Conjugate z̄ = 2 - 3i. Modulus |z| = √(4 + 9) = √13.

If z = 2 + 3i, Find z̄ and |z|

Question

Given $z = 2 + 3i$ , find:

The conjugate $\bar{z}$
The modulus $|z|$

Solution — Step by Step

Write $z$ in standard form $a + bi$ . Here, $a = 2$ (real part) and $b = 3$ (imaginary part). This separation is the foundation for everything that follows.

The conjugate simply flips the sign of the imaginary part. If $z = a + bi$ , then $\bar{z} = a - bi$ .

\bar{z} = 2 - 3i

The modulus $|z|$ is the distance of the complex number from the origin on the Argand plane. Why? Because $z = a + bi$ maps to the point $(a, b)$ , and distance from origin is $\sqrt{a^2 + b^2}$ by Pythagoras.

|z| = \sqrt{a^2 + b^2} = \sqrt{(2)^2 + (3)^2}

|z| = \sqrt{4 + 9} = \sqrt{13}

Since $\sqrt{13}$ has no perfect square factors, this is already fully simplified.

Final answers:

\bar{z} = 2 - 3i \qquad |z| = \sqrt{13}

Why This Works

The conjugate $\bar{z}$ reflects the point across the real axis on the Argand plane. The real part stays fixed; only the imaginary part changes sign. Geometrically, $z$ and $\bar{z}$ are mirror images of each other about the x-axis.

The modulus formula comes directly from the Pythagorean theorem. We treat the complex number as a position vector — $a$ units along the real axis, $b$ units along the imaginary axis. The length of that vector is $\sqrt{a^2 + b^2}$ .

One elegant check: $z \cdot \bar{z} = (2 + 3i)(2 - 3i) = 4 + 9 = 13 = |z|^2$ . This identity — that the product of a complex number with its conjugate equals the square of its modulus — appears repeatedly in JEE problems and is worth memorising.

Alternative Method

Using the $z \cdot \bar{z} = |z|^2$ identity to verify:

$(2 + 3i)(2 - 3i) = 2^2 - (3i)^2 = 4 - 9i^2 = 4 - 9(-1) = 4 + 9 = 13$

So $|z|^2 = 13$ , giving $|z| = \sqrt{13}$ . This is a useful cross-check during exams — if you’ve already found $\bar{z}$ , multiply out to confirm $|z|^2$ .

In MCQs where the full solution isn’t required, compute $a^2 + b^2$ first. If it’s a perfect square (like 25, 36, 169), the modulus is a clean integer. If not (like 13), leave it as a surd — never approximate $\sqrt{13} \approx 3.6$ in CBSE or JEE unless the question explicitly asks for a decimal.

Common Mistake

Students often write $\bar{z} = -2 + 3i$ , flipping the real part instead of the imaginary part. The conjugate only changes the sign of $bi$ . The real part $a$ is completely untouched.

Another slip: computing $|z| = \sqrt{2 + 3} = \sqrt{5}$ by adding $a$ and $b$ directly instead of squaring them first. Always square both parts: $\sqrt{a^2 + b^2}$ , never $\sqrt{a + b}$ .

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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