Modulus and conjugate

Every complex number $z=a+ib$ carries two natural "shadows": its conjugate $\bar z=a-ib$ , the mirror image obtained by flipping the sign of the imaginary part, and its modulus $|z|=\sqrt{a^2+b^2}$ , the non-negative real number measuring its distance from the origin. Neither is exotic — they are read straight off $a$ and $b$ — yet together they unlock the single most useful trick in this chapter: multiplying by the conjugate turns $z\bar z$ into the purely real number $|z|^2$ , which is exactly the move that clears $i$ out of a denominator and makes dividing complex numbers routine. This article, part of CBSE Class 11's "Complex Numbers and Quadratic Equations" (Unit II: Algebra), builds the definitions, proves the identity $z\bar z=|z|^2$ and the standard properties that follow from it, and works two exam-style problems — a direct modulus/conjugate check and a full division — so the mechanics become second nature before the next chapter leans on them.

The core insight

The conjugate $\bar z=a-ib$ reflects $z=a+ib$ across the real axis while the modulus $|z|=\sqrt{a^2+b^2}$ measures its distance from the origin, and the identity $z\bar z=|z|^2$ turns multiplication by the conjugate into the standard tool for dividing complex numbers.

The core idea

For $z=a+ib$ with $a,b\in\mathbb{R}$ , the conjugate simply reverses the sign attached to $i$ , leaving the real part untouched.

\bar z = a-ib

where: $a=\operatorname{Re}(z)$ is the real part and $b=\operatorname{Im}(z)$ is the imaginary part of $z$ . The modulus is the length of the line joining $z$ to the origin in the Argand plane, so it is always a non-negative real number, never a complex one.

|z|=\sqrt{a^2+b^2}

where: $|z|=0$ exactly when $a=b=0$ , i.e. when $z=0$ . Multiplying a complex number by its own conjugate always cancels the imaginary part and leaves a real, non-negative result — this is the single fact the whole topic hinges on.

z\bar z=(a+ib)(a-ib)=a^2-(ib)^2=a^2+b^2=|z|^2

Reading $z$ and $\bar z$ as mirror-image points also explains a handful of properties that are easy to remember once seen geometrically rather than memorised as a list: conjugating twice undoes itself ( $\overline{\bar z}=z$ ), conjugation distributes over sums and products ( $\overline{z_1+z_2}=\bar z_1+\bar z_2$ , $\overline{z_1z_2}=\bar z_1\bar z_2$ ), the modulus is multiplicative and divisive ( $|z_1z_2|=|z_1||z_2|$ and $|z_1/z_2|=|z_1|/|z_2|$ for $z_2\ne0$ ), and a number is real exactly when it equals its own conjugate ( $z=\bar z\iff\operatorname{Im}(z)=0$ , since reflecting a point on the real axis leaves it fixed). The multiplicative property $|z_1z_2|=|z_1||z_2|$ is worth seeing proved once rather than just quoted, because the same trick — leaning on $z\bar z=|z|^2$ instead of expanding everything in terms of $a$ and $b$ — reappears throughout the chapter. Since $|z|^2=z\bar z$ for any complex number, $|z_1z_2|^2=(z_1z_2)\overline{(z_1z_2)}=z_1z_2\bar z_1\bar z_2=(z_1\bar z_1)(z_2\bar z_2)=|z_1|^2|z_2|^2$ , and since both sides are non-negative, taking square roots gives $|z_1z_2|=|z_1||z_2|$ directly, with no need to substitute $a,b,c,d$ and multiply out four terms. Two more identities are useful for rewriting an expression in terms of $z$ and $\bar z$ alone, without tracking $a$ and $b$ separately: adding $z=a+ib$ and $\bar z=a-ib$ gives $2a$ , and subtracting gives $2ib$ , so

\operatorname{Re}(z)=\frac{z+\bar z}{2}, \qquad \operatorname{Im}(z)=\frac{z-\bar z}{2i}.

where: these recover the real and imaginary parts of any $z$ purely from $z$ and $\bar z$ , which is occasionally faster than reading $a$ and $b$ off a messy expression by inspection.

Figure

An Argand diagram showing a point z above the real axis and its conjugate z bar reflected below it, both joined to the origin by radii of equal length labelled the modulus

The derivation

Derivation AWhy z times z bar equals the square of the modulus, and why that clears i from a denominator

The goal of this derivation is to establish the identity $z\bar z=|z|^2$ from the raw definitions of conjugate and modulus for $z=a+ib$ , and then to show precisely why this one algebraic fact is the standard method for dividing complex numbers — the same reasoning that later underlies rationalising expressions with surds, but here applied to $i$ instead of $\sqrt2$ .

Step 1. Write $z=a+ib$ and its conjugate $\bar z=a-ib$ , then multiply them as an ordinary product of two binomials, treating $i$ as an unknown for now: $z\bar z=(a+ib)(a-ib)=a\cdot a-a\cdot ib+ib\cdot a-ib\cdot ib=a^2-a^2b i^{0}\ldots$ — more directly, this is a difference of squares, $(a+ib)(a-ib)=a^2-(ib)^2$ .

Step 2. Simplify $(ib)^2=i^2b^2$ , and substitute the defining property $i^2=-1$ : $(ib)^2=-b^2$ . So $z\bar z=a^2-(-b^2)=a^2+b^2$ . Every imaginary term has cancelled, leaving a sum of two real squares.

Step 3. Compare this with the definition of modulus, $|z|=\sqrt{a^2+b^2}$ , so $|z|^2=a^2+b^2$ . Since Step 2 gave exactly $z\bar z=a^2+b^2$ , the two expressions agree term for term: $z\bar z=|z|^2$ . This holds for every complex number, with no restriction on $a$ or $b$ .

Step 4. Apply this to division. To simplify $1/z$ for $z\ne0$ , multiply numerator and denominator by $\bar z$ : $\dfrac{1}{z}=\dfrac{\bar z}{z\bar z}=\dfrac{\bar z}{|z|^2}$ . The denominator $|z|^2$ is now a plain real number, so $\bar z/|z|^2$ is already in the standard form $a'+ib'$ after dividing each part of $\bar z$ by the real number $|z|^2$ — the same trick extends to $z_1/z_2$ by multiplying top and bottom by the conjugate of $z_2$ . Result: $z\bar z=|z|^2=a^2+b^2$ for every $z=a+ib$ , and consequently $1/z=\bar z/|z|^2$ for $z\ne0$ .

Worked example

Problem: For $z=-3+4i$ , find $\bar z$ and $|z|$ , verify $z\bar z=|z|^2$ , and use that check to divide $\dfrac{2-i}{3+2i}$ into standard form.

Step 1. Conjugate: flip the sign of the imaginary part only. $\bar z=-3-4i$ .

Step 2. Modulus: $|z|=\sqrt{(-3)^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$ .

Step 3. Verify the identity: $z\bar z=(-3+4i)(-3-4i)=(-3)^2-(4i)^2=9-16i^2=9+16=25=5^2=|z|^2$ . Checks out. $\bar z=-3-4i$ and $|z|=5$ .

Step 4 (division, reusing the same tool). To simplify $\dfrac{2-i}{3+2i}$ , multiply top and bottom by the conjugate of the denominator, $3-2i$ : $\dfrac{2-i}{3+2i}\times\dfrac{3-2i}{3-2i}=\dfrac{(2-i)(3-2i)}{(3+2i)(3-2i)}$ . The numerator is $6-4i-3i+2i^2=6-7i-2=4-7i$ . The denominator uses $z\bar z=|z|^2$ with $z=3+2i$ , so it is just $3^2+2^2=13$ , no need to expand by hand.

Step 5. So $\dfrac{2-i}{3+2i}=\dfrac{4-7i}{13}=\dfrac{4}{13}-\dfrac{7}{13}i$ . Final answer: $\dfrac{4}{13}-\dfrac{7}{13}i$ , and its modulus is $\left|\dfrac{4}{13}-\dfrac{7}{13}i\right|=\dfrac{\sqrt{4^2+7^2}}{13}=\dfrac{\sqrt{65}}{13}$ .

Step 6 (independent check on Step 1–3, using the Re/Im shortcuts). $\operatorname{Re}(z)=\dfrac{z+\bar z}{2}=\dfrac{(-3+4i)+(-3-4i)}{2}=\dfrac{-6}{2}=-3$ , correctly reproducing the real part of $z=-3+4i$ , and $\operatorname{Im}(z)=\dfrac{z-\bar z}{2i}=\dfrac{8i}{2i}=4$ , correctly reproducing the imaginary part. Both shortcuts agree with the original $z$ , confirming that $\bar z=-3-4i$ was computed correctly before it was ever used.

Competitive stretch

Cross-check the modulus using $|z_1/z_2|=|z_1|/|z_2|$ instead of expanding: $|2-i|=\sqrt5$ and $|3+2i|=\sqrt{13}$ , so $|z_1/z_2|=\sqrt5/\sqrt{13}=\sqrt{5/13}=\sqrt{65}/13$ — matching Step 5 without recomputing $(4/13)^2+(7/13)^2$ from scratch. Being able to switch between "expand everything" and "use a known property" is a fast way to double-check answers under exam time pressure.

Common traps

Flipping the sign of the real part instead of the imaginary part — e.g. writing $\overline{2-3i}$ as $-2+3i$ instead of the correct $2+3i$ . Only the $i$ -term's sign changes.
Computing $a^2+b^2$ and stopping there, forgetting the square root that turns it into $|z|$ — this slip is common precisely because $z\bar z=a^2+b^2$ (no square root) is also used constantly in this chapter.
Treating $|z|$ and $\bar z$ as interchangeable — $|z|$ is a single non-negative real number, $\bar z$ is another complex number; they answer different questions.
Assuming $z+\bar z$ or $z\bar z$ could end up with a leftover $i$ term — both are always real ( $z+\bar z=2a$ , $z\bar z=a^2+b^2$ ), so any final answer for either should be checked for a stray imaginary part.
Multiplying a fraction by $\bar z$ but forgetting to also multiply the numerator by it — the trick only works because you multiply top and bottom by the same conjugate, which is really multiplying by $1$ in disguise; skipping the numerator changes the value of the expression.

Ask the tutor: if a sign slip in a conjugate calculation, the modulus-vs- $z\bar z$ mix-up, or the division-by-conjugate trick with a new pair of numbers would help you lock this in, ask the AI tutor to work through it with you step by step.

Why does multiplying by the conjugate always remove i from the denominator?

Because $z\bar z=a^2+b^2$ is a difference of squares in disguise: the cross terms $-aib$ and $+aib$ in $(a+ib)(a-ib)$ cancel exactly, and $(ib)^2=-b^2$ turns the remaining imaginary piece into a real one. So the denominator becomes the plain real number $|z|^2$ , no matter what $a$ and $b$ are — this is guaranteed by the identity, not a coincidence for particular numbers.

Is the modulus the same as just squaring a and b and adding them?

No — that gives $z\bar z=a^2+b^2$ , which is real but is not yet the modulus. The modulus is the square root of that quantity, $|z|=\sqrt{a^2+b^2}$ . Students often stop one step early because $a^2+b^2$ shows up so often (it equals $|z|^2$ and also $z\bar z$ ) that it is easy to forget the final square root is needed to actually get $|z|$ .

Do I always need the conjugate to divide two complex numbers, or only sometimes?

You need it whenever the denominator is not already a real number. If you are dividing by a purely real number, ordinary division of each part works directly. But whenever the denominator has a nonzero imaginary part, like $3+2i$ , multiplying numerator and denominator by that denominator's conjugate is the standard way to force the denominator to become real, using exactly the identity $z\bar z=|z|^2$ .