Complex number a+bi — real part, imaginary part: Exercises

Before we begin, one shared picture. Every complex number $z = a + bi$ is a point in a flat plane: walk $a$ steps sideways (right if positive, left if negative) and $b$ steps up-or-down (up if positive, down if negative). This is the Argand diagram — see Argand diagram & modulus. Keep this picture in mind; several problems below live on it.

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

Level 1 — Recognition

(Can you read off the parts and use the basic definition?)

Recall Solution 1.1

WHAT we do: match $z$ to the template $a + bi$ literally. Here $a = - 6$ (the part with no $i$ ) and $b = 11$ (the real number multiplying $i$ ). WHY the sign stays with $a$ : the minus belongs to the $- 6$ , and $Re$ is just that whole real chunk. $Re (z) = - 6, Im (z) = 11. ✅$ Note $Im (z) = 11$ , not $11 i$ .

Recall Solution 1.2

Write it fully as $z = 0 + 7 i$ . WHY: there is no sideways step, so $a = 0$ . $Re (z) = 0, Im (z) = 7.$ Because $a = 0$ but $b \neq = 0$ , this is purely imaginary — it sits exactly on the vertical (imaginary) axis. ✅

Recall Solution 1.3

The tool: powers of $i$ repeat every 4 steps because $i^{4} = (i^{2})^{2} = (- 1)^{2} = 1$ . So only the remainder $n mod 4$ matters.

$i^{3} = i^{2} \cdot i = (- 1) \cdot i = - i .$
$6 mod 4 = 2$ , so $i^{6} = i^{2} = - 1.$ $i^{3} = - i, i^{6} = - 1. ✅$

Level 2 — Application

(Apply the arithmetic rules — add, multiply, split equalities.)

Recall Solution 2.1

WHAT: add real parts to real parts, imaginary to imaginary — exactly like grouping "sideways steps" and "up steps" separately. $(4 + (- 1)) + (- 3 + 8) i = 3 + 5 i . ✅$ WHAT IT LOOKS LIKE: on the Argand diagram this is placing the two arrows tip-to-tail; see Addition and multiplication of complex numbers.

Recall Solution 2.2

WHAT: multiply out all four products (FOIL), then use $i^{2} = - 1$ . $= 2 \cdot 3 + 2 \cdot (- 4 i) + 5 i \cdot 3 + 5 i \cdot (- 4 i)$ $= 6 - 8 i + 15 i - 20 i^{2} .$ WHY the last term flips sign: $- 20 i^{2} = - 20 (- 1) = + 20$ . Regroup real and imaginary: $= (6 + 20) + (- 8 + 15) i = 26 + 7 i . ✅$

Recall Solution 2.3

The tool — equality rule: two complex numbers are equal exactly when their real parts match AND their imaginary parts match. WHY: if the imaginary parts differed, $i$ would have to equal a real fraction, which is impossible since $i^{2} = - 1 < 0$ . $3 x - 1 = 8 ⟹ x = 3, y + 4 = - 2 ⟹ y = - 6.$ $x = 3, y = - 6. ✅$

Level 3 — Analysis

(Reason about structure: cycles, conjugates, why signs behave as they do.)

Recall Solution 3.1

For $i^{2025}$ : find $2025 mod 4$ . Since $2024 = 4 \times 506$ is divisible by 4, the remainder is $1$ . $i^{2025} = i^{1} = i .$ For $i^{- 1}$ : a negative power means "reciprocal". Multiply top and bottom by $i$ (a mini use of the conjugate idea): $i^{- 1} = \frac{1}{i} = \frac{1}{i} \cdot \frac{i}{i} = \frac{i}{i ^{2}} = \frac{i}{- 1} = - i .$ WHY this trick: we don't like $i$ in a denominator, so we make the denominator real ( $i^{2} = - 1$ ). See Complex conjugate for the general version. $i^{2025} = i, i^{- 1} = - i . ✅$

Recall Solution 3.2

Here $z = 5 - 12 i$ , so $\overset{z}{ˉ} = 5 + 12 i$ (flip the sign of the $- 12$ ). Sum: $z + \overset{z}{ˉ} = (5 - 12 i) + (5 + 12 i) = 10.$ The imaginary parts cancel — the sum of a number and its conjugate is always twice the real part, a real number. Product: $z \overset{z}{ˉ} = (5 - 12 i) (5 + 12 i) = 25 + 60 i - 60 i - 144 i^{2} = 25 + 144 = 169.$ WHY it comes out real and positive: this is $a^{2} + b^{2} = 5^{2} + 1 2^{2}$ , which is $∣ z ∣^{2}$ , the squared distance from the origin. $z + \overset{z}{ˉ} = 10, z \overset{z}{ˉ} = 169. ✅$

Recall Solution 3.3

WHY be careful: the rule $a b = ab$ only holds for non-negative reals. For negatives we must convert to $i$ -form first. $- 16 = 4 i, - 25 = 5 i .$ $4 i \cdot 5 i = 20 i^{2} = - 20. ✅$

Level 4 — Synthesis

(Combine equality-splitting, multiplication, and algebra to solve for unknowns.)

Recall Solution 4.1

WHAT: expand the left side into $A + B i$ form, then use the equality rule to get two real equations. $(a + bi) (1 + i) = a + ai + bi + b i^{2} = (a - b) + (a + b) i .$ Match to $3 + 5 i$ : $a - b = 3, a + b = 5.$ WHY two equations: one complex equation always splits into a real pair. Add them: $2 a = 8 \Rightarrow a = 4$ . Then $b = a - 3 = 1$ . $a = 4, b = 1. ✅$ Check: $(4 + i) (1 + i) = 4 + 4 i + i + i^{2} = 3 + 5 i .$ ✔

Recall Solution 4.2

The tool — multiply by the conjugate of the bottom. WHY: $(2 - i) (2 + i) = 4 - i^{2} = 5$ is a real number, clearing $i$ from the denominator. $\frac{3 + i}{2 - i} \cdot \frac{2 + i}{2 + i} = \frac{( 3 + i ) ( 2 + i )}{5} .$ Numerator: $6 + 3 i + 2 i + i^{2} = 6 + 5 i - 1 = 5 + 5 i .$ $= \frac{5 + 5 i}{5} = 1 + i . ✅$

Recall Solution 4.3

WHY $i$ is needed: no real number squares to a negative. We write $- 49 = 49 \cdot (- 1)$ and use $- 1 = i$ . $x = \pm - 49 = \pm 7 i .$ Two roots: $x = 7 i$ and $x = - 7 i$ . Check: $(7 i)^{2} = 49 i^{2} = - 49$ ✔, and $(- 7 i)^{2} = 49 i^{2} = - 49$ ✔. $x = 7 i or x = - 7 i . ✅$ This is the doorway to Quadratic equations with complex roots.

Level 5 — Mastery

(Full multi-step problems mixing every idea, including geometry.)

Recall Solution 5.1

Step 1 — find $z^{2}$ . $z^{2} = (1 + i)^{2} = 1 + 2 i + i^{2} = 1 + 2 i - 1 = 2 i .$ Step 2 — square that. $z^{4} = (z^{2})^{2} = (2 i)^{2} = 4 i^{2} = - 4.$ WHAT IT LOOKS LIKE: $z^{4} = - 4 = - 4 + 0 i$ is a point $4$ units left of the origin on the real axis. Each multiplication by $z = 1 + i$ rotated the point $4 5^{\circ}$ counter-clockwise and stretched it — four such turns give $18 0^{\circ}$ , landing on the negative real axis. (Full story in Polar form and Euler's formula.) $z^{4} = - 4. ✅$

Recall Solution 5.2

Step 1 — expand the left side. $(x + y i)^{2} = x^{2} + 2 x y i + y^{2} i^{2} = (x^{2} - y^{2}) + (2 x y) i .$ Step 2 — split by the equality rule. $x^{2} - y^{2} = 3 (A), 2 x y = 4 (B) .$ Step 3 — from (B): $y = \frac{2}{x}$ (valid since $x = 0$ would force $0 = 4$ in B, impossible). Substitute into (A): $x^{2} - \frac{4}{x ^{2}} = 3 ⟹ x^{4} - 3 x^{2} - 4 = 0.$ Step 4 — let $u = x^{2}$ : $u^{2} - 3 u - 4 = 0 \Rightarrow (u - 4) (u + 1) = 0$ . Since $u = x^{2} \geq 0$ (real $x$ ), we keep $u = 4$ , so $x^{2} = 4$ , $x = \pm 2$ . Step 5 — back-substitute. If $x = 2$ , then $y = 2/2 = 1$ . If $x = - 2$ , then $y = 2/ (- 2) = - 1$ . $(x, y) = (2, 1) or (- 2, - 1) . ✅$ Check: $(2 + i)^{2} = 4 + 4 i + i^{2} = 3 + 4 i$ ✔.

Recall Solution 5.3

Step 1 — encode the first condition. $Re = Im$ means $a = b$ , so write $z = a + ai = a (1 + i)$ . Step 2 — square. $z^{2} = a^{2} (1 + i)^{2} = a^{2} \cdot 2 i = 2 a^{2} i .$ Step 3 — match to $18 i$ . We need $2 a^{2} = 18$ , i.e. $a^{2} = 9$ , so $a = \pm 3$ . Step 4 — write both answers. $z = 3 + 3 i or z = - 3 - 3 i . ✅$ Check: $(3 + 3 i)^{2} = 9 + 18 i + 9 i^{2} = 18 i$ ✔. And $(- 3 - 3 i)^{2}$ gives the same, since squaring kills the overall sign.

Recap ladder

Recall Self-test: one line per level (click to reveal)

L1 — $Im (- 6 + 11 i)$ ::: $11$ (a real number, not $11 i$ ) L2 — $(2 + 5 i) (3 - 4 i)$ ::: $26 + 7 i$ L3 — $z \overset{z}{ˉ}$ for $z = 5 - 12 i$ ::: $169$ L4 — $\frac{3 + i}{2 - i}$ ::: $1 + i$ L5 — solve $(x + y i)^{2} = 3 + 4 i$ ::: $(x, y) = (2, 1)$ or $(- 2, - 1)$

Connections

Argand diagram & modulus — the plane every point above lives in; $∣ z ∣ = a^{2} + b^{2}$ appeared in 3.2.
Complex conjugate — the engine behind division (4.2) and $z \overset{z}{ˉ}$ (3.2).
Addition and multiplication of complex numbers — the L2 arithmetic rules.
Polar form and Euler's formula — the rotation view of 5.1 and the shortcut for 5.2.
Quadratic equations with complex roots — where $x^{2} = - 49$ (4.3) generalises.