Algebraic operations — add, subtract, multiply, divide (rectangular and polar): Worked examples

This is the practice ground for the parent topic on adding, subtracting, multiplying, and dividing complex numbers. The parent showed you the rules. Here we hunt down every case those rules can face — every quadrant, every awkward sign, zeros, degenerate inputs, and a couple of exam traps — and work each one to the ground.

The scenario matrix

Every worked example below is tagged with the cell it covers. Together they hit all of these:

Cell	What makes it tricky	Example
A. Add across quadrants	components with mixed signs, result lands in a new quadrant	Ex 1
B. Argument in Quadrant III	plain $arctan$ lies here — must fix by sign	Ex 2
C. Multiply (rect + polar agree)	scale-and-spin; both routes must match	Ex 3
D. Multiply by $i$ (pure rotation)	modulus $1$ , angle $9 0^{\circ}$ — degenerate "no stretch"	Ex 4
E. Divide via conjugate	denominator with sign, split cleanly	Ex 5
F. Zero / degenerate inputs	dividing by zero, adding zero, real-only numbers	Ex 6
G. Limiting / repeated multiply (powers)	angle stacks each time → De Moivre's Theorem	Ex 7
H. Word problem (AC circuit)	complex numbers as real-world quantities	Ex 8
I. Exam twist (find $z$ from an equation)	reverse-engineer a complex unknown	Ex 9

Ex 1 — Cell A · Adding across quadrants

Worked example Add

$z_{1} = 4 + 3 i$ and $z_{2} = - 6 - 7 i$ . Forecast: guess the quadrant of the answer before reading on. (One arrow points up-right, the other down-left and longer — where do we end up?)

Step 1. Add the real (horizontal) parts: $4 + (- 6) = - 2$ . Why this step? Horizontal reach just piles up — sideways travel adds to sideways travel, exactly like Vectors in 2D. The axes never mix.

Step 2. Add the imaginary (vertical) parts: $3 + (- 7) = - 4$ . Why this step? Same reason, one axis over. Up-travel adds to up-travel.

Step 3. Combine: $z_{1} + z_{2} = - 2 - 4 i$ . Why this step? Both parts negative → the tip lands down-left, in Quadrant III. See the tip-to-tail picture below.

Verify: subtract back — $(- 2 - 4 i) - (- 6 - 7 i) = (- 2 + 6) + (- 4 + 7) i = 4 + 3 i = z_{1}$ . ✓ Undoing addition returns the first arrow.

The figure below draws $z_{1}$ from the origin, lays $z_{2}$ tip-to-tail onto its head, and marks the red resultant arrow that reaches the same final point — the geometric meaning of "add the components."

Figure — Algebraic operations — add, subtract, multiply, divide (rectangular and polar)

Ex 2 — Cell B · Argument in Quadrant III (the $arctan$ trap)

Worked example Find the modulus and argument of

$z = - 1 - i$ (so $a = - 1$ , $b = - 1$ ). Forecast: its tail is at the origin and its tip is down-left. Roughly what angle from the positive real axis is that — closer to $4 5^{\circ}$ or past $18 0^{\circ}$ ?

Step 1. Modulus: $r = a^{2} + b^{2} = (- 1)^{2} + (- 1)^{2} = 2 \approx 1.414$ . Why this step? $r$ is just the arrow's length by Pythagoras — signs vanish under squaring, so this part never cares about quadrant.

Step 2. The naive angle: $arctan \frac{b}{a} = arctan \frac{- 1}{- 1} = arctan (1) = 4 5^{\circ}$ . Why this step? To expose the trap. $tan θ = b / a$ is genuinely true, but $tan$ repeats every $18 0^{\circ}$ , so $arctan$ (which only ever answers in $(- 9 0^{\circ}, 9 0^{\circ})$ ) can't tell $4 5^{\circ}$ apart from $22 5^{\circ}$ . It picked the wrong one.

Step 3. Fix by the signs of $a$ and $b$ : $a = - 1 < 0$ and $b = - 1 < 0$ means Quadrant III, so add $18 0^{\circ}$ : $4 5^{\circ} + 18 0^{\circ} = 22 5^{\circ}$ (equivalently $- 13 5^{\circ}$ ). Why this step? When $a < 0$ the true arrow points into the left half-plane; $arctan$ folded it into the right half-plane. Adding $18 0^{\circ}$ flips it back where the signs say it belongs. This is exactly what $atan2 (b, a)$ does automatically. See Modulus and Argument.

Verify: rebuild from polar — $a = r cos 22 5^{\circ} = 2 \cdot (- \frac{1}{2}) = - 1$ ✓ and $b = r sin 22 5^{\circ} = 2 \cdot (- \frac{1}{2}) = - 1$ ✓. The naive $4 5^{\circ}$ would have given $+ 1, + 1$ — the wrong point.

The figure makes the trap visible: the red arrow is the true $z = - 1 - i$ pointing down-left at $22 5^{\circ}$ , while the dashed black arrow is where naive $arctan$ mistakenly places it (up-right at $4 5^{\circ}$ ). The two differ by exactly the $18 0^{\circ}$ that the sign of $a$ tells you to add.

Formula The full sign-to-quadrant table for

$θ = ar g (z)$ Compute the reference angle $φ = arctan \frac{b}{a}$ (always in $[0^{\circ}, 9 0^{\circ})$ ), then read off the true argument by the signs of $a, b$ — this is precisely what $atan2 (b, a)$ builds in:

Quadrant	sign of $a$	sign of $b$	true $θ$
I	$a > 0$	$b > 0$	$φ$
II	$a < 0$	$b > 0$	$18 0^{\circ} - φ$
III	$a < 0$	$b < 0$	$18 0^{\circ} + φ$
IV	$a > 0$	$b < 0$	$- φ$
on $+$ real axis	$a > 0$	$b = 0$	$0^{\circ}$
on $-$ real axis	$a < 0$	$b = 0$	$18 0^{\circ}$
on $+$ imag axis	$a = 0$	$b > 0$	$9 0^{\circ}$
on $-$ imag axis	$a = 0$	$b < 0$	$- 9 0^{\circ}$

Note $a = 0$ is the case where $b / a$ is undefined — you cannot use $arctan$ at all; the axis cases above cover it directly.

Ex 3 — Cell C · Multiply, two routes must agree

Worked example Compute

$(2 + 2 i) (1 + 3 i)$ in rectangular and in polar, and check they match. Forecast: the second factor points at $6 0^{\circ}$ . Will the product's angle be more or less than the first factor's angle?

Step 1. Rectangular (FOIL, then $i^{2} = - 1$ ): $2 \cdot 1 + 23 i + 2 i + 23 i^{2} = 2 + 23 i + 2 i - 23 = (2 - 23) + (2 + 23) i .$ Why this step? Distribute all four products; the term $23 i^{2} = - 23$ slides into the real part — that's the sign-flip that makes multiplication non-obvious.

Step 2. Polar of each factor. $z_{1} = 2 + 2 i$ : $r_{1} = 4 + 4 = 22$ , angle $= 4 5^{\circ}$ (Quadrant I). $z_{2} = 1 + 3 i$ : $r_{2} = 1 + 3 = 2$ , angle $= arctan (3 /1) = 6 0^{\circ}$ . Why this step? Polar exposes the geometry: length and heading of each arrow.

Step 3. Multiply moduli, add angles. $r = 22 \cdot 2 = 42, θ = 4 5^{\circ} + 6 0^{\circ} = 10 5^{\circ} .$ Why this step? This is the whole point of polar (see Euler's formula): $r_{1} e^{i α} \cdot r_{2} e^{i β} = r_{1} r_{2} e^{i (α + β)}$ . Stretch by the other length, spin by the other angle. So yes — the angle grew.

Step 4. Convert polar back to rectangular to compare: $42 cos 10 5^{\circ} \approx - 1.464, 42 sin 10 5^{\circ} \approx 5.464.$ Why this step? The two routes only "agree" if they name the same arrow, but they currently speak different languages — Step 1 gives components, Step 3 gives length-and-angle. Converting the polar result back to components ( $a = r cos θ$ , $b = r sin θ$ ) puts both answers in the same units so we can compare them number for number.

Verify: the rectangular answer numerically is $2 - 23 \approx - 1.464$ and $2 + 23 \approx 5.464$ — matching Step 4 exactly. Geometry (spin-and-stretch) and algebra (FOIL) are the same operation wearing different clothes.

The figure shows both factor arrows ( $z_{1}$ at $4 5^{\circ}$ , $z_{2}$ at $6 0^{\circ}$ ) and the red product arrow standing at $4 5^{\circ} + 6 0^{\circ} = 10 5^{\circ}$ — you can literally see the angles adding.

Ex 4 — Cell D · Multiplying by $i$ is a pure $9 0^{\circ}$ turn

Worked example Compute

$i \cdot (3 + i)$ and describe what happened geometrically. Forecast: $i$ has length $1$ . If multiplying scales by the other length, this multiply won't stretch at all. So what's left for it to do?

Step 1. Rectangular: $i (3 + i) = 3 i + i^{2} = - 1 + 3 i$ . Why this step? Straight distribution; $i^{2} = - 1$ becomes the real part $- 1$ .

Step 2. Read it in polar. $i$ has $r = 1$ , angle $9 0^{\circ}$ . Multiplying by it means: scale by $1$ (no stretch — the degenerate case) and rotate by $9 0^{\circ}$ . Why this step? This isolates rotation with the stretch switched off, so we can see rotation alone.

Step 3. Check the turn. Start point $3 + i$ sits at angle $arctan (1/3) \approx 18.4 3^{\circ}$ . Add $9 0^{\circ}$ → $108.4 3^{\circ}$ . The point $- 1 + 3 i$ has $a = - 1 < 0$ , $b = 3 > 0$ (Quadrant II), so by the sign table its angle is $18 0^{\circ} - arctan (3/1) \approx 108.4 3^{\circ}$ . ✓ Why this step? Confirms multiply-by- $i$ = quarter-turn anticlockwise, no length change. The lengths match too: $9 + 1 = 1 + 9 = 10$ .

Verify: apply $i$ four times: $i^{4} = 1$ , so four quarter-turns = full circle back to start. Indeed $i \cdot i \cdot i \cdot i (3 + i) = 3 + i$ . ✓

The figure shows the original arrow $3 + i$ in black and the red rotated result $- 1 + 3 i$ , with the $9 0^{\circ}$ arc between them — same length, quarter-turn anticlockwise.

Ex 5 — Cell E · Division by the conjugate

Worked example Compute

$\frac{5 + 5 i}{3 - 4 i}$ . Forecast: the denominator has length $5$ . Dividing shrinks the numerator's length by that factor and rotates it. Will the answer be bigger or smaller than $1$ in length?

Step 1. Multiply top and bottom by the conjugate $3 + 4 i$ . Why this step? $(3 - 4 i) (3 + 4 i) = 3^{2} + 4^{2} = 25$ , a real number — we can't "divide by an arrow," so we turn the denominator real first. See Complex conjugate. Multiplying by $\frac{3 + 4 i}{3 + 4 i} = 1$ never changes the value — but you must multiply the whole numerator too.

Step 2. Numerator: $(5 + 5 i) (3 + 4 i) = 15 + 20 i + 15 i + 20 i^{2} = 15 + 35 i - 20 = - 5 + 35 i$ . Why this step? FOIL again; $20 i^{2} = - 20$ lands in the real part.

Step 3. Divide by the real $25$ : $\frac{- 5 + 35 i}{25} = - 0.2 + 1.4 i .$ Why this step? Now that the denominator is a plain number, real and imaginary parts split cleanly.

Step 4. Sanity via polar. $∣5 + 5 i ∣ = 52$ , $∣3 - 4 i ∣ = 5$ , so the answer's length is $\frac{5 2}{5} = 2 \approx 1.414$ . And $(- 0.2)^{2} + 1. 4^{2} = 0.04 + 1.96 = 2$ . ✓ Bigger than $1$ , as forecast.

Verify: multiply back — $(- 0.2 + 1.4 i) (3 - 4 i) = - 0.6 + 0.8 i + 4.2 i - 5.6 i^{2} = - 0.6 + 5 i + 5.6 = 5 + 5 i$ . ✓

Ex 6 — Cell F · Zero and degenerate inputs

Worked example Handle each edge case: (a)

$(7 - 2 i) + 0$ , (b) $(4 + 0 i) (5 + 0 i)$ , (c) $\frac{2 + 3 i}{0}$ . Forecast: which one is a trick with no valid answer?

(a) Adding zero. $0 = 0 + 0 i$ is the zero arrow (a point at the origin). $(7 - 2 i) + (0 + 0 i) = 7 - 2 i$ . Unchanged — adding the zero arrow moves nothing. Why this step? Confirms $0$ is the additive identity even here.

(b) Real times real. $4 + 0 i$ and $5 + 0 i$ are ordinary numbers lying on the real axis (angle $0^{\circ}$ ). Product: $(4) (5) + (imag terms all 0) = 20$ . Angles $0^{\circ} + 0^{\circ} = 0^{\circ}$ — stays on the real axis. Why this step? Shows complex arithmetic contains ordinary arithmetic as the special case $b = d = 0$ ; nothing breaks.

(c) Dividing by zero. $\frac{2 + 3 i}{0}$ is undefined. There is no conjugate trick: the conjugate of $0$ is $0$ , and $0 \cdot 0 = 0$ , so the denominator stays $0$ . Length-wise you'd need $\frac{r _{1}}{0}$ , which blows up. Why this step? Every division example silently assumes $z_{2} \neq = 0$ ; this is the boundary where the method has no answer.

Verify: (a) subtract $0$ back → $7 - 2 i$ ✓. (b) $20$ is real, imaginary part $0$ ✓. (c) no numeric check — undefined is the correct "answer."

Ex 7 — Cell G · Repeated multiply → powers stack the angle

Worked example Compute

$(1 + i)^{4}$ using De Moivre's Theorem. Forecast: each multiply adds $4 5^{\circ}$ . After four, where does the angle point, and is that on an axis?

Step 1. Polar form of the base. $1 + i$ : $r = 2$ , $θ = 4 5^{\circ}$ . Why this step? Powers are just repeated multiplication, and multiplication is cleanest on the "clock."

Step 2. Apply De Moivre: $(r cis θ)^{n} = r^{n} cis (n θ)$ . $(2)^{4} cis (4 \cdot 4 5^{\circ}) = 4 cis (18 0^{\circ}) .$ Why this step? Each factor multiplies the length by $2$ and adds $4 5^{\circ}$ ; doing it $4$ times multiplies length by $(2)^{4} = 4$ and adds $4 \times 4 5^{\circ} = 18 0^{\circ}$ . The angle landed exactly on the negative real axis.

Step 3. Convert back: $4 cis 18 0^{\circ} = 4 (cos 18 0^{\circ} + i sin 18 0^{\circ}) = 4 (- 1 + 0 i) = - 4$ . Why this step? At $18 0^{\circ}$ the arrow lies flat on the real axis pointing left — a real number.

Verify (rectangular, the long way): $(1 + i)^{2} = 2 i$ , so $(1 + i)^{4} = (2 i)^{2} = 4 i^{2} = - 4$ . ✓ The clock and the algebra agree.

The figure plots the successive powers $n = 1, 2, 3, 4$ : each black arrow is $4 5^{\circ}$ further round and $2$ times longer than the last, and the final red arrow at $18 0^{\circ}$ is $(1 + i)^{4} = - 4$ sitting on the negative real axis.

Ex 8 — Cell H · Word problem (AC circuit impedance)

Worked example In an AC circuit, a resistor and inductor in series have impedance

$Z_{1} = 3 + 4 i$ ohms, and a second branch has $Z_{2} = 6 + 8 i$ ohms. In series, impedances add. Find the total impedance in rectangular form, and its magnitude (the quantity an ohmmeter-style measurement reports). Forecast: the two branches point the same direction ( $Z_{2} = 2 Z_{1}$ ). Will the total's angle differ from either branch's?

Step 1. Series = add (Cell A logic, real units): $Z = Z_{1} + Z_{2} = (3 + 6) + (4 + 8) i = 9 + 12 i Ω$ . Why this step? Series impedances add exactly like arrows tip-to-tail — the real part (resistance) and imaginary part (reactance) each pile up independently.

Step 2. Magnitude: $∣ Z ∣ = 9^{2} + 1 2^{2} = 81 + 144 = 225 = 15 Ω$ . Why this step? The magnitude is the arrow's length — physically, the total opposition to current flow. Units stay ohms throughout.

Step 3. Angle (phase): $ar g Z = arctan (12/9) \approx 53.1 3^{\circ}$ , same as each branch since $Z_{2} = 2 Z_{1}$ (same direction, just longer). Why this step? Confirms the forecast: adding parallel-pointing arrows keeps the direction, only the length grows.

Verify: $9^{2} + 1 2^{2} = 225 = 1 5^{2}$ ✓ (a $3$ – $4$ – $5$ triangle scaled by $3$ ). Units: ohms + ohms = ohms. ✓

Ex 9 — Cell I · Exam twist: solve for an unknown $z$

Recall Quick self-test

Which cell does "add $4 + 3 i$ and $- 6 - 7 i$ " belong to, and why is rectangular the right form? ::: Cell A (add across quadrants); rectangular because addition just piles up components independently. In Ex 2, why did $arctan$ give $4 5^{\circ}$ instead of $22 5^{\circ}$ ? ::: $tan$ repeats every $18 0^{\circ}$ ; $arctan$ only answers in $(- 9 0^{\circ}, 9 0^{\circ})$ , so with $a < 0$ you must add $18 0^{\circ}$ . For a point in Quadrant II ( $a < 0$ , $b > 0$ ) with reference angle $φ$ , what is the true argument? ::: $18 0^{\circ} - φ$ . What does multiplying by $i$ do geometrically? ::: Rotates $9 0^{\circ}$ anticlockwise with no change in length (Cell D). Why is $\frac{2 + 3 i}{0}$ undefined? ::: The conjugate of $0$ is $0$ ; the denominator stays $0$ , and length divides by zero (Cell F). After four multiplications by $1 + i$ , why is the result real? ::: The angle stacks $4 \times 4 5^{\circ} = 18 0^{\circ}$ , landing on the real axis (Cell G, De Moivre).

Connections

Parent · Hinglish — the rules these examples exercise.
Complex plane (Argand diagram) — every figure here lives on it.
Modulus and Argument — Ex 2's quadrant fix.
Complex conjugate — the engine of Ex 5 and Ex 9.
Euler's formula — why polar multiply/divide is trivial (Ex 3, 7).
De Moivre's Theorem — powers stack the angle (Ex 7).
Vectors in 2D — addition is vector addition (Ex 1, 8).