 i$ , so $\frac{1}{z} = \frac{1 - 3 i}{4} = \frac{1}{4} - \frac{3}{4} i .$ Step 3 (geometry). $∣ z ∣ = 2$ , so $\frac{1}{z} = \frac{1}{∣ z ∣} = \frac{1}{2}$ . Why? Property $∣ \overset{z}{ˉ} ∣ = ∣ z ∣$ gives $\frac{1}{z} = \frac{∣ z ˉ ∣}{∣ z ∣ ^{2}} = \frac{1}{∣ z ∣}$ . Also the argument flips sign: $z$ has argument $+ 6 0^{\circ}$ , and $1/ z$ has argument $- 6 0^{\circ}$ . Why? Taking a conjugate reflects across the real axis (negates the argument); dividing by a positive real $∣ z ∣^{2}$ doesn't rotate.

Look at the figure: $z$ (plum) sits on the circle of radius $2$ at $+ 6 0^{\circ}$ ; $\overset{z}{ˉ}$ (teal, dotted) is its mirror below the real axis; $1/ z$ (orange) is the shrunk mirror, radius $\frac{1}{2}$ , at $- 6 0^{\circ}$ . Reciprocal = reflect, then shrink to radius $1/∣ z ∣$ .

Verify: $∣1/ z ∣ = 1/2$ and $z \cdot (1/ z) = 1$ . See =VERIFY=.

Example 6 — Cell F: zero / degenerate input

Example 7 — Cell G: real-world (AC circuit impedance)

Worked example In an AC circuit, two impedances

$Z_{1} = 3 + 4 i$ (ohms) and $Z_{2} = 1 - 2 i$ (ohms) are in parallel. The combined impedance is $Z = \frac{Z _{1} Z _{2}}{Z _{1} + Z _{2}}$ . Find $Z$ . Forecast: the answer is a complex number of ohms. Guess whether its real (resistive) part is bigger or smaller than $1 Ω$ .

Step 1. Numerator $Z_{1} Z_{2} = (3 + 4 i) (1 - 2 i) = 3 - 6 i + 4 i - 8 i^{2} = 3 - 2 i + 8 = 11 - 2 i$ . Why this step? Parallel formula needs product on top. Step 2. Denominator $Z_{1} + Z_{2} = (3 + 1) + (4 - 2) i = 4 + 2 i$ . Why? Add real parts, add imaginary parts. Step 3. Now divide: multiply top and bottom by $\overline{4 + 2 i} = 4 - 2 i$ . Why this step? Turn the complex denominator real via $z \overset{z}{ˉ} = ∣ z ∣^{2}$ . $Z = \frac{( 11 - 2 i ) ( 4 - 2 i )}{( 4 + 2 i ) ( 4 - 2 i )}$ Step 4. Denominator $= 4^{2} + 2^{2} = 20$ . Numerator $= (11) (4) + (11) (- 2 i) + (- 2 i) (4) + (- 2 i) (- 2 i) = 44 - 22 i - 8 i + 4 i^{2} = 44 - 30 i - 4 = 40 - 30 i$ . Why? $4 i^{2} = - 4$ . Answer: $Z = \frac{40 - 30 i}{20} = 2 - 1.5 i$ ohms. Resistance $2 Ω$ , reactance $- 1.5 Ω$ (capacitive).

Units check: every quantity is in ohms; ratio $\frac{Ω \cdot Ω}{Ω} = Ω$ . ✓ Verify: $(4 + 2 i) (2 - 1.5 i)$ must equal $11 - 2 i$ . See =VERIFY=.

Example 8 — Cell H: exam twist (solve an equation with a conjugate)

Example 9 — Cell I: property verification across a quotient

Recall Which cell was hardest?

Cell that has NO valid answer? ::: Cell F — division by zero is undefined; $z \overset{z}{ˉ} = 0$ only when $z = 0$ .
Conjugate of a real number? ::: itself (real $\Rightarrow z = \overset{z}{ˉ}$ ).
Conjugate of a purely imaginary number $k i$ ? ::: $- k i$ .
Geometric effect of taking $1/ z$ ? ::: reflect across real axis, then shrink to radius $1/∣ z ∣$ .
A single complex equation = how many real equations? ::: two (real part and imaginary part).

Active recall

What is

\frac{4 + 3 i}{2 + i}

\frac{11}{5} + \frac{2}{5} i

What is

\frac{1 + 2 i}{- 3 + i}

- \frac{1}{10} - \frac{7}{10} i

What is

\frac{5 - 2 i}{4 i}

- \frac{1}{2} - \frac{5}{4} i

Reciprocal of

1 + 3 i

\frac{1}{4} - \frac{3}{4} i

(modulus

\frac{1}{2}

, argument

- 6 0^{\circ}

)

\frac{2 + i}{0}

defined?

No — undefined;

z \overset{z}{ˉ} = 0

only for

z = 0

Parallel impedance of

3 + 4 i

and

1 - 2 i

2 - 1.5 i

ohms.

Solve

\frac{x + i y}{1 - i} = 3 + i

x = 4, y = - 2

Connections

Complex conjugate — properties, applications in division (index 3.5.9)
Complex Numbers — Modulus and Argument
Argand Diagram and Geometric Representation
Polar and Exponential Form
Roots of Complex Numbers
Rationalising Denominators (Real Numbers)
Quadratic Equations with Complex Roots

Worked examples — Complex conjugate — properties, applications in division

The scenario matrix

Example 1 — Cell A: friendly Quadrant I denominator

Example 2 — Cell B: mixed-sign denominator (all quadrants)

Example 3 — Cell C: purely imaginary denominator

Example 4 — Cell D: real denominator (the degenerate "conjugate does nothing")

Example 5 — Cell E: reciprocal and its geometry

Example 6 — Cell F: zero / degenerate input

Example 7 — Cell G: real-world (AC circuit impedance)

Example 8 — Cell H: exam twist (solve an equation with a conjugate)

Example 9 — Cell I: property verification across a quotient

Active recall

Connections

Scenario Map