Complex number a+bi — real part, imaginary part: Worked examples

The scenario matrix

Here is every kind of thing a question about $a + bi$ can be. Each row is a "cell"; every worked example below is tagged with the cell it covers.

Cell	Case class	What makes it tricky	Example
C1	Both parts positive, straight read-off	none — the warm-up	Ex 1
C2	Negative imaginary part (sign traps)	the minus belongs to $b$	Ex 2
C3	Degenerate: purely real / purely imaginary / zero	one part is $0$	Ex 3
C4	High power of $i$ (cyclic, incl. negative exponent)	remainder mod 4, negative $n$	Ex 4
C5	Product returned to $A + B i$ (all sign combos)	replacing $i^{2} = - 1$ , regrouping	Ex 5
C6	The $- a$ trap (limiting/illegal move)	$a b = ab$ fails	Ex 6
C7	Equality rule → two real equations	splitting one equation into two	Ex 7
C8	Real-world word problem (AC circuit / phase)	translating words into $a + bi$	Ex 8
C9	Exam-style twist: unknown inside a product	solve for real $x, y$ after multiplying	Ex 9

Example 1 — Straight read-off `[C1]`

Example 2 — Negative imaginary part `[C2]`

Example 3 — Degenerate inputs `[C3]`

Figure — Complex number a+bi — real part, imaginary part

Example 4 — Powers of $i$ , including a negative exponent `[C4]`

Worked example Compute

$i^{50}$ and $i^{- 7}$ . Forecast: both should collapse to one of ${1, i, - 1, - i}$ . Which?

Powers of $i$ cycle every 4: $i^{1} = i, i^{2} = - 1, i^{3} = - i, i^{4} = 1$ , then repeat. So $i^{n} = i^{(n mod 4)}$ . Why this step? $i^{n} = i^{4 q + r} = (i^{4})^{q} i^{r} = 1^{q} i^{r} = i^{r}$ — only the remainder $r = n mod 4$ survives.
$50 mod 4 = 2$ (since $48$ is divisible by 4), so $i^{50} = i^{2} = - 1$ . Why this step? Discard whole loops of 4; two extra steps land on $- 1$ .
For $i^{- 7}$ : add multiples of 4 to make the exponent land in ${0, 1, 2, 3}$ . $- 7 + 8 = 1$ , so $i^{- 7} = i^{1} = i$ . Why this step? $i^{4} = 1$ , so multiplying by $i^{4}$ (or $i^{8}$ ) changes nothing — it just re-labels a negative power as a positive one in the same cycle.

Verify: $i^{50} \cdot i^{- 7} \cdot i^{- 43}$ should be $i^{0} = 1$ . Numerically $(- 1) \cdot i \cdot i^{(- 43 mod 4 = 1)} = (- 1) (i) (i) = (- 1) (i^{2}) = (- 1) (- 1) = 1$ . ✅ (See figure: the four powers as clock positions ticking counter-clockwise.)

Example 5 — Product back to $A + B i$ , mixed signs `[C5]`

Worked example Compute

$(- 2 + 3 i) (4 - 5 i)$ . Forecast: guess whether the real part comes out positive or negative before computing.

Distribute (FOIL) all four products: $(- 2) (4) + (- 2) (- 5 i) + (3 i) (4) + (3 i) (- 5 i) = - 8 + 10 i + 12 i - 15 i^{2} .$ Why this step? Multiplication in $C$ is ordinary distribution — the only new fact is what happens to $i^{2}$ .
Replace $i^{2} = - 1$ : $- 15 i^{2} = - 15 (- 1) = + 15$ . Why this step? This single substitution is the entire reason products stay inside $C$ (closure). See Addition and multiplication of complex numbers.
Regroup real and imaginary: $(- 8 + 15) + (10 + 12) i = 7 + 22 i$ . Why this step? Real chunks together, $i$ -chunks together — the answer must be a single $A + B i$ .

Verify (general formula): $(a + bi) (c + d i) = (a c - b d) + (a d + b c) i$ with $a = - 2, b = 3, c = 4, d = - 5$ : $a c - b d = - 8 - (- 15) = 7$ ; $a d + b c = 10 + 12 = 22$ . Gives $7 + 22 i$ . ✅

Example 6 — The $- a$ trap (illegal move) `[C6]`

Example 7 — Equality rule → two real equations `[C7]`

Example 8 — Word problem: two AC voltages `[C8]`

Worked example In an AC circuit each voltage is written as a complex number

$a + bi$ (real part = "in-phase", imaginary part = "out-of-phase"). Source A gives $30 + 40 i$ volts, source B gives $10 - 25 i$ volts, wired in series. Find the combined voltage and its magnitude. Forecast: in series the voltages add; guess the sign of the combined imaginary part first.

Add real to real and imaginary to imaginary: $(30 + 10) + (40 - 25) i = 40 + 15 i$ volts. Why this step? Series voltages superpose linearly; addition of $a + bi$ is componentwise, matching the physics.
Magnitude $∣ z ∣ = a^{2} + b^{2} = 4 0^{2} + 1 5^{2} = 1600 + 225 = 1825$ . Why this step? The magnitude is the true "size" of the voltage phasor — the straight-line distance from the origin on the Argand diagram & modulus.
$1825 = 25 \cdot 73 = 573 \approx 42.72$ volts. Why this step? Pull out the perfect square $25$ to simplify; the decimal is the measured peak amplitude.

Verify (units + sanity): all terms are volts, so the sum is volts and $∣ z ∣$ is volts. ✅ The magnitude $42.72$ V is larger than either individual real part ( $30$ , $10$ ) but the parts partially cancel out-of-phase — physically reasonable.

Example 9 — Exam twist: unknowns inside a product `[C9]`

Worked example Find real

$x, y$ such that $(x + y i) (2 - i) = 3 + 4 i$ . Forecast: you will multiply out first, then use the equality rule from Ex 7.

Expand the left side: $(x + y i) (2 - i) = 2 x - x i + 2 y i - y i^{2}$ . Why this step? We must reach a single $A + B i$ before we can match parts.
Replace $i^{2} = - 1$ and regroup: $- y i^{2} = + y$ , so $= (2 x + y) + (2 y - x) i$ . Why this step? Collect real chunk $(2 x + y)$ and imaginary chunk $(2 y - x)$ separately.
Apply equality rule to $(2 x + y) + (2 y - x) i = 3 + 4 i$ : $2 x + y = 3, - x + 2 y = 4.$ Why this step? One complex equation splits into two real linear equations (exactly as in Ex 7).
Solve the system. From the first, $y = 3 - 2 x$ . Substitute: $- x + 2 (3 - 2 x) = 4 \Rightarrow - x + 6 - 4 x = 4 \Rightarrow - 5 x = - 2 \Rightarrow x = \frac{2}{5}$ . Then $y = 3 - 2 \cdot \frac{2}{5} = \frac{11}{5}$ . Why this step? Standard substitution; the fractions are fine — nothing forces $x, y$ to be integers.

Verify: with $x = \frac{2}{5}, y = \frac{11}{5}$ : real part $2 x + y = \frac{4}{5} + \frac{11}{5} = \frac{15}{5} = 3$ ✅; imaginary part $2 y - x = \frac{22}{5} - \frac{2}{5} = \frac{20}{5} = 4$ ✅. So $(x + y i) (2 - i) = 3 + 4 i$ .

Recall

Recall Which cell of the matrix does each skill belong to? (click)

Negative imaginary part is which cell? ::: C2 (the minus lives in $b$ ) Simplifying $i^{- 7}$ is which cell? ::: C4 (powers cycle, add multiples of 4) Why does $- 2 - 8 = - 4$ not $+ 4$ ? ::: C6 — convert to $i$ first; the shortcut loses a factor $i^{2} = - 1$ One complex equation gives how many real equations? ::: Two (equality rule, cell C7/C9)

Connections

3.5.02 Complex number a+bi — real part, imaginary part (Hinglish) — the parent topic these examples drill.
Argand diagram & modulus — the plotting and $∣ z ∣ = a^{2} + b^{2}$ used in Ex 3 and Ex 8.
Addition and multiplication of complex numbers — the arithmetic behind Ex 5 and Ex 9.
Complex conjugate — why the sign of $Im$ (Ex 2) matters.
Polar form and Euler's formula — the "phasor" view underlying the AC word problem (Ex 8).
Quadratic equations with complex roots — where the $- a$ move (Ex 6) shows up for real.