If z = x + iy, xy ≠ 0, Satisfies the Equation z² + iz̄ = 0, Then |z²| Is Equal To

JEE Main 2024 — 30 January, Shift 1. Previous Year Question.

Problem

If $z=x+iy$, $xy\ne0$, satisfies the equation $z^2+i\bar z=0$, then $|z^2|$ is equal to:

Key insight. Expanding $z^2$ and $i\bar z$ in terms of $x$ and $y$ turns one complex equation into two real equations — separating real and imaginary parts. The condition $xy\ne0$ is what rules out the trivial (and invalid) solution branch, forcing a specific value of $y$.

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Approach

Substitute $z=x+iy$ into $z^2+i\bar z=0$ and expand both terms using $i^2=-1$. Collect the real and imaginary parts separately — each must equal zero on its own, since the whole expression equals zero. The imaginary-part equation factors neatly and, together with $xy\ne0$, pins down $y$ exactly; substituting that back into the real-part equation gives $x^2$, and from there $|z^2|$ follows directly.

Solution

Step 1 — Expand z² and iz̄

$$z^2 = (x+iy)^2 = x^2-y^2+2ixy$$

$$\bar z = x-iy \implies i\bar z = i(x-iy) = ix-i^2y = y+ix$$

Step 2 — Add and separate real/imaginary parts

$$z^2+i\bar z = (x^2-y^2+y) + i(2xy+x) = 0$$

Setting both parts to zero:

$$\text{Real: } x^2-y^2+y=0 \quad\text{…(1)}$$

$$\text{Imaginary: } 2xy+x=0 \quad\text{…(2)}$$

Step 3 — Solve the imaginary-part equation

$$x(2y+1) = 0$$

Since $xy\ne0$, we know $x\ne0$, so the other factor must vanish:

$$2y+1=0 \implies y=-\frac12$$

Step 4 — Substitute into the real-part equation

$$x^2-\left(-\frac12\right)^2+\left(-\frac12\right) = 0 \implies x^2-\frac14-\frac12=0 \implies x^2=\frac34$$

Step 5 — Compute z²

$$z^2 = x^2-y^2+2ixy = \left(\frac34-\frac14\right)+2ix\left(-\frac12\right) = \frac12-ix$$

Step 6 — Compute |z²|

$$|z^2|^2 = \left(\frac12\right)^2+x^2 = \frac14+\frac34 = 1 \implies |z^2|=1$$

Answer

Common mistakes

• Accepting x=0 as a solution. The imaginary-part equation $x(2y+1)=0$ has $x=0$ as one branch, but this is exactly ruled out by the given condition $xy\ne0$ — without that condition being applied, it’s easy to keep both branches and get a spurious extra case.
• Sign errors expanding $i\bar z$. $i(x-iy)=ix-i^2y=ix+y$, since $-i^2=1$ — losing track of this double negative is a common place to introduce a sign error that flips the real part of the final equation.
• Forgetting that $|z^2|$ doesn’t need the sign of $x$. Since $|z^2|^2=\left(\frac12\right)^2+x^2$ depends only on $x^2$ (not on whether $x=\sqrt{3}/2$ or $x=-\sqrt3/2$), there’s no need to determine the sign of $x$ to answer the question — both cases give the same modulus.

Practise next

• If $z=x+iy$, $xy\ne0$, satisfies $z^2+2i\bar z=0$, find $|z^2|$.