How do I know when a complex expression is real?

A complex number w is real exactly when it equals its own conjugate: w = w-bar. This is equivalent to its imaginary part being zero. It is a condition you impose and solve, not a property you read off one piece of the expression. For a quotient especially, a real numerator does not make the quotient real: 5/(1 + 2i) has a real numerator 5 but equals 1 - 2i, which is not real.

Is z-bar = 1/z always true?

No. z-bar = 1/z means z times z-bar = 1, that is |z| squared = 1, so it holds only on the unit circle |z| = 1. For z = 1 + 2i, |z| squared = 5, so z-bar = 1 - 2i while 1/z = (1 - 2i)/5, which are not equal. Swapping z-bar for 1/z inside an equation and then multiplying through by z quietly assumes |z| = 1 and can drop the z = 0 solution, which is exactly what sinks the naive count in JEE Advanced 2022 Paper 1 Q5.

How do I make a quotient like (a + bz + cz squared)/(a - bz + cz squared) real?

Impose w = w-bar and cross-multiply. For w = (3 + 5z + 12z squared)/(3 - 5z + 12z squared) the difference of the cross-products factors to 10(z - z-bar)(3 - 12|z| squared). For a non-real z the factor z - z-bar is non-zero, so 3 - 12|z| squared = 0, giving |z| squared = 1/4. This is the mechanism behind JEE Advanced 2022 Paper 1 Q4, whose coefficients 2, 3, 4 give |z| squared = 1/2.

What is the Im = 0 gate for a parameterised expression?

When an expression depends on an angle, realness is the single equation Im(w) = 0. For w = (3 + 2i sin theta)/(1 - i cos theta), multiplying by the conjugate of the denominator gives a real denominator and a numerator whose imaginary part is 3 cos theta + 2 sin theta. Setting that to zero gives tan theta = -3/2. JEE Advanced 2023 Paper 1 Q11 is exactly this gate, with tan theta = -1/2, and the whole question lives behind it.

JEE Advanced Maths · Complex Numbers

Is an expression real when it looks real?

Not by inspection. A complex number $w$ is real exactly when it equals its own conjugate, $w = \bar w$ , equivalently when its imaginary part is zero. That is a condition you impose and solve, not something you read off a numerator or a single piece. A real numerator does not make a quotient real: $w = 5/(1+2i)$ has the real numerator $5$ yet equals $1 - 2i$ , which is not real.

This trips up JEE Advanced across 2022 and 2023. The clean statement to hold is: realness is the equation $w = \bar w$ , and $\bar z = 1/z$ is legal only when $|z| = 1$ .

Ready to fix this? The Complex Numbers lesson works through this misconception and the others in Complex Numbers, one altitude at a time.

How to spot it in your own work

You called a quotient real because its numerator was real, without imposing any equation.
You replaced $\bar z$ with $1/z$ without checking $|z| = 1$ .
You divided an equation through by $z$ and never checked whether $z = 0$ was a lost solution.
You decided a parameterised expression was real by eyeballing its real part instead of setting $\mathrm{Im}\,w = 0$ .

An exam question that triggers it

Here is a question in the style of JEE Advanced 2022 Paper 1 Q4, which tests this misconception head-on:

For some complex $z$ , let

w = \frac{3 + 5z + 12z^2}{3 - 5z + 12z^2}.

Given that $w$ is real for a non-real $z$ , find $|z|^2$ .

The misconception answer never writes an equation: it declares the quotient real because the pieces look real, or forces $|z| = 1$ out of the $\bar z = 1/z$ reflex. The correct move imposes the gate $w = \bar w$ , i.e. $N\bar D = \bar N D$ , and cross-multiplies. The difference factors cleanly:

N\bar D - \bar N D = 10\,(z - \bar z)\,(3 - 12|z|^2).

For a non-real $z$ the factor $z - \bar z$ is non-zero, so $3 - 12|z|^2 = 0$ , giving $|z|^2 = \tfrac{1}{4}$ . The genuine 2022 paper used coefficients $2, 3, 4$ and produced $|z|^2 = \tfrac{1}{2}$ . The gate is the whole question.

Why students fall for this

The belief underneath is that $a + bi$ is two numbers glued together, so realness ought to be visible on one of them. Nordlander and Nordlander (2012) document exactly this ontology: students treat the real and imaginary parts as separate objects rather than one number, so "is it real?" feels like a look at a part rather than a single statement about the whole. Mutambara and Tsakeni (2022) record conjugation itself being modelled loosely as "negate something", which is why $\bar z = 1/z$ gets treated as a free swap.

The instinct half-works, which is why it survives. When a number is already written as $a + bi$ , real really does mean $b = 0$ , and you can read that off. The habit only breaks when $w$ arrives as a quotient or a parameterised expression, where nothing is split and nothing is real until you make it so. What dislodges it is imposing $w = \bar w$ once and watching a real-looking expression fail the test.

The fix: Realness is one equation: w = w-bar

$w$ is real if and only if $w = \bar w$ . This is the same as $\mathrm{Im}\,w = 0$ . You impose it and solve it; you never read realness off a numerator or a single term. For a quotient $w = N/D$ , the condition is $N\bar D = \bar N D$ , which you cross-multiply and simplify to a locus.

The two guards that go with it: $\bar z = 1/z$ means $z\bar z = 1$ , so it holds only when $|z| = 1$ ; and dividing an equation through by $z$ can drop the $z = 0$ solution, so check it separately. On a parameterised expression the same gate becomes $\mathrm{Im}\,w = 0$ , which you solve for the parameter.

Worked example

Watch inspection fail once, then use the equation that does not. Take $w = \dfrac{5}{1 + 2i}$ , whose numerator is real.

Split it. Rationalise by the conjugate of the denominator: $w = \dfrac{5(1 - 2i)}{(1 + 2i)(1 - 2i)} = \dfrac{5(1 - 2i)}{1 + 4} = 1 - 2i$ .
Apply the gate. $w = 1 - 2i$ and $\bar w = 1 + 2i$ . Since $1 - 2i \ne 1 + 2i$ , $w \ne \bar w$ , so $w$ is not real. The real numerator was no guide.
The parameterised gate. For $w = \dfrac{3 + 2i\sin\theta}{1 - i\cos\theta}$ , multiply by $1 + i\cos\theta$ . The imaginary part of the numerator is $3\cos\theta + 2\sin\theta$ , so $\mathrm{Im}\,w = 0$ gives $\tan\theta = -\tfrac{3}{2}$ .
The domain guard. For $z = 1 + 2i$ , $|z|^2 = 5$ , so $\bar z = 1 - 2i$ while $1/z = (1 - 2i)/5$ . They differ, because $\bar z = 1/z$ holds only on $|z| = 1$ .
Conclude.
$w \text{ real} \iff w = \bar w \iff \mathrm{Im}\,w = 0, \quad \text{imposed and solved, never read off.}$

The same gate settles JEE Advanced 2022 Paper 1 Q4 (a quotient forced real gives $|z|^2$ ), 2023 Paper 1 Q11 (the $\mathrm{Im} = 0$ gate $\tan\theta = -\tfrac{1}{2}$ ), and 2022 Paper 1 Q5 (where the $\bar z = 1/z$ shortcut drops the $z = 0$ root).

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Common questions

Why is a real numerator not enough for a quotient to be real?: Because dividing by a complex denominator mixes an imaginary part back in. For $5/(1 + 2i)$ , the numerator is $5$ but the value is $1 - 2i$ . Realness is a statement about the whole number $w$ , captured by $w = \bar w$ , not a property of the top alone.
When exactly does $\bar z = 1/z$ hold?: Exactly on the unit circle. $\bar z = 1/z$ rearranges to $z\bar z = 1$ , i.e. $|z|^2 = 1$ . Off the circle it is simply false, and using it as a free identity is a common source of wrong answers.
What is the $\mathrm{Im} = 0$ gate?: It is the same condition $w = \bar w$ written as $\mathrm{Im}\,w = 0$ , most useful when $w$ depends on a parameter. You compute the imaginary part and set it to zero, which solves for the parameter. For the angle case above it gives $\tan\theta = -\tfrac{3}{2}$ .
How can dividing by $z$ lose a solution?: If an equation is satisfied by $z = 0$ and you divide both sides by $z$ (often while swapping $\bar z$ for $1/z$ ), you silently discard $z = 0$ and can undercount the roots. Always test $z = 0$ separately before dividing. This is the miscount JEE Advanced 2022 Paper 1 Q5 is built to catch.

Related misconceptions

Is |z|² the same as z²?No: |z|² = z times its conjugate is a real, non-negative size, while z² is a point in the plane; they agree only when z is real.
Modulus and arg conditions are lociAn equation in z carves out a whole curve, not a few points; the Apollonius circle from |z - a| = k|z - b| is not centred between the two points.

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