How many complex numbers satisfy a single modulus equation?

Infinitely many, in general. A complex number z carries two real numbers, its real part x and its imaginary part y. A single modulus or argument condition is one real equation on those two unknowns, so it leaves one degree of freedom: a curve. For example |z squared + z + 1| = 1 is one real equation and its solution set is a whole curve, not the four points you get by wrongly splitting it into z squared + z + 1 = plus or minus 1. This is exactly what JEE Advanced 2020 Paper 1 Q9 is built to catch.

When does an equation in z actually give finitely many points?

When it is really two real equations at once. A polynomial set to zero, like z cubed = 27, matches a real part and an imaginary part simultaneously, so it pins down finitely many points (here the three cube roots 3, 3 omega, 3 omega squared). A full complex equation like z squared = 5 - 12i does the same, giving the two points 3 - 2i and -3 + 2i. The count-the-degree habit is correct for these, and only for these.

Where is the centre of the Apollonius circle from |z - a| = k|z - b|?

Not at the midpoint of a and b. For k not equal to 1 the locus is a circle, and its centre lies on the line through a and b but outside the segment joining them. Find it from the two points that divide the segment in ratio k to 1 internally and externally: those two points are the ends of a diameter. For |z - 3| = 2|z - 12| the internal point is 9 and the external point is 21, so the centre is 15 and the radius is 6. The midpoint 7.5 is wrong, and this is the wrong-answer structure of JEE Advanced 2025 Paper 1 Q7.

What does an argument condition like arg((z - a)/(z - b)) = phi describe?

An arc of a circle passing through a and b, on which the segment from b to a subtends the fixed directed angle phi. It is a curve, infinitely many z, and the sign of phi selects which side of the line through a and b the arc sits on. Arguments are read in the principal range from -pi to pi.

JEE Advanced Maths · Complex Numbers

Does an equation in z pin down a few points?

Not in general. A complex $z$ carries two real numbers, its real part $x$ and its imaginary part $y$ . A single modulus or argument condition is one real equation on those two unknowns, so it leaves one degree of freedom: a curve, not a finite set. The polynomial-root habit ("degree $n$ gives $n$ points") is imported from real algebra and only fits a genuine polynomial set to zero.

This trips up JEE Advanced across 2020 and 2025. The clean statement to hold is: count the real equations. Two of them pin down points; one of them carves a curve. And an Apollonius circle from $|z - a| = k|z - b|$ is centred outside the segment joining $a$ and $b$ , never at the midpoint.

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 counted the solutions of $|z^2 + 1| = 2$ or $|z^2 + z + 1| = 1$ by treating the modulus bars like an equals sign.
You split $|w| = 1$ into $w = \pm 1$ instead of a whole circle of values.
You agreed $|z - a| = k|z - b|$ was a circle, then put its centre at the midpoint of $a$ and $b$ .
You reported an argument condition $\arg((z-a)/(z-b)) = \varphi$ as a pair of points rather than an arc.

An exam question that triggers it

Here is a question in the style of JEE Advanced 2025 Paper 1 Q7, which tests this misconception head-on:

Find the centre and radius of the locus of complex numbers

|z - 3| = 2\,|z - 12|.

The misconception answer either reads this as a couple of points, or agrees it is a circle and drops the centre at the midpoint $7.5$ of $3$ and $12$ . The correct move finds the two points that divide the segment from $3$ to $12$ in ratio $2 : 1$ , internally and externally, because those two points are the ends of a diameter.

On the real axis, $|x - 3| = 2|x - 12|$ gives the internal point $x - 3 = 2(12 - x) \Rightarrow x = 9$ and the external point $x - 3 = 2(x - 12) \Rightarrow x = 21$ . So

\text{centre} = \frac{9 + 21}{2} = 15, \qquad \text{radius} = \frac{21 - 9}{2} = 6.

The centre $15$ lies outside the segment $[3, 12]$ , and the midpoint $7.5$ is nowhere near it. The set is a circle, infinitely many $z$ .

Why students fall for this

The belief underneath is that $z$ is a single algebraic unknown just like a real $x$ , so an equation "should" pin it down to a short list. The Fundamental Theorem of Algebra reinforces this: a genuine polynomial of degree $n$ really does have $n$ roots, so the count-the-degree move works often enough to feel safe.

It breaks the instant the equation is a single modulus or argument condition, because a complex $z$ has two real degrees of freedom and one real equation removes only one of them. What dislodges the habit is writing $z = x + iy$ once and counting: a polynomial set to zero matches a real part and an imaginary part, two real equations, so finitely many points; a single modulus condition is one real equation, so a curve.

The fix: Count the real equations, not the degree

One real equation on a complex $z$ carves out a curve; count roots only when the equation genuinely matches two complex quantities. Write $z = x + iy$ . A polynomial $p(z) = 0$ is two real equations (real part zero and imaginary part zero), so it pins down $\deg p$ points. A single condition $|z - c| = r$ or $\arg(\cdots) = \varphi$ is one real equation, so it leaves a curve.

For an Apollonius circle $|z - a| = k|z - b|$ with $k \ne 1$ , do not guess the centre. Find the internal and external division points of the segment $[a, b]$ in ratio $k : 1$ ; they are the ends of a diameter, so the centre is their midpoint and the radius is half their separation. The centre lies on the line through $a$ and $b$ but outside the segment.

Worked example

Count once, watch the count leak, then use the dimension test that does not. Take $|z^2 + z + 1| = 1$ , the surface of JEE Advanced 2020 Paper 1 Q9.

Try to count. The reflex splits it into $z^2 + z + 1 = 1$ (giving $z = 0, -1$ ) and $z^2 + z + 1 = -1$ (giving two more), for "four points". But $|w| = 1$ is the whole unit circle of values, not $w = \pm 1$ .
Write it in real and imaginary parts. With $z = x + iy$ ,
$z^2 + z + 1 = (x^2 - y^2 + x + 1) + i\,(2xy + y).$
Count real equations. $|z^2 + z + 1| = 1$ becomes $(x^2 - y^2 + x + 1)^2 + (2xy + y)^2 = 1$ , one real equation on the two unknowns $x$ and $y$ : a curve, infinitely many $z$ . The points $z = 0$ and $z = -1$ are just two of them.
Contrast a genuine polynomial. $z^3 = 27$ matches two real quantities at once, so it is two real equations and gives exactly three points, $3,\ 3\omega,\ 3\omega^2$ . The habit is right here.
Locate an Apollonius circle. For $|z - 1| = 2|z - 7|$ , the division points are $5$ and $13$ , so the centre is $9$ and the radius is $4$ (not the midpoint $4$ of $1$ and $7$ ).
Conclude.
$\text{one real equation} \iff \text{a curve}; \quad \text{two real equations} \iff \text{finitely many points.}$

The same reasoning settles JEE Advanced 2020 Paper 1 Q9 (a curve, not four points) and 2025 Paper 1 Q7 (an Apollonius circle whose centre is outside the segment, not the midpoint).

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Common questions

Why is $|z^2 + z + 1| = 1$ not four points?: Because $|w| = 1$ means $w$ lies on the whole unit circle, not that $w = \pm 1$ . Writing $z = x + iy$ turns the condition into a single real equation on $x$ and $y$ , whose solution set is a curve. Splitting into $z^2 + z + 1 = \pm 1$ only samples a few points of that curve.
How do I find the centre of an Apollonius circle?: Take the two points dividing the segment $[a, b]$ in ratio $k : 1$ , internally and externally. They are the ends of a diameter, so the centre is their midpoint and the radius is half their distance apart. For $|z - 3| = 2|z - 12|$ the points are $9$ and $21$ , giving centre $15$ and radius $6$ .
Is the centre ever the midpoint of $a$ and $b$ ?: Only in the degenerate case $k = 1$ , where $|z - a| = |z - b|$ is not a circle at all but the perpendicular bisector of $a$ and $b$ . For $k \ne 1$ the locus is a genuine circle whose centre lies on the line through $a$ and $b$ but outside the segment joining them.
What shape is an argument condition?: $\arg((z - a)/(z - b)) = \varphi$ is an arc of a circle through $a$ and $b$ , on which the segment from $b$ to $a$ subtends the fixed directed angle $\varphi$ . It is a curve, and the sign of $\varphi$ chooses which side of the line the arc sits on. Arguments are read in the principal range $(-\pi, \pi]$ .

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.
Is arg z always arctan(y/x)?No: arctan(y/x) only gives the argument in the right half-plane; elsewhere you place z by its quadrant and read the principal value in (-pi, pi], and a modulus-1 factor like -1 shifts the argument by pi rather than leaving it alone.

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