Complex number a+bi — real part, imaginary part: Question bank

True or false — justify

Each line states a claim. Decide true or false first, then reveal the reasoning.

The imaginary part of

3 + 4 i

is the number

4 i

False. By definition

Im (z)

is the real coefficient

b

, so

Im (3 + 4 i) = 4

— a plain real number. Formulas like

∣ z ∣ = a^{2} + b^{2}

need a real

b

, not

4 i

Every real number is also a complex number.

True. A real number

a

is just

a + 0 i

, i.e. a complex number with imaginary part

0

. So

R

sits inside

C

Every complex number is also a real number.

False. Only those with

Im (z) = 0

are real. Something like

2 i

has no place on the real line, so

C

is strictly bigger than

R

0

is a purely imaginary number.

Debatable/False by usual convention.

0 = 0 + 0 i

has both parts zero; most texts call "purely imaginary" the numbers

bi

with

b \neq = 0

, excluding

0

. It is both purely real and purely imaginary in a trivial sense, but treating it as a special "zero" case avoids confusion.

i

is bigger than

0

False — the question is meaningless. Complex numbers have no ordering that respects arithmetic. If

i > 0

then

i \cdot i > 0

would force

- 1 > 0

; if

i < 0

then

(- i) (- i) > 0

also gives

- 1 > 0

. Either way a contradiction, so "

<

" simply does not extend to

C

z^{2}

is a negative real number, then

z

must be purely imaginary.

True. Writing

z = a + bi

z^{2} = (a^{2} - b^{2}) + 2 ab i

. For this to be a negative real, the imaginary part

2 ab = 0

and the real part

a^{2} - b^{2} < 0

. If

a = 0

we get

- b^{2} < 0

✔; if

b = 0

we'd get

a^{2} \geq 0

✘. So

a = 0

: purely imaginary.

i^{2} = - 1

means

i = - 1

is a single well-defined number.

Careful — mostly false as stated. Both

i

and

- i

square to

- 1

, so "

- 1

" is ambiguous the way

4

isn't. We choose

i

to be one of the two roots and build everything consistently; the symbol

- 1

hides that choice.

Two complex numbers can be equal even if their imaginary parts differ.

False.

a + bi = c + d i

forces

a = c

and

b = d

separately. If

b \neq = d

you could solve for

i

as a real fraction

\frac{a - c}{d - b}

, contradicting

i^{2} = - 1 < 0

Spot the error

Each line contains a plausible-looking argument with one broken step. Name the flaw.

- 4 \cdot - 9 = (- 4) (- 9) = 36 = 6

Flaw: the rule

a b = ab

only holds for non-negative reals. Convert first:

- 4 = 2 i

- 9 = 3 i

, so the product is

6 i^{2} = - 6

, not

6

i^{2} = - 1

, so

i = \frac{- 1}{i}

, and since

\frac{- 1}{i}

is negative,

i < 0

Flaw: "

\frac{- 1}{i}

is negative" assumes complex numbers carry a sign/order. They don't. In fact

\frac{- 1}{i} = \frac{- 1}{i} \cdot \frac{i}{i} = \frac{- i}{- 1} = i

, consistent, and no sign can be attached.

Im (2 - 5 i) = 5

because the imaginary part is the size of the vertical component."

Flaw: the imaginary part keeps its sign.

b = - 5

, so

Im (2 - 5 i) = - 5

. The size

∣ b ∣ = 5

is a different quantity (Argand diagram & modulus).

"Since

i^{2} = - 1

and

i^{4} = 1

, we have

i^{2.5} = i^{2} \cdot i^{0.5} = - i

, a real times

i

Flaw: the period-4 cycle

i^{n} = i^{n mod 4}

is proved only for integer

n

, using

i^{4} = 1

. Fractional powers open a whole can of multi-valued worms and can't be split this casually.

(x + 2) + (y - 1) i = 5

has no solution because the right side has no imaginary part."

Flaw:

5 = 5 + 0 i

. Equality splits into

x + 2 = 5

and

y - 1 = 0

, giving

x = 3, y = 1

. A missing term just means that part is zero, not absent.

3 + 4 i

and

4 + 3 i

are the same number reordered."

Flaw: the

a

and

b

slots are not interchangeable — one lands on the real axis, the other on the imaginary axis. As points

(3, 4)

and

(4, 3)

are different locations on the Argand diagram.

Why questions

Explain the reason, not just the fact.

Why did mathematicians invent

i

instead of just declaring

x^{2} = - 1

has "no solution"?

Because leaving gaps makes algebra inconsistent: many polynomial and physics equations only close up if roots always exist. Adding one symbol with

i^{2} = - 1

makes

C

algebraically complete (Quadratic equations with complex roots).

Why is one complex equation worth two real equations?

The equality rule

a + bi = c + d i ⟺ a = c, b = d

splits it into independent real and imaginary parts. Matching two axes separately gives two constraints from one statement.

Why must the product

(a + bi) (c + d i)

still be of the form

A + B i

Because expanding gives

(a c - b d) + (a d + b c) i

— the

i^{2}

collapses to

- 1

, a real number, so nothing escapes the plane. This closure is what proves

C

is a self-consistent number system (Addition and multiplication of complex numbers).

Why do powers of

i

repeat every 4 and not every 2 or 3?

Because

i^{4} = (i^{2})^{2} = (- 1)^{2} = 1

is the first power returning to

1

, and

i^{1}, i^{2}, i^{3}

are all distinct (

i, - 1, - i

). Geometrically each multiply by

i

is a

9 0^{\circ}

turn, and four turns complete a full circle.

Why can't we order complex numbers like we order the real line?

A useful order would need

x^{2} \geq 0

for all

x

, but

i^{2} = - 1 < 0

breaks that no matter whether we tried

i > 0

i < 0

. The plane simply has no consistent "left-to-right" ranking.

Why do we plot the imaginary part

b

(a real number) on a vertical real axis?

Because

b

is real — it's the real coefficient of

i

. The vertical axis measures "how many steps of

i

", so its tick marks are ordinary real numbers even though the direction represents the imaginary unit.

Edge cases

The degenerate and boundary inputs the definition still has to handle.

What are

Re

and

Im

of the number

0

Both are

0

, since

0 = 0 + 0 i

. It is the unique complex number lying on both axes at once — the origin of the Argand diagram.

- 7

a complex number, and if so what is its imaginary part?

Yes:

- 7 = - 7 + 0 i

, a purely real complex number with

Im = 0

. Real numbers are the horizontal slice of

C

What is

Re (5 i)

0

. Here

5 i = 0 + 5 i

, so it is purely imaginary: it sits on the vertical axis with zero real part.

What is

i^{0}

, and does the period-4 rule cover it?

i^{0} = 1

(any nonzero base to the power

0

). It fits the cycle as the "

r = 0

" case:

i^{4} = i^{0} = 1

, the corner where the clock resets.

Does

Im (z) = 0

tell you

z = 0

No. It only tells you

z

is real.

z = - 7

has

Im = 0

but is far from zero. You need both

Re (z) = 0

and

Im (z) = 0

to conclude

z = 0

a + bi = 0

, what must

a

and

b

be?

Both

0

. Apply the equality rule against

0 + 0 i

: real matches real (

a = 0

) and imaginary matches imaginary (

b = 0

Can a complex number equal its own square?

Yes, in exactly two real cases:

z = 0

and

z = 1

(both satisfy

z^{2} = z

). No genuinely non-real

z

works, since

z^{2} = z \Rightarrow z (z - 1) = 0

forces

z \in {0, 1}

, both real.

Where to go next

Argand diagram & modulus — turns the sign of $b$ and the size $∣ z ∣$ into geometry.
Complex conjugate — the reflection $\overset{z}{ˉ} = a - bi$ that many of these traps quietly use.
Addition and multiplication of complex numbers — the closure argument behind the "Why" section.
Polar form and Euler's formula — where the $9 0^{\circ}$ -turn view of $i$ becomes exact.
Quadratic equations with complex roots — the original reason $i$ was worth inventing.