Modulus and Conjugate of a Complex Number

Two Structural Tools

For $z = a + ib$:

Key Point (Definitions): The modulus of z is the non-negative real number $|z| = \sqrt{a^2 + b^2}$. The conjugate of z is the complex number $\bar{z} = a - ib$ (flip the sign of the imaginary part only).

Examples: $|3 + i| = \sqrt{9 + 1} = \sqrt{10}$; $|2 - 5i| = \sqrt{4 + 25} = \sqrt{29}$; $\overline{3 + i} = 3 - i$; $\overline{2 - 5i} = 2 + 5i$; $\overline{-3i - 5} = 3i - 5$.

The workhorse identity

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

A complex number times its own conjugate is a real number — the squared modulus. Immediate payoff: the multiplicative inverse in one line,

$$z^{-1} = \frac{1}{z} = \frac{\bar{z}}{z\bar{z}} = \frac{\bar{z}}{|z|^2}$$

[Board Important] Two instant tests worth quoting: z is REAL $\iff z = \bar{z}$; z is PURELY IMAGINARY $\iff z + \bar{z} = 0$ (with z $\neq$ 0). Also $z + \bar{z} = 2\,\text{Re}\,z$ and $z - \bar{z} = 2i\,\text{Im}\,z$.

The Property Toolkit

For any complex numbers $z_1$, $z_2$ (with $z_2 \neq 0$ where it appears in a denominator):

Modulus rules:

• $|z_1 z_2| = |z_1|\,|z_2|$ — moduli multiply
• $\left|\dfrac{z_1}{z_2}\right| = \dfrac{|z_1|}{|z_2|}$ — and divide
• $|z| = |\bar{z}| = |-z|$; and $|z| = 0 \iff z = 0$

Conjugate rules:

• $\overline{z_1 z_2} = \bar{z}_1\,\bar{z}_2$ and $\overline{\left(\dfrac{z_1}{z_2}\right)} = \dfrac{\bar{z}_1}{\bar{z}_2}$
• $\overline{z_1 \pm z_2} = \bar{z}_1 \pm \bar{z}_2$
• $\overline{\bar{z}} = z$ (conjugating twice restores z)

The practical meaning: conjugates and moduli pass through products and quotients term by term. To find $\left|\frac{(3-2i)(2+3i)}{(1+2i)(2-i)}\right|$, never expand — compute $\frac{\sqrt{13}\sqrt{13}}{\sqrt{5}\sqrt{5}} = \frac{13}{5}$ directly.

[JEE Tip] Modulus questions about horrendous quotients are gifts: apply $|z_1z_2| = |z_1||z_2|$ factor-wise and skip all expansion. Conjugate questions distribute the bar the same way. Expansion is only for questions demanding the actual a + ib form.

Solved Examples

Example 1: Multiplicative inverse

Find the multiplicative inverse of $2 - 3i$.

Solution:

Step 1 — Write the conjugate and the squared modulus. $\bar{z} = 2 + 3i$ and $|z|^2 = 2^2 + (-3)^2 = 13$.

Step 2 — Apply the formula. $z^{-1} = \frac{\bar{z}}{|z|^2} = \frac{2 + 3i}{13} = \frac{2}{13} + \frac{3}{13}i$.

Step 3 — Check by multiplying. $(2 - 3i)\left(\frac{2 + 3i}{13}\right) = \frac{4 + 9}{13} = 1$ ✓.

Takeaway: Inverse = conjugate over modulus-squared, verified by one multiplication.

Example 2: A quotient to standard form

Express $\frac{5 + \sqrt{2}i}{1 - \sqrt{2}i}$ in the form $a + ib$.

Solution:

Step 1 — Multiply top and bottom by the conjugate of the denominator. $\frac{(5 + \sqrt{2}i)(1 + \sqrt{2}i)}{(1 - \sqrt{2}i)(1 + \sqrt{2}i)}$.

Step 2 — The denominator turns real. $1 - (\sqrt{2}i)^2 = 1 + 2 = 3$.

Step 3 — Expand the numerator. $5 + 5\sqrt{2}i + \sqrt{2}i + 2i^2 = 3 + 6\sqrt{2}i$.

Step 4 — Divide. $\frac{3 + 6\sqrt{2}i}{3} = 1 + 2\sqrt{2}i$.

Takeaway: The denominator $1 - (\sqrt{2}i)^2 = 3$ is real by design — that is the conjugate trick working.

Example 3: Inverse of −i

Find the multiplicative inverse of $-i$.

Solution:

Step 1 — Conjugate and modulus. $\overline{-i} = i$ and $|-i|^2 = 0^2 + (-1)^2 = 1$.

Step 2 — Apply the formula. $z^{-1} = \frac{i}{1} = i$.

Step 3 — Check. $(-i)(i) = -i^2 = 1$ ✓.

Takeaway: Unit-modulus numbers have inverse = conjugate.

Example 4: Two inverses with surds

Find the multiplicative inverses of (i) $4 - 3i$ (ii) $\sqrt{5} + 3i$.

Solution:

Step 1 — (i) Modulus squared. $|z|^2 = 16 + 9 = 25$.

Step 2 — Formula. $z^{-1} = \frac{4 + 3i}{25} = \frac{4}{25} + \frac{3}{25}i$.

Step 3 — (ii) Modulus squared. $|z|^2 = 5 + 9 = 14$.

Step 4 — Formula. $z^{-1} = \frac{\sqrt{5} - 3i}{14} = \frac{\sqrt{5}}{14} - \frac{3}{14}i$.

Takeaway: The formula handles surds without complaint.

Example 5: Modulus of a quotient, no expansion

Find $\left|\frac{(3 - 2i)(2 + 3i)}{(1 + 2i)(2 - i)}\right|$.

Solution:

Step 1 — Take moduli factor-wise. $|3 - 2i| = \sqrt{13}$, $|2 + 3i| = \sqrt{13}$, $|1 + 2i| = \sqrt{5}$, $|2 - i| = \sqrt{5}$.

Step 2 — Multiply and divide. $\frac{\sqrt{13} \cdot \sqrt{13}}{\sqrt{5} \cdot \sqrt{5}} = \frac{13}{5}$.

Takeaway: Four multiplications avoided entirely — $|z_1z_2| = |z_1||z_2|$ is the fastest tool in the chapter.

Example 6: Conjugate of a quotient

Find the conjugate of $\frac{(3 - 2i)(2 + 3i)}{(1 + 2i)(2 - i)}$.

Solution:

Step 1 — Expand top and bottom. Top: $(3 - 2i)(2 + 3i) = 6 + 9i - 4i - 6i^2 = 12 + 5i$. Bottom: $(1 + 2i)(2 - i) = 2 - i + 4i - 2i^2 = 4 + 3i$.

Step 2 — Rationalise the quotient. $\frac{12 + 5i}{4 + 3i} \times \frac{4 - 3i}{4 - 3i} = \frac{48 - 36i + 20i + 15}{16 + 9} = \frac{63 - 16i}{25}$.

Step 3 — Flip the imaginary sign. $\overline{\left(\frac{63 - 16i}{25}\right)} = \frac{63}{25} + \frac{16}{25}i$.

Takeaway: Here the a + ib form IS needed, so expansion is unavoidable; the conjugate then just flips one sign.

Example 7: The x² + y² = 1 classic

If $x + iy = \frac{a + ib}{a - ib}$, prove that $x^2 + y^2 = 1$.

Solution:

Step 1 — Take moduli of both sides. $|x + iy| = \frac{|a + ib|}{|a - ib|}$.

Step 2 — Conjugates share a modulus. Both are $\sqrt{a^2 + b^2}$, so the quotient is 1: $|x + iy| = 1$.

Step 3 — Square. $x^2 + y^2 = 1$. ∎

Takeaway: Two lines by moduli (the expansion route takes ten). $|z| = 1$ statements are always modulus statements in disguise.

Example 8: The generalised version

If $x - iy = \sqrt{\frac{a - ib}{c - id}}$, prove that $(x^2 + y^2)^2 = \frac{a^2 + b^2}{c^2 + d^2}$.

Solution:

Step 1 — Clear the square root. $(x - iy)^2 = \frac{a - ib}{c - id}$.

Step 2 — Take moduli. $|x - iy|^2 = \frac{|a - ib|}{|c - id|}$, i.e. $x^2 + y^2 = \frac{\sqrt{a^2 + b^2}}{\sqrt{c^2 + d^2}}$.

Step 3 — Square again. $(x^2 + y^2)^2 = \frac{a^2 + b^2}{c^2 + d^2}$. ∎

Takeaway: The same modulus trick, applied after clearing the square root.

Example 9: Product of four factors

If $(a + ib)(c + id)(e + if)(g + ih) = A + iB$, show that $(a^2 + b^2)(c^2 + d^2)(e^2 + f^2)(g^2 + h^2) = A^2 + B^2$.

Solution:

Step 1 — Take moduli of both sides. $|a + ib|\,|c + id|\,|e + if|\,|g + ih| = |A + iB|$, since moduli multiply across any number of factors.

Step 2 — Square both sides. Each $|\cdot|^2$ becomes its sum of squares: $(a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2) = A^2 + B^2$. ∎

Takeaway: The identity $|z_1z_2| = |z_1||z_2|$ extends to any number of factors; squaring finishes it.

Example 10: Modulus of a difference of quotients

Find the modulus of $\frac{1 + i}{1 - i} - \frac{1 - i}{1 + i}$.

Solution:

Step 1 — Recognise the standard quotients. $\frac{1+i}{1-i} = i$ and $\frac{1-i}{1+i} = -i$.

Step 2 — Subtract. $i - (-i) = 2i$.

Step 3 — Take the modulus. $|2i| = 2$.

Takeaway: Recognising the two standard quotients turns this into mental arithmetic.

Example 11: Conjugate detective work

Find real numbers x and y if $(x - iy)(3 + 5i)$ is the conjugate of $-6 - 24i$.

Solution:

Step 1 — Write down the target. $\overline{-6 - 24i} = -6 + 24i$.

Step 2 — Expand the left side. $(x - iy)(3 + 5i) = 3x + 5xi - 3yi - 5yi^2 = (3x + 5y) + i(5x - 3y)$.

Step 3 — Match real and imaginary parts. $3x + 5y = -6$ and $5x - 3y = 24$.

Step 4 — Solve the system. Multiply the first by 3 and the second by 5: $9x + 15y = -18$ and $25x - 15y = 120$; adding, $34x = 102$, so $x = 3$, and then $y = \frac{-6 - 9}{5} = -3$.

Step 5 — Check. $(3 + 3i)(3 + 5i) = 9 + 15i + 9i - 15 = -6 + 24i$ ✓.

Takeaway: "Is the conjugate of" just fixes the right-hand side; then it is equality of complex numbers again.

Example 12: A real-quotient condition

Find real y such that $\frac{1 + iy}{1 - iy}$ has modulus 1, and show this holds for every real y.

Solution:

Step 1 — Take moduli. $\left|\frac{1 + iy}{1 - iy}\right| = \frac{\sqrt{1 + y^2}}{\sqrt{1 + y^2}} = 1$ for ALL real y.

Step 2 — See why. Numerator and denominator are conjugates of each other, and conjugates share a modulus.

Takeaway: A quotient of conjugates always lies on the unit circle — a fact reused in the β−α problem below.

Example 13: The unit-modulus quotient

If $\alpha$ and $\beta$ are different complex numbers with $|\beta| = 1$, find $\left|\frac{\beta - \alpha}{1 - \bar{\alpha}\beta}\right|$.

Solution:

Step 1 — Replace the 1 in the denominator. $|\beta| = 1$ means $\beta\bar{\beta} = 1$, so $1 - \bar{\alpha}\beta = \beta\bar{\beta} - \bar{\alpha}\beta$.

Step 2 — Factor out β. $= \beta(\bar{\beta} - \bar{\alpha}) = \beta\,\overline{(\beta - \alpha)}$.

Step 3 — Take moduli. $\left|\frac{\beta - \alpha}{\beta\overline{(\beta - \alpha)}}\right| = \frac{|\beta - \alpha|}{|\beta|\,|\overline{\beta - \alpha}|} = \frac{|\beta - \alpha|}{1 \cdot |\beta - \alpha|} = 1$.

Final Answer: 1.

Takeaway: A JEE favourite — factor β out of the denominator using $\beta\bar{\beta} = 1$, and the moduli cancel.

Example 14: Counting solutions

Find the number of non-zero integral solutions of $|1 - i|^x = 2^x$.

Solution:

Step 1 — Compute the modulus. $|1 - i| = \sqrt{1 + 1} = \sqrt{2} = 2^{1/2}$.

Step 2 — Rewrite with equal bases. $2^{x/2} = 2^x$.

Step 3 — Equate exponents. $\frac{x}{2} = x$ gives $x = 0$ — but the question demands NON-ZERO solutions.

Step 4 — Conclude. There are no non-zero integral solutions: the count is 0.

Takeaway: Reduce to equal bases, and read the question's fine print ("non-zero").