Algebra of complex numbers

Real numbers alone cannot solve $x^2+1=0$ — no real number squares to $-1$ . Complex numbers fix this by inventing a new symbol $i$ with $i^2=-1$ , then building numbers of the form $a+ib$ out of it. What makes this more than a trick is that these numbers behave exactly like the real numbers you already know: you can add, subtract, multiply and divide them, and the same commutative and associative laws hold. This section covers that algebra — the four operations on $z=a+ib$ , and the identities and inverses that make the complex numbers a genuine number system, in line with CBSE Class 11's "Complex Numbers and Quadratic Equations" chapter. The one new habit to learn is: expand exactly as you would with real binomials, and only at the last step replace $i^2$ by $-1$ . Division needs one extra trick — multiplying by the conjugate — because you cannot simplify a fraction with an imaginary denominator term by term the way you can with real fractions.

The core insight

Complex numbers $a+ib$ add, subtract, multiply and divide using ordinary real-number algebra plus the single rule $i^2=-1$ , with division requiring multiplication by the conjugate to clear the imaginary denominator.

The core idea

Let $z_1=a+ib$ and $z_2=c+id$ , with $a,b,c,d\in\mathbb{R}$ . All four operations reduce to real-number arithmetic on the parts, plus the rule $i^2=-1$ wherever it shows up.

Addition and subtraction combine the real parts and imaginary parts separately — no different from combining like terms in algebra.

z_1+z_2=(a+c)+i(b+d), \qquad z_1-z_2=(a-c)+i(b-d)

where: $a,c$ are the real parts and $b,d$ are the imaginary parts of $z_1,z_2$ .

Multiplication expands like two binomials, then substitutes $i^2=-1$ once, at the end.

z_1z_2=(a+ib)(c+id)=(ac-bd)+i(ad+bc)

where: the term $i^2bd$ that appears mid-expansion becomes $-bd$ , which is why the real part picks up a minus sign ( $ac-bd$ ) even though multiplication itself has no minus signs in it.

Division cannot be done term by term. Instead, multiply numerator and denominator by the conjugate of the denominator, $\overline{c+id}=c-id$ , which turns the denominator into the real number $c^2+d^2$ .

\frac{z_1}{z_2}=\frac{(a+ib)(c-id)}{(c+id)(c-id)}=\frac{(ac+bd)+i(bc-ad)}{c^2+d^2}, \qquad z_2\ne 0

where: $(c+id)(c-id)=c^2-i^2d^2=c^2+d^2$ is always real and non-negative — this is the entire reason the conjugate trick works. These four operations obey the same laws real-number arithmetic obeys: closure ( $z_1+z_2$ and $z_1z_2$ are complex numbers again), commutativity and associativity for both $+$ and $\times$ , an additive identity $0=0+i0$ , a multiplicative identity $1=1+i0$ , an additive inverse $-z=-a-ib$ , and — for $z\ne0$ — a multiplicative inverse $z^{-1}=\dfrac{a-ib}{a^2+b^2}$ . Notice $z^{-1}$ is nothing but $\dfrac1z=\dfrac{1}{a+ib}$ worked out by the same conjugate trick used for division; there is really only one technique to learn, not two. This full package of properties is exactly why complex numbers count as a genuine number system, not just useful notation. It helps to see multiplication as an organised version of the FOIL pattern you already use for real binomials: first terms $ac$ , outer terms $iad$ , inner terms $ibc$ , last terms $i^2bd$ . Keeping the four partial products separate before combining anything is the single biggest habit that prevents sign slips, because it delays the moment you have to think about $i^2$ until the very last step, when there is only one substitution left to make. The same discipline pays off for powers of $i$ : since $i^2=-1$ , it follows that $i^3=i^2\cdot i=-i$ and $i^4=(i^2)^2=1$ , after which the pattern $i,-1,-i,1$ repeats every four powers — a fact used constantly when simplifying expressions that involve $i$ raised to some power.

Figure

Four boxes showing addition subtraction multiplication and division of two complex numbers with the results in standard form

Derivation AMultiplying two complex numbers and identifying the real and imaginary parts

The goal is to show, with nothing more than ordinary binomial expansion and the single substitution $i^2=-1$ , that the product of $z_1=a+ib$ and $z_2=c+id$ always lands in standard form $(ac-bd)+i(ad+bc)$ , so that "multiply two complex numbers" reduces to a formula you can apply directly instead of re-expanding every time.

Step 1. Expand $(a+ib)(c+id)$ exactly like two real binomials $(a+b)(c+d)$ , keeping all four partial products separate and not yet simplifying anything: $ac+iad+ibc+i^2bd$ .

Step 2. Substitute $i^2=-1$ in the one term where it appears: $i^2bd$ becomes $-bd$ . The expression is now $ac+iad+ibc-bd$ .

Step 3. Group the terms with no $i$ (the real part) separately from the terms with a single factor of $i$ (the imaginary part): real part $=ac-bd$ , imaginary part $=ad+bc$ (the coefficient of $i$ , not $i(ad+bc)$ itself).

Step 4. Write the result in standard form $a'+ib'$ : $z_1z_2=(ac-bd)+i(ad+bc)$ . This is the general formula; every specific multiplication is this same four-step process with numbers plugged in for $a,b,c,d$ .

Worked example

Problem: If $z_1=2+3i$ and $z_2=1-4i$ , find $z_1+z_2$ , $z_1-z_2$ , $z_1z_2$ , and $\dfrac{z_1}{z_2}$ .

Step 1 — addition and subtraction. $z_1+z_2=(2+1)+i(3-4)=3-i$ . $z_1-z_2=(2-1)+i(3-(-4))=1+7i$ .

Step 2 — multiplication. Expand as in the derivation above: $z_1z_2=(2+3i)(1-4i)=2-8i+3i-12i^2=2-5i-12(-1)=2-5i+12=14-5i$ .

Step 3 — division. Multiply numerator and denominator by the conjugate of $z_2=1-4i$ , which is $1+4i$ : $\dfrac{2+3i}{1-4i}\times\dfrac{1+4i}{1+4i}=\dfrac{(2+3i)(1+4i)}{1^2+4^2}$ . The numerator is $2+8i+3i+12i^2=2+11i-12=-10+11i$ , and the denominator is $1+16=17$ . So $\dfrac{z_1}{z_2}=\dfrac{-10+11i}{17}=-\dfrac{10}{17}+\dfrac{11}{17}i$ .

Step 4 — check the division. Multiply the quotient back by $z_2$ : $\left(-\tfrac{10}{17}+\tfrac{11}{17}i\right)(1-4i)=-\tfrac{10}{17}+\tfrac{40}{17}i+\tfrac{11}{17}i-\tfrac{44}{17}i^2=-\tfrac{10}{17}+\tfrac{51}{17}i+\tfrac{44}{17}=\tfrac{34}{17}+\tfrac{51}{17}i=2+3i$ , which matches $z_1$ exactly.

Answer: $z_1+z_2=3-i$ , $z_1-z_2=1+7i$ , $z_1z_2=14-5i$ , and $\dfrac{z_1}{z_2}=-\dfrac{10}{17}+\dfrac{11}{17}i$ .

Competitive stretch

Find $z^{-1}$ for $z=3+4i$ using $z^{-1}=\dfrac{a-ib}{a^2+b^2}$ . Here $a=3,b=4$ , so $z^{-1}=\dfrac{3-4i}{9+16}=\dfrac{3-4i}{25}=\dfrac{3}{25}-\dfrac{4}{25}i$ . Checking: $(3+4i)\left(\tfrac{3}{25}-\tfrac{4}{25}i\right)=\tfrac{9}{25}-\tfrac{12}{25}i+\tfrac{12}{25}i-\tfrac{16}{25}i^2=\tfrac{9}{25}+\tfrac{16}{25}=1$ . Since $z^{-1}=1/z$ , this is really the division technique applied with numerator $1$ .

Common traps

Forgetting to replace $i^2$ with $-1$ during multiplication — $3i\cdot4i=12i^2$ must become $-12$ , not stay $12i^2$ or flip to $+12$ .
Trying to divide term by term instead of multiplying by the conjugate first — $\dfrac{a+ib}{c+id}$ cannot be split the way real fractions can.
Writing the conjugate with the wrong sign — the conjugate of $c+id$ is $c-id$ (flip only the imaginary sign), not $-c+id$ or $-c-id$ .
Confusing $(a+ib)^2$ with $a^2+b^2$ — the correct expansion is $(a+ib)^2=(a^2-b^2)+2iab$ , not a sum of squares.
Losing track of which quantity is "the imaginary part" — for $z=a+ib$ the imaginary part is the real number $b$ , not $ib$ ; saying the imaginary part of $3+5i$ is $5i$ instead of $5$ is a common slip in definitions-based questions.

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Why does multiplying by the conjugate always make the denominator real?

Because $(c+id)(c-id)$ is a difference of squares, $c^2-(id)^2=c^2-i^2d^2$ , and $i^2=-1$ turns the minus into a plus: $c^2+d^2$ . That is a sum of two real squares, always real and non-negative (zero only when $c=d=0$ , which is excluded since $z_2\ne0$ ). So the conjugate is not a random trick — it is the one multiplier guaranteed to cancel every $i$ from the denominator, using the same difference-of-squares identity you already know from real algebra.

Is z times its conjugate the same as z squared?

No, and mixing these up is a common slip. $z\cdot\overline z=(a+ib)(a-ib)=a^2+b^2$ , a real number equal to the square of the modulus, $|z|^2$ . But $z^2=(a+ib)^2=(a^2-b^2)+2iab$ , which is generally still a complex number. The conjugate product is what makes division work because it is real; squaring $z$ by itself gives you another complex number, not a real one, so it cannot clear an imaginary denominator.

Do the commutative and associative laws really need to be checked separately for complex numbers?

Conceptually you could just trust that they hold, but stating them explicitly matters because it is what makes complex numbers a genuine extension of the real number system rather than a new, unrelated set of symbols. Each law follows directly from the corresponding real-number law applied to the real and imaginary parts separately — for instance $z_1+z_2=(a+c)+i(b+d)=(c+a)+i(d+b)=z_2+z_1$ because real addition is commutative. So you are not proving something new each time, you are transferring a familiar real-number fact into the $a+ib$ framework.

Why does the multiplicative inverse formula have a2 plus b2 in the denominator and not something else?

Because $z^{-1}$ is built by the same conjugate trick as division: $z^{-1}=\dfrac1z=\dfrac{1}{a+ib}\times\dfrac{a-ib}{a-ib}=\dfrac{a-ib}{a^2+b^2}$ , and the denominator $a^2+b^2$ appears because that is exactly what $z\cdot\overline z$ equals. It is never zero unless $a=b=0$ , which is precisely why $z^{-1}$ only fails to exist for $z=0$ — matching the familiar real-number rule that only $0$ has no reciprocal. If you ever forget the formula, just rebuild it from $1/z$ times the conjugate over itself rather than memorising it as a separate fact.

Algebra of complex numbers

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads