Algebra of Complex Numbers

Addition, Subtraction and Their Laws

For $z_1 = a + ib$ and $z_2 = c + id$:

$$z_1 + z_2 = (a + c) + i(b + d)$$

Add real parts together, imaginary parts together. Example: $(2 + i3) + (-6 + i5) = -4 + i8$.

Addition obeys all the familiar laws: closure (the sum is again complex), commutativity ($z_1 + z_2 = z_2 + z_1$), associativity, the additive identity $0 = 0 + i0$ (with $z + 0 = z$), and the additive inverse $-z = -a + i(-b)$ (with $z + (-z) = 0$).

Subtraction is addition of the inverse:

$$z_1 - z_2 = z_1 + (-z_2) = (a - c) + i(b - d)$$

Example: $(6 + 3i) - (2 - i) = 4 + 4i$, while $(2 - i) - (6 + 3i) = -4 - 4i$ — order matters, exactly as with reals.

[Board Important] These property names (closure, commutative law, associative law, existence of identity/inverse) are quotable theory — one of them routinely appears as a 1-mark "state the law" question.

Multiplication and Division

Multiplication

$$z_1 z_2 = (ac - bd) + i(ad + bc)$$

In practice, nobody memorises this — just expand with the distributive law (FOIL) and replace $i^2$ by $-1$:

$$(3 + i5)(2 + i6) = 6 + 18i + 10i + 30i^2 = 6 + 28i - 30 = -24 + 28i$$

Multiplication is closed, commutative, associative, distributes over addition, has identity $1 = 1 + i0$, and every NON-ZERO z has a multiplicative inverse:

$$z^{-1} = \frac{a}{a^2 + b^2} + i\,\frac{-b}{a^2 + b^2}$$

(the next section will reveal this as $\frac{\bar{z}}{|z|^2}$).

Division

$$\frac{z_1}{z_2} = z_1 \cdot \frac{1}{z_2} \qquad (z_2 \neq 0)$$

Worked model: $\frac{6 + 3i}{2 - i} = (6 + 3i) \cdot \frac{2 + i}{5} = \frac{(12 - 3) + i(6 + 6)}{5} = \frac{9 + 12i}{5}$.

[JEE Tip] Never divide by expanding blindly — multiply top and bottom by the conjugate of the denominator (Section 3's technique). It converts every division into one real division.

Identities — Real Algebra Survives

Because complex multiplication obeys the same laws as real multiplication, the classical identities carry over verbatim. The first follows from the distributive and commutative laws:

$$(z_1 + z_2)^2 = z_1^2 + 2z_1z_2 + z_2^2$$

and lists its siblings:

• $(z_1 - z_2)^2 = z_1^2 - 2z_1z_2 + z_2^2$
• $(z_1 + z_2)^3 = z_1^3 + 3z_1^2z_2 + 3z_1z_2^2 + z_2^3$
• $(z_1 - z_2)^3 = z_1^3 - 3z_1^2z_2 + 3z_1z_2^2 - z_2^3$
• $z_1^2 - z_2^2 = (z_1 + z_2)(z_1 - z_2)$

In fact every polynomial identity true for all reals remains true for all complex numbers.

One new phenomenon with no real counterpart: sums of squares now FACTOR —

$$a^2 + b^2 = (a + ib)(a - ib)$$

This is why $x^2 + 1 = (x + i)(x - i)$, and it will power the modulus identity $z\bar{z} = |z|^2$.

[JEE Tip] Binomial expansions of $(1 + i)^n$ combine with $i$-cycles for quick evaluations: $(1 + i)^2 = 2i$, so $(1 + i)^4 = -4$, $(1 + i)^8 = 16$. Memorise $(1 \pm i)^2 = \pm 2i$ — it collapses whole problems.

Solved Examples

Example 1: Products with i

Express in the form $a + ib$: (i) $(-5i)\left(\frac{1}{8}i\right)$ (ii) $(-i)(2i)\left(-\frac{1}{8}i\right)^3$.

Solution:

Step 1 — (i) Multiply coefficients and powers of i separately. $(-5)\left(\frac{1}{8}\right) = -\frac{5}{8}$ and $i \cdot i = i^2 = -1$.

Step 2 — Combine. $-\frac{5}{8} \cdot (-1) = \frac{5}{8} = \frac{5}{8} + i0$.

Step 3 — (ii) Handle the first pair. $(-i)(2i) = -2i^2 = 2$.

Step 4 — Cube the third factor. $\left(-\frac{1}{8}i\right)^3 = -\frac{1}{512}i^3 = -\frac{1}{512}(-i) = \frac{1}{512}i$.

Step 5 — Multiply. $2 \times \frac{1}{512}i = \frac{1}{256}i = 0 + \frac{1}{256}i$.

Takeaway: Collect the numeric coefficients, then reduce the power of i separately.

Example 2: A cube

Express $(5 - 3i)^3$ in the form $a + ib$.

Solution:

Step 1 — Set up the binomial expansion. $(5 - 3i)^3 = 5^3 - 3 \cdot 5^2(3i) + 3 \cdot 5(3i)^2 - (3i)^3$.

Step 2 — Evaluate the powers of 3i. $(3i)^2 = 9i^2 = -9$ and $(3i)^3 = 27i^3 = -27i$.

Step 3 — Substitute. $125 - 225i + 15(-9) - (-27i) = 125 - 225i - 135 + 27i$.

Step 4 — Collect. Real: $125 - 135 = -10$; imaginary: $-225 + 27 = -198$: answer $-10 - 198i$.

Takeaway: $(3i)^2 = -9$ and $(3i)^3 = -27i$ — powers of the whole term 3i, signs from the i-cycle.

Example 3: Surds and i together

Express $(-\sqrt{3} + \sqrt{-2})(2\sqrt{3} - i)$ in the form $a + ib$.

Solution:

Step 1 — Convert the negative radicand first. $\sqrt{-2} = \sqrt{2}i$, so the first factor is $-\sqrt{3} + \sqrt{2}i$.

Step 2 — Expand with FOIL. $(-\sqrt{3})(2\sqrt{3}) + (-\sqrt{3})(-i) + (\sqrt{2}i)(2\sqrt{3}) + (\sqrt{2}i)(-i) = -6 + \sqrt{3}i + 2\sqrt{6}i - \sqrt{2}i^2$.

Step 3 — Replace $i^2$ and collect. $-\sqrt{2}i^2 = +\sqrt{2}$: answer $(\sqrt{2} - 6) + (\sqrt{3} + 2\sqrt{6})i$.

Takeaway: i-form first, FOIL second, collect real/imaginary last — the invariable order.

Example 4: Linear combinations

Express $3(7 + i7) + i(7 + i7)$ in the form $a + ib$.

Solution:

Step 1 — Expand both products. $21 + 21i + 7i + 7i^2$.

Step 2 — Replace $i^2$ and collect. $21 + 28i - 7 = 14 + 28i$.

Step 3 — Cross-check by factoring. $(3 + i)(7 + 7i) = 7(3 + i)(1 + i) = 7(3 + 3i + i - 1) = 7(2 + 4i) = 14 + 28i$ ✓.

Takeaway: Factoring the common bracket first is the slicker route — both give the same answer.

Example 5: Subtraction chain

Express $(1 - i) - (-1 + i6)$ in the form $a + ib$.

Solution:

Step 1 — Distribute the minus over BOTH parts. $(1 - i) + 1 - 6i$.

Step 2 — Collect. Real: $1 + 1 = 2$; imaginary: $-1 - 6 = -7$: answer $2 - 7i$.

Takeaway: The minus flips BOTH parts of the second number.

Example 6: Fractional parts

Express $\left(\frac{1}{5} + i\frac{2}{5}\right) - \left(4 + i\frac{5}{2}\right)$ in the form $a + ib$.

Solution:

Step 1 — Subtract the real parts. $\frac{1}{5} - 4 = \frac{1 - 20}{5} = -\frac{19}{5}$.

Step 2 — Subtract the imaginary parts. $\frac{2}{5} - \frac{5}{2} = \frac{4 - 25}{10} = -\frac{21}{10}$.

Step 3 — Assemble. $-\frac{19}{5} - \frac{21}{10}i$.

Takeaway: Work each part as ordinary fraction arithmetic; the i never mixes in.

Example 7: The fourth power of (1 − i)

Express $(1 - i)^4$ in the form $a + ib$.

Solution:

Step 1 — Square once. $(1 - i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i$.

Step 2 — Square again. $(1 - i)^4 = (-2i)^2 = 4i^2 = -4$.

Step 3 — Standard form. $-4 + 0i$.

Takeaway: Square-then-square beats the full binomial — and $(1 \pm i)^2 = \pm 2i$ is the key shortcut of the chapter.

Example 8: A cube with fractions

Express $\left(\frac{1}{3} + 3i\right)^3$ in the form $a + ib$.

Solution:

Step 1 — Set up the binomial. $\left(\frac{1}{3}\right)^3 + 3\left(\frac{1}{3}\right)^2(3i) + 3 \cdot \frac{1}{3}(3i)^2 + (3i)^3$.

Step 2 — Evaluate term by term. $\frac{1}{27} + 3 \cdot \frac{1}{9} \cdot 3i + (3i)^2 + (3i)^3 = \frac{1}{27} + i + (-9) + (-27i)$.

Step 3 — Collect. Real: $\frac{1}{27} - 9 = \frac{1 - 243}{27} = -\frac{242}{27}$; imaginary: $1 - 27 = -26$: answer $-\frac{242}{27} - 26i$.

Takeaway: Fractions stay in the real part; keep them exact.

Example 9: Another cube

Express $\left(-2 - \frac{1}{3}i\right)^3$ in the form $a + ib$.

Solution:

Step 1 — Pull out the minus. $\left(-2 - \frac{1}{3}i\right)^3 = -\left(2 + \frac{1}{3}i\right)^3$, since $(-1)^3 = -1$.

Step 2 — Binomial on the bracket. $8 + 3(4)\frac{i}{3} + 3(2)\frac{i^2}{9} + \frac{i^3}{27} = 8 + 4i - \frac{2}{3} - \frac{i}{27}$.

Step 3 — Collect the bracket. Real: $8 - \frac{2}{3} = \frac{22}{3}$; imaginary: $4 - \frac{1}{27} = \frac{108 - 1}{27} = \frac{107}{27}$: bracket $= \frac{22}{3} + \frac{107}{27}i$.

Step 4 — Negate. Answer $= -\frac{22}{3} - \frac{107}{27}i$.

Takeaway: Pulling out $(-1)^3 = -1$ first keeps every binomial term positive-signed and slip-free.

Example 10: Verifying an identity numerically

Verify $z_1^2 - z_2^2 = (z_1 + z_2)(z_1 - z_2)$ for $z_1 = 2 + i$, $z_2 = 1 - i$.

Solution:

Step 1 — Compute the left side. $z_1^2 = (2 + i)^2 = 4 + 4i + i^2 = 3 + 4i$; $z_2^2 = (1 - i)^2 = -2i$. LHS $= (3 + 4i) - (-2i) = 3 + 6i$.

Step 2 — Compute the right side. $z_1 + z_2 = 3 + 0i = 3$ and $z_1 - z_2 = 1 + 2i$: RHS $= 3(1 + 2i) = 3 + 6i$.

Step 3 — Compare. LHS = RHS ✓.

Takeaway: Identities proved abstractly can always be spot-checked with small numbers — a habit that catches expansion errors.

Example 11: Division worked in full

Express $\frac{6 + 3i}{2 - i}$ in the form $a + ib$.

Solution:

Step 1 — Multiply top and bottom by the conjugate of the denominator. $\frac{(6 + 3i)(2 + i)}{(2 - i)(2 + i)}$.

Step 2 — Expand the denominator. $(2 - i)(2 + i) = 4 - i^2 = 4 + 1 = 5$ — a positive real, as always.

Step 3 — Expand the numerator. $12 + 6i + 6i + 3i^2 = 12 + 12i - 3 = 9 + 12i$.

Step 4 — Divide. $\frac{9 + 12i}{5} = \frac{9}{5} + \frac{12}{5}i$.

Takeaway: The denominator became $2^2 + 1^2 = 5$ — always a positive real; that is the whole point of the conjugate trick.

Example 12: Nested arithmetic

Express $\left[\left(\frac{1}{3} + i\frac{7}{3}\right) + \left(4 + i\frac{1}{3}\right)\right] - \left(-\frac{4}{3} + i\right)$ in the form $a + ib$.

Solution:

Step 1 — Do the inner sum first. $\left(\frac{1}{3} + 4\right) + i\left(\frac{7}{3} + \frac{1}{3}\right) = \frac{13}{3} + \frac{8}{3}i$.

Step 2 — Subtract, flipping both parts. $\left(\frac{13}{3} + \frac{4}{3}\right) + i\left(\frac{8}{3} - 1\right)$.

Step 3 — Collect. $\frac{17}{3} + \frac{5}{3}i$.

Takeaway: Brackets first, one operation at a time; the parts never interact.

Example 13: Powers via (1 + i)² = 2i

Evaluate $\left(\frac{1 + i}{1 - i}\right)^{3}$, and then $(1 + i)^6$.

Solution:

Step 1 — Simplify the base. $\frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i} = \frac{(1 + i)^2}{1 + 1} = \frac{2i}{2} = i$.

Step 2 — Cube. $i^3 = -i$.

Step 3 — Second part: nest the square. $(1 + i)^6 = \left[(1 + i)^2\right]^3 = (2i)^3 = 8i^3 = -8i$.

Takeaway: $\frac{1+i}{1-i} = i$ and $\frac{1-i}{1+i} = -i$ — two mini-results JEE assumes you know on sight.

Example 14: Real and imaginary parts of a product

For any two complex numbers, prove that Re($z_1 z_2$) = Re $z_1$ Re $z_2$ $-$ Im $z_1$ Im $z_2$.

Solution:

Step 1 — Name the parts. $z_1 = a + ib$, $z_2 = c + id$, so Re $z_1$ = a, Im $z_1$ = b, Re $z_2$ = c, Im $z_2$ = d.

Step 2 — Multiply. $z_1 z_2 = (ac - bd) + i(ad + bc)$.

Step 3 — Read off the real part. Re($z_1z_2$) = $ac - bd$ = Re $z_1$Re $z_2$ $-$ Im $z_1$Im $z_2$. ∎

Takeaway: The product formula IS the proof; write it and point.