Algebraic operations — add, subtract, multiply, divide (rectangular and polar): Question bank

Before you start, recall the two languages this whole topic speaks:

Recall The two forms, in one glance

Rectangular $z = a + bi$ : named by its shadows $a$ (right) and $b$ (up).
Polar $z = r (cos θ + i sin θ) = r e^{i θ}$ : named by length $r = ∣ z ∣$ and angle $θ = ar g z$ .
Add/subtract → rectangular. Multiply/divide → polar. ("Add in the box, multiply on the clock.")

True or false — justify

True or false:

ar g (z_{1} z_{2}) = ar g (z_{1}) + ar g (z_{2})

is always exactly true.

False as an equality of principal values. The angles genuinely add, but

ar g

is usually kept in

(- 18 0^{\circ}, 18 0^{\circ}]

, so the sum may fall outside that window and need a

\pm 36 0^{\circ}

adjustment to land back in range. The geometric rotation is correct; only the label on the angle can differ by a full turn.

True or false:

∣ z_{1} + z_{2} ∣ = ∣ z_{1} ∣ + ∣ z_{2} ∣

for all complex numbers.

False in general. Lengths only add when the two arrows point the same direction (same argument); otherwise the tip-to-tail triangle makes the sum shorter — this is the triangle inequality

∣ z_{1} + z_{2} ∣ \leq ∣ z_{1} ∣ + ∣ z_{2} ∣

, with equality only when arguments match.

True or false: multiplying

z

i

rotates it

9 0^{\circ}

counter-clockwise without changing its length.

True.

i = 1 \cdot (cos 9 0^{\circ} + i sin 9 0^{\circ})

, so it has modulus

1

and argument

9 0^{\circ}

; multiplying scales length by

1

(no stretch) and adds

9 0^{\circ}

to the angle (pure spin).

True or false:

z + \overset{z}{ˉ}

is always a real number.

True. If

z = a + bi

then

\overset{z}{ˉ} = a - bi

, so

z + \overset{z}{ˉ} = 2 a

— the imaginary parts cancel. Geometrically the arrow and its mirror-in-the-real-axis add tip-to-tail to land exactly on the real axis.

True or false:

z \cdot \overset{z}{ˉ}

is always real and non-negative.

True.

(a + bi) (a - bi) = a^{2} + b^{2} = ∣ z ∣^{2} \geq 0

. This is why the conjugate is the key to division — it converts any denominator into a plain non-negative real number. See Complex conjugate.

True or false: if

z_{1} = z_{2}

, then

ar g (z_{1}) = ar g (z_{2})

Not necessarily — the arguments are equal up to a multiple of

36 0^{\circ}

. The same arrow can be labelled

3 0^{\circ}

39 0^{\circ}

- 33 0^{\circ}

; these are the same direction. Equal numbers have equal modulus and equal argument modulo a full turn.

True or false: for any

z \neq = 0

\frac{1}{z}

has argument

- ar g (z)

True. Division subtracts angles, and

1

has argument

0^{\circ}

, so

ar g (1/ z) = 0 - ar g (z) = - ar g (z)

. Its length is

1/∣ z ∣

: the reciprocal flips the angle across the real axis and inverts the length.

True or false: subtraction is easiest in polar form.

False. Subtraction, like addition, works on independent axes — you subtract real parts and imaginary parts. Moduli and angles have no simple "difference" rule, so convert to rectangular first (Vectors in 2D behaves identically).

Spot the error

Student writes: "

ar g (- 1 - i) = arctan (\frac{- 1}{- 1}) = arctan (1) = 4 5^{\circ}

." Find the error.

The point

- 1 - i

sits in quadrant III (both parts negative), but plain

arctan

only returns angles in

(- 9 0^{\circ}, 9 0^{\circ})

, always quadrant I/IV. Because

tan

repeats every

18 0^{\circ}

, you must add

18 0^{\circ}

when

a < 0

: the true argument is

4 5^{\circ} + 18 0^{\circ} = 22 5^{\circ}

(or

- 13 5^{\circ}

). See Modulus and Argument.

Student writes: "

(2 + 3 i) (1 + 4 i) = 2 + 12 i

." Spot the mistake.

They multiplied the real parts and the imaginary parts separately, treating

i

like a second independent axis you can't mix. Multiplication needs FOIL plus

i^{2} = - 1

(2 + 3 i) (1 + 4 i) = 2 + 8 i + 3 i + 12 i^{2} = (2 - 12) + (11) i = - 10 + 11 i

Student computes

\frac{a + bi}{c + d i}

by writing

\frac{a + bi}{c + d i} \cdot \frac{1}{c - d i}

. What went wrong?

They multiplied only the denominator by the conjugate, not the numerator, so they multiplied the fraction by

\frac{1}{c - d i}

instead of by

\frac{c - d i}{c - d i} = 1

. That changes the value. The legal move multiplies top and bottom by

c - d i

Student says "

z_{1} z_{2} = r_{1} r_{2} cis (θ_{1} θ_{2})

." Find the flaw.

The angles are added, not multiplied:

z_{1} z_{2} = r_{1} r_{2} cis (θ_{1} + θ_{2})

. This comes straight from the angle-sum identities for

cos

and

sin

(equivalently

e^{i α} e^{i β} = e^{i (α + β)}

, see Euler's formula).

Student writes "

i^{2} = 1

, so

(1 + i)^{2} = 1 + 2 i + 1 = 2 + 2 i

." Spot the error.

The defining property is

i^{2} = - 1

, not

+ 1

. Correctly

(1 + i)^{2} = 1 + 2 i + i^{2} = 1 + 2 i - 1 = 2 i

— a pure

9 0^{\circ}

rotation-and-stretch, matching the polar view (angle

4 5^{\circ} \to 9 0^{\circ}

, length

2 \to 2

Student converts

z = 3 cis (12 0^{\circ})

to rectangular as "

a = 3 cos 12 0^{\circ} = 1.5

." What is wrong?

cos 12 0^{\circ} = - \frac{1}{2}

, not

+ \frac{1}{2}

, so

a = 3 (- \frac{1}{2}) = - 1.5

. In quadrant II the horizontal shadow points left, so the real part must be negative — dropping the sign puts the arrow in the wrong quadrant.

Student says: "To add

2 cis 3 0^{\circ}

and

2 cis 6 0^{\circ}

, just add:

4 cis 9 0^{\circ}

." Spot the error.

You cannot add moduli or arguments for a sum — those rules belong to multiplication. Convert each to rectangular, add components, then (if wanted) convert back. Adding the angles here would even give the wrong direction.

Why questions

Why is rectangular form the natural home for addition?

Because the two axes (real/horizontal, imaginary/vertical) are independent — the sideways reaches pile up separately from the upward reaches, so you simply add real parts and imaginary parts, exactly like vector components.

Why does the term

a c - b d

(with a minus) show up in the real part of

(a + bi) (c + d i)

The product

bi \cdot d i = b d i^{2} = b d (- 1) = - b d

, and being real it lands in the real part. The minus is the fingerprint of

i^{2} = - 1

Why does multiplying by a complex number of modulus

1

never change the length of

z

Multiplication scales length by the other number's modulus. If that modulus is

1

, the stretch factor is

1

— pure rotation, no size change. Such numbers, like

cos θ + i sin θ

, are exactly the "spin without stretch" operators.

Why does the conjugate trick turn a complex denominator into a real number?

Because

(c + d i) (c - d i) = c^{2} - (d i)^{2} = c^{2} + d^{2}

, a sum of squares with the

i

cancelled. Geometrically the arrow times its mirror-image gives

∣ z ∣^{2}

, always real and non-negative.

Why does dividing complex numbers subtract the arguments?

Division undoes multiplication. If multiplying adds angles, then dividing must remove the added angle, i.e. subtract it:

ar g (z_{1} / z_{2}) = ar g z_{1} - ar g z_{2}

. In exponentials,

\frac{e ^{i α}}{e ^{i β}} = e^{i (α - β)}

Why do the angle-sum trig identities appear when you multiply two polar numbers?

When you FOIL

(cos α + i sin α) (cos β + i sin β)

, the real part becomes

cos α cos β - sin α sin β

and the imaginary part

sin α cos β + cos α sin β

— precisely

cos (α + β)

and

sin (α + β)

. That is why angles add.

Why can the same arrow have infinitely many arguments?

A direction is unchanged by adding whole turns, so

θ

θ + 36 0^{\circ}

θ - 36 0^{\circ}

, … all describe the same arrow. We pick a single principal value (usually

(- 18 0^{\circ}, 18 0^{\circ}]

) to make

ar g

well-defined.

Edge cases

What is

ar g (0)

Undefined. The zero arrow has length

0

and no direction, so no angle points anywhere. Its modulus is

0

but its argument is meaningless.

What happens if you try to divide by

z_{2} = 0

in polar form?

You cannot:

\frac{r _{1}}{r _{2}}

needs

r_{2} \neq = 0

, and

ar g 0

is undefined, so both pieces of the polar rule break. Division by

0

is forbidden for complex numbers just as for reals.

What is the argument of a positive real number like

z = 5

$0^{\circ}$ . It lies on the positive real axis, pointing straight right, so its angle from that axis is zero. Multiplying by it is a pure stretch (by

5

) with no rotation.

What is the argument of a negative real number like

z = - 4

$18 0^{\circ}$ (or equivalently

- 18 0^{\circ}

depending on convention). It points straight left along the real axis; here

a < 0

b = 0

, exactly the case where naive

arctan (0/ a) = 0^{\circ}

fails and the sign of

a

forces the

18 0^{\circ}

correction.

What is the argument of a pure imaginary number like

z = 2 i

$9 0^{\circ}$ . It points straight up the imaginary axis. Note

arctan (b / a)

can't even be formed here since

a = 0

(division by zero) — you must read the angle from the geometry.

What is the argument of

z = - 3 i

$- 9 0^{\circ}$ (i.e.

27 0^{\circ}

). It points straight down;

a = 0

again blocks the

arctan

formula, so use the picture: on the negative imaginary axis the angle is

- 9 0^{\circ}

∣ z ∣ = 1

, what geometric shape do all such

z

trace, and what does multiplying by one do?

They form the unit circle (all arrows of length

1

). Multiplying by any of them rotates without stretching — these are the pure rotation operators, e.g.

e^{i θ}

from Euler's formula.

What is

(cis θ)^{n}

for the degenerate case

n = 0

$1$ . By De Moivre's Theorem,

(cis θ)^{n} = cis (n θ)

, so

n = 0

gives

cis (0) = cos 0 + i sin 0 = 1

— no rotation, unit length, the identity.

Connections

Hinglish version of the parent
Complex plane (Argand diagram) — the arrows every trap is about.
Modulus and Argument — source of the quadrant / $ar g$ traps.
Complex conjugate — the division-trap items.
Euler's formula — why angles add on multiplication.
De Moivre's Theorem — the power edge case.
Vectors in 2D — addition/subtraction mirror.