Chapter 4 · Complex Numbers and Quadratic Equations

4.2 Conjugate, Modulus and Division

Change the sign of the imaginary part and you get the conjugate $\bar z$. Multiply $z$ by it and every $i$ disappears, leaving the real number $|z|^2$ — one fact that gives the modulus, the inverse $z^{-1}$ and a recipe for dividing complex numbers.

In Section 4.1 we learnt to add, subtract and multiply complex numbers. Division is the missing operation: what is $\dfrac{3+4i}{1-2i}$ in the form $a+ib$? You have met this kind of problem before. In Class 9 and 10, to simplify $\dfrac{1}{3-\sqrt2}$ you multiplied the numerator and the denominator by $3+\sqrt2$, because $(3-\sqrt2)(3+\sqrt2)=9-2=7$ has no surd left in it. Complex numbers work the same way: $(1-2i)(1+2i)=1-4i^2=5$ has no $i$ left in it.

The partner $1+2i$ of $1-2i$ is called its conjugate, and the real number you get by multiplying a complex number by its conjugate measures its size — the square of its modulus. In this section we define the conjugate and the modulus, collect their properties, and use them to find the multiplicative inverse of every non-zero complex number and to divide any complex number by another. The same ideas let engineers compute currents in AC circuits and the gain of audio filters.

The Conjugate of a Complex Number

📘 Definition 4.6 — Conjugate

The conjugate of the complex number $z=a+ib$ ($a,b\in\mathbb{R}$) is $$\bar z=a-ib.$$ It has the same real part as $z$ and the opposite imaginary part. For example, $\overline{3+4i}=3-4i$, $\overline{-2i}=2i$ and $\bar 5=5$.

📐 Theorem 4.5 — Properties of the conjugate

For all complex numbers $z=a+ib$, $z_1$, $z_2$:

$\overline{\bar z}=z$;
$z+\bar z=2\operatorname{Re}z$ and $z-\bar z=2i\operatorname{Im}z$;
$z=\bar z$ if and only if $z$ is real;
$\overline{z_1+z_2}=\bar z_1+\bar z_2$, $\ \overline{z_1-z_2}=\bar z_1-\bar z_2$ and $\ \overline{z_1z_2}=\bar z_1\,\bar z_2$;
$z\bar z=a^2+b^2$, a real number that is never negative.

Proof of two of them. $z\bar z=(a+ib)(a-ib)=a^2-(ib)^2$ $=a^2-i^2b^2=a^2+b^2$. For the product, let $z_1=a+ib$ and $z_2=c+id$. Then $z_1z_2=(ac-bd)+i(ad+bc)$, so $\overline{z_1z_2}=(ac-bd)-i(ad+bc)$. On the other hand $\bar z_1\bar z_2=(a-ib)(c-id)$ $=(ac-bd)+i(-ad-bc)$ — the same number. The remaining properties follow directly from the definition. $\blacksquare$

✏️ Example 1 — Working with conjugates

(a) For $z=3-4i$, find $\bar z$, $z+\bar z$, $z-\bar z$ and $z\bar z$.
(b) Check that $\overline{z_1z_2}=\bar z_1\,\bar z_2$ for $z_1=1+2i$ and $z_2=3-i$.
(c) Find the complex number $z$ such that $2z+\bar z=6-2i$.

💡 Show solution▶

(a) $\bar z=3+4i$; $z+\bar z=6=2\operatorname{Re}z$; $z-\bar z=-8i=2i\operatorname{Im}z$; $z\bar z=3^2+4^2=25$.

(b) $z_1z_2=3-i+6i-2i^2=5+5i$, so $\overline{z_1z_2}=5-5i$. Also $\bar z_1\bar z_2=(1-2i)(3+i)=3+i-6i-2i^2=5-5i$ ✓.

The Modulus of a Complex Number

📘 Definition 4.7 — Modulus

The modulus of $z=a+ib$ ($a,b\in\mathbb{R}$) is the non-negative real number $$|z|=\sqrt{a^2+b^2}.$$ By Theorem 4.5, $z\bar z=|z|^2$. For a real number $z=a$, $|z|=\sqrt{a^2}$ is the usual modulus $|a|$ of Section 2.3.

Think of $a$ and $b$ as the two legs of a right-angled triangle: by Pythagoras' theorem, $|z|$ is the length of the hypotenuse. So $|3+4i|=\sqrt{9+16}=5$, $|1-i|=\sqrt2$ and $|2i|=2$. Notice that $i$ is not included when squaring $b$: $|a+ib|=\sqrt{a^2+b^2}$, never $\sqrt{a^2+(ib)^2}$.

📐 Theorem 4.6 — Properties of the modulus

For all complex numbers $z$, $z_1$, $z_2$:

$|z|\ge0$, and $|z|=0$ if and only if $z=0$;
$|\bar z|=|z|=|-z|$;
$|z_1z_2|=|z_1|\,|z_2|$ — the modulus of a product is the product of the moduli.

Proof. $a^2+b^2\ge0$, with equality only when $a=b=0$. Next, $\bar z=a-ib$ and $-z=-a-ib$ both have modulus $\sqrt{a^2+b^2}$. For the product, use $w\bar w=|w|^2$ and Theorem 4.5: $$\begin{aligned}|z_1z_2|^2&=(z_1z_2)\overline{(z_1z_2)}=z_1z_2\,\bar z_1\bar z_2\\&=(z_1\bar z_1)(z_2\bar z_2)=|z_1|^2|z_2|^2.\end{aligned}$$ Both $|z_1z_2|$ and $|z_1||z_2|$ are non-negative, so taking square roots gives $|z_1z_2|=|z_1|\,|z_2|$. $\blacksquare$

✏️ Example 2 — Moduli, and the product rule

(a) Find $|{-5+12i}|$, $|1-i|$, $|-7|$ and $|\sqrt3+i|$.
(b) Find $|(3+4i)(5-12i)|$ without multiplying out, then check by multiplying.

💡 Show solution▶

(a) $\sqrt{25+144}=13$; $\ \sqrt{1+1}=\sqrt2$; $\ \sqrt{49}=7$; $\ \sqrt{3+1}=2$.

(b) Product rule $|3+4i|\,|5-12i|=5\times13=65$.

(b) Check $(3+4i)(5-12i)=15-36i+20i-48i^2=63-16i$, and $\sqrt{63^2+16^2}=\sqrt{3969+256}=\sqrt{4225}=65$ ✓.

Each complex number $a+ib$ is drawn as a right-angled triangle: a red leg of length $|a|$, a blue leg of length $|b|$, and the hypotenuse $|z|$. The dashed mirror triangle is the conjugate $\bar z$ — same legs, same hypotenuse. All three triangles use the same scale. Watch the product rule: the hypotenuse of $z_1z_2$ is always the hypotenuse of $z_1$ times the hypotenuse of $z_2$.

Numbers:

$|z_1|$—

$|z_2|$—

$|z_1|\,|z_2|$—

$z_1z_2$ and $|z_1z_2|$—

The Multiplicative Inverse and Division

For a non-zero real number $x$, the inverse $\dfrac1x$ is the number whose product with $x$ is $1$. To find such a number for $z=a+ib\ne0$, start from $z\bar z=|z|^2$ and divide both sides by the non-zero real number $|z|^2$: $$z\cdot\dfrac{\bar z}{|z|^2}=1.$$

📘 Definition 4.8 — Multiplicative inverse and quotient

The multiplicative inverse of a non-zero complex number $z=a+ib$ is $$z^{-1}=\dfrac1z=\dfrac{\bar z}{|z|^2},$$ that is, $$z^{-1}=\dfrac{a}{a^2+b^2}-i\,\dfrac{b}{a^2+b^2}.$$ For $z_2\ne0$, the quotient of $z_1$ by $z_2$ is $\dfrac{z_1}{z_2}=z_1\,z_2^{-1}$.

📐 Theorem 4.7 — The inverse and the quotient

Let $z,z_1,z_2\in\mathbb{C}$ with $z\ne0$ and $z_2\ne0$. Then:

$z\,z^{-1}=1$, and $z^{-1}$ is the only complex number with this property;
$\left|\dfrac{z_1}{z_2}\right|=\dfrac{|z_1|}{|z_2|}$ and $\overline{\left(\dfrac{z_1}{z_2}\right)}=\dfrac{\bar z_1}{\bar z_2}$.

Proof. $z\cdot\dfrac{\bar z}{|z|^2}=\dfrac{z\bar z}{|z|^2}=\dfrac{|z|^2}{|z|^2}=1$. If also $zw=1$, then $w=w\,(z\,z^{-1})=(wz)\,z^{-1}=z^{-1}$, so the inverse is unique. For the quotient, put $q=\dfrac{z_1}{z_2}$, so that $z_1=q\,z_2$. Theorem 4.6 gives $|z_1|=|q|\,|z_2|$, and Theorem 4.5 gives $\bar z_1=\bar q\,\bar z_2$; divide by $|z_2|\ne0$ and by $\bar z_2\ne0$ respectively. $\blacksquare$

🧭 Problem-solving strategy — Dividing complex numbers

Multiply the numerator and the denominator by the conjugate of the denominator: $\dfrac{a+ib}{c+id}=\dfrac{(a+ib)(c-id)}{(c+id)(c-id)}$.
The denominator becomes the real number $c^2+d^2$.
Expand the numerator with $i^2=-1$, then divide its real and imaginary parts separately by $c^2+d^2$.
Check by multiplying your answer by the original denominator.

✏️ Example 3 — Multiplicative inverses

Find the multiplicative inverse of (a) $4-3i$ (b) $i$ (c) $\sqrt2-i\sqrt7$.

💡 Show solution▶

(a) $\bar z=4+3i$ and $|z|^2=16+9=25$, so $z^{-1}=\dfrac{4+3i}{25}=\dfrac{4}{25}+\dfrac{3}{25}i$.

(a) Check $(4-3i)\cdot\dfrac{4+3i}{25}=\dfrac{16+9}{25}=1$ ✓.

(b) $\bar z=-i$, $|z|^2=1$: $i^{-1}=-i$ — as found in Section 4.1.

✏️ Example 4 — Division

Express in the form $a+ib$: (a) $\dfrac{3+4i}{1-2i}$ (b) $\dfrac{1+i}{1-i}$, and hence $\left(\dfrac{1+i}{1-i}\right)^{100}$ (c) $\dfrac{5}{2+i}$.

💡 Show solution▶

(a) Multiply top and bottom by $1+2i$: $\dfrac{(3+4i)(1+2i)}{1^2+2^2}=\dfrac{3+6i+4i+8i^2}{5}$ $=\dfrac{-5+10i}{5}=-1+2i$.

(a) Check $(-1+2i)(1-2i)=-1+2i+2i-4i^2=3+4i$ ✓.

(b) $\dfrac{(1+i)(1+i)}{(1-i)(1+i)}=\dfrac{2i}{2}=i$. Then $\left(\dfrac{1+i}{1-i}\right)^{100}=i^{100}=1$, since $100=4\times25$.

✏️ Example 5 — Moduli without multiplying out

Find $\left|\dfrac{(3+4i)(1-i)}{2+i\sqrt5}\right|$.

💡 Show solution▶

Rules By Theorems 4.6 and 4.7, the modulus of a product or quotient is the product or quotient of the moduli: the answer is $\dfrac{|3+4i|\,|1-i|}{|2+i\sqrt5|}$.

Moduli $|3+4i|=5$, $\ |1-i|=\sqrt2$, $\ |2+i\sqrt5|=\sqrt{4+5}=3$.

Answer $\dfrac{5\sqrt2}{3}\approx2.357$ — found without a single complex multiplication.

✏️ Example 6 — A modulus identity for quotients

Let $a,b,c,d,x,y$ be real, with $c$ and $d$ not both zero. Prove that if $\dfrac{a+ib}{c+id}=x+iy$, then $x^2+y^2=\dfrac{a^2+b^2}{c^2+d^2}$.

💡 Show solution▶

Idea Take the modulus of both sides — that turns one complex equation into one real equation.

Modulus $|x+iy|=\left|\dfrac{a+ib}{c+id}\right|=\dfrac{|a+ib|}{|c+id|}$ by Theorem 4.7.

Square $x^2+y^2=|x+iy|^2=\dfrac{a^2+b^2}{c^2+d^2}$. $\blacksquare$

Check $\dfrac{3+4i}{1-2i}=-1+2i$ (Example 4): $x^2+y^2=1+4=5$ and $\dfrac{9+16}{1+4}=5$ ✓.

⚠️ Common mistakes with conjugates and moduli

1. The conjugate changes only the imaginary sign. $\overline{-2+5i}=-2-5i$, not $2-5i$ and not $2+5i$.

2. No $i$ inside the modulus formula. $|3+4i|=\sqrt{9+16}=5$. Writing $\sqrt{3^2+(4i)^2}=\sqrt{9-16}$ is wrong.

3. $z\bar z$ is not $z^2$. $(3+4i)(3-4i)=25$, but $(3+4i)^2=-7+24i$. Only for real $z$ are they equal.

4. The modulus of a sum is not the sum of the moduli. $|3+4i|=5$, yet $|3|+|4i|=7$. Products and quotients split ($|z_1z_2|=|z_1||z_2|$); sums do not.

5. Multiply top and bottom by the conjugate. Multiplying only the denominator changes the value of the fraction.

🚀 Beyond the board syllabus (JEE Main · JEE Advanced)

In JEE Main and JEE Advanced 2026 (both list quadratic equations and their formation from given roots); not in the CBSE year-end exam. Both use conjugates to study the roots of quadratic equations with real coefficients.

Conjugate roots. Let $ax^2+bx+c=0$ with $a,b,c$ real and $a\ne0$. If $\alpha$ is a root, take conjugates of $a\alpha^2+b\alpha+c=0$: since $\bar a=a$, $\bar b=b$, $\bar c=c$ and $\overline{\alpha^2}=\bar\alpha^2$ (Theorem 4.5), we get $a\bar\alpha^2+b\bar\alpha+c=0$. So non-real roots come in conjugate pairs $p\pm iq$ — exactly when the discriminant $b^2-4ac$ is negative. Their sum is $2p$ and their product is $p^2+q^2=|\alpha|^2$, so the quadratic with leading coefficient $1$ is $x^2-2px+(p^2+q^2)=0$.

Worked example. Find the quadratic equation with real coefficients and leading coefficient $1$ that has $3-2i$ as a root.
Solution. The other root is $3+2i$; sum $=6$, product $=9+4=13$. The equation is $x^2-6x+13=0$. Check: $(3-2i)^2-6(3-2i)+13$ $=(5-12i)-18+12i+13=0$ ✓.

Exercises

Conjugate and modulus. In parts 1–4 find the conjugate and the modulus.

1 $5-12i$
2 $-3i$
3 $-\sqrt3+i$
4 $(2+i)^2$
5 For $z=4+3i$, find $z+\bar z$, $z-\bar z$ and $z\bar z$.
6 Verify that $\overline{z_1z_2}=\bar z_1\,\bar z_2$ for $z_1=2+i$ and $z_2=3-2i$.
7 Find the complex number $z$ with $z+2\bar z=9-4i$.
8 Prove that for every complex number $z$, $z\bar z$ is real and non-negative, and $z-\bar z$ is purely imaginary or zero.

💡 Show solutions▶

1 $5+12i$; $\sqrt{25+144}=13$.

2 $3i$; $3$.

3 $-\sqrt3-i$; $\sqrt{3+1}=2$.

4 $(2+i)^2=3+4i$: conjugate $3-4i$, modulus $5$. (Or $|2+i|^2=5$ directly, by Theorem 4.6.)

5 $z+\bar z=8$, $z-\bar z=6i$, $z\bar z=16+9=25$.

6 $z_1z_2=6-4i+3i-2i^2=8-i$, so $\overline{z_1z_2}=8+i$; $\bar z_1\bar z_2=(2-i)(3+2i)=6+4i-3i-2i^2=8+i$ ✓.

7 With $z=a+ib$: $a+ib+2a-2ib=3a-ib=9-4i$, so $a=3$, $b=4$: $z=3+4i$. Check: $(3+4i)+(6-8i)=9-4i$ ✓.

8 With $z=a+ib$: $z\bar z=a^2+b^2$, real and $\ge0$; $z-\bar z=2ib$, which is $0$ if $b=0$ and purely imaginary otherwise.

Inverses and quotients. Find the multiplicative inverse in parts 1–3; express parts 4–10 in the form $a+ib$.

1 $3+4i$
2 $-i$
3 $\sqrt5+3i$
4 $\dfrac{2+3i}{1+i}$
5 $\dfrac{4-3i}{2+i}$
6 $\dfrac{1}{(1-i)^2}$
7 $\dfrac{1-i}{1+i}$
8 $\left(\dfrac{1+i}{1-i}\right)^3-\left(\dfrac{1-i}{1+i}\right)^3$
9 $\dfrac{5+2i}{2-5i}$
10 $\dfrac{3+2i}{2-i}+\dfrac{1-i}{1+2i}$

Check: divide any two complex numbers

$z_1$:+$i$$z_2$:+$i$

$z_1/z_2$—

$1/z_2$—

$|z_1/z_2|^2$—

💡 Show solutions▶

1 $\dfrac{3-4i}{25}=\dfrac{3}{25}-\dfrac{4}{25}i$.

2 $\dfrac{\overline{-i}}{|-i|^2}=\dfrac{i}{1}=i$. Check: $(-i)(i)=-i^2=1$ ✓.

3 $|z|^2=5+9=14$: $\dfrac{\sqrt5-3i}{14}=\dfrac{\sqrt5}{14}-\dfrac{3}{14}i$.

4 $\dfrac{(2+3i)(1-i)}{2}=\dfrac{2-2i+3i-3i^2}{2}$ $=\dfrac{5+i}{2}=\dfrac52+\dfrac12i$.

5 $\dfrac{(4-3i)(2-i)}{5}=\dfrac{8-4i-6i+3i^2}{5}$ $=\dfrac{5-10i}{5}=1-2i$.

6 $(1-i)^2=-2i$, so the number is $\dfrac{1}{-2i}=\dfrac{i}{-2i\cdot i}=\dfrac{i}{2}=0+\dfrac12i$.

7 $\dfrac{(1-i)^2}{(1+i)(1-i)}=\dfrac{-2i}{2}=-i$.

8 By Example 4(b) and part 7 the two fractions are $i$ and $-i$, so the value is $i^3-(-i)^3=-i-i=-2i$.

9 $\dfrac{(5+2i)(2+5i)}{29}=\dfrac{10+25i+4i+10i^2}{29}$ $=\dfrac{29i}{29}=i$.

10 $\dfrac{(3+2i)(2+i)}{5}=\dfrac{4+7i}{5}$ and $\dfrac{(1-i)(1-2i)}{5}=\dfrac{-1-3i}{5}$; the sum is $\dfrac{3+4i}{5}=\dfrac35+\dfrac45i$.

III

Using the properties.

1 Find $|(1+i)(2-i)(3+4i)|$ without multiplying out.
2 Find $\left|\dfrac{6+8i}{1-i\sqrt3}\right|$.
3 If $|z|=3$, find $z\bar z$, $|\bar z|$, $|-2z|$ and $|z^2|$.
4 For $z_1=1+2i$ and $z_2=3-i$, compute $|z_1z_2|$ and $|z_1|\,|z_2|$ separately and compare.
5 Show by an example that $|z_1+z_2|$ need not equal $|z_1|+|z_2|$.
6 Find real $x$, $y$ with $(x+iy)(2-i)=1+3i$, and check that $x^2+y^2=\dfrac{|1+3i|^2}{|2-i|^2}$.
7 If $(1+i)z=(1-i)\bar z$, show that $z=k(1-i)$ for some real number $k$.
8 Find all complex numbers $z$ with $\bar z=z^2$.

💡 Show solutions▶

1 $\sqrt2\cdot\sqrt5\cdot5=5\sqrt{10}$.

2 $\dfrac{|6+8i|}{|1-i\sqrt3|}=\dfrac{10}{2}=5$.

3 $z\bar z=|z|^2=9$; $|\bar z|=3$; $|-2z|=|-2|\,|z|=6$; $|z^2|=|z|^2=9$.

4 $z_1z_2=3-i+6i-2i^2=5+5i$, $|z_1z_2|=\sqrt{50}=5\sqrt2$. $|z_1||z_2|=\sqrt5\cdot\sqrt{10}=\sqrt{50}=5\sqrt2$ — equal ✓.

5 $z_1=3$, $z_2=4i$: $|z_1+z_2|=|3+4i|=5$ but $|z_1|+|z_2|=7$.

6 $x+iy=\dfrac{1+3i}{2-i}=\dfrac{(1+3i)(2+i)}{5}=\dfrac{-1+7i}{5}$: $x=-\tfrac15$, $y=\tfrac75$. Then $x^2+y^2=\tfrac{1+49}{25}=2$ and $\tfrac{10}{5}=2$ ✓.

7 Put $z=a+ib$. Left: $(1+i)(a+ib)=(a-b)+i(a+b)$. Right: $(1-i)(a-ib)=(a-b)-i(a+b)$. Equal imaginary parts give $a+b=-(a+b)$, so $b=-a$ and $z=a-ia=a(1-i)$; take $k=a$.

8 With $z=a+ib$: $a-ib=(a^2-b^2)+2abi$, so $a=a^2-b^2$ and $-b=2ab$, i.e. $b(2a+1)=0$. If $b=0$: $a=a^2$, so $a=0$ or $1$. If $a=-\tfrac12$: $b^2=\tfrac14+\tfrac12=\tfrac34$, $b=\pm\tfrac{\sqrt3}{2}$. Solutions: $0$, $1$, $-\tfrac12\pm\tfrac{\sqrt3}{2}i$.

Applications. In AC circuits the voltage $V$ (volts), current $I$ (amperes) and impedance $Z$ (ohms) are complex numbers with $V=IZ$.

1 Two components with $Z_1=3+4i\ \Omega$ and $Z_2=3-4i\ \Omega$ are connected in parallel; the combined impedance is $Z=\dfrac{Z_1Z_2}{Z_1+Z_2}$. Find $Z$.
2 A voltage $V=10$ V is applied across $Z=3+4i\ \Omega$. Find the current $I=\dfrac VZ$ and its size $|I|$, and check that $|I|=\dfrac{|V|}{|Z|}$.
3 Audio filter. A simple low-pass filter multiplies a signal by $H=\dfrac{1}{1+ix}$, where $x$ grows with the frequency. Show that $|H|=\dfrac{1}{\sqrt{1+x^2}}$, and find $|H|$ when $x=1$ and when $x=3$.

💡 Show solutions▶

1 $Z_1Z_2=|Z_1|^2=25$ (conjugates) and $Z_1+Z_2=6$, so $Z=\dfrac{25}{6}\approx4.17\ \Omega$ — a purely real impedance.

2 $I=\dfrac{10}{3+4i}=\dfrac{10(3-4i)}{25}=1.2-1.6i$ A. $|I|=\sqrt{1.44+2.56}=2$ A, and $\dfrac{|V|}{|Z|}=\dfrac{10}{5}=2$ A ✓.

3 $|H|=\dfrac{|1|}{|1+ix|}=\dfrac{1}{\sqrt{1+x^2}}$ by Theorem 4.7. For $x=1$: $\tfrac{1}{\sqrt2}\approx0.707$; for $x=3$: $\tfrac{1}{\sqrt{10}}\approx0.316$ — high frequencies are reduced more.

🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case studyJEE MainJEE Advanced

How this is examined: Board papers ask for an inverse or a conjugate (1 mark), a quotient in the form $a+ib$ (2–3 marks), and modulus results proved with $|z_1z_2|=|z_1||z_2|$. JEE combines $z\bar z=|z|^2$ with conditions such as $|z|=1$ — then $\bar z=\dfrac1z$, a fact worth remembering.

E1. Board · 1 mark The multiplicative inverse of $2-3i$ is (A) $2+3i$ (B) $\dfrac{2+3i}{13}$ (C) $\dfrac{2+3i}{\sqrt{13}}$ (D) $\dfrac{-2+3i}{13}$

💡 Show solution▶

Idea $z^{-1}=\dfrac{\bar z}{|z|^2}$ with $\bar z=2+3i$ and $|z|^2=4+9=13$.

Trap (A) is only the conjugate; (C) divides by $|z|$ instead of $|z|^2$ — check: $(2-3i)\cdot\dfrac{2+3i}{\sqrt{13}}=\sqrt{13}\ne1$.

Answer (B).

E2. Board · 2 marks Express $\dfrac{7+i}{(2+i)^2}$ in the form $a+ib$.

💡 Show solution▶

Denominator $(2+i)^2=4+4i+i^2=3+4i$. (½ mark)

Divide $\dfrac{(7+i)(3-4i)}{3^2+4^2}=\dfrac{21-28i+3i-4i^2}{25}$ $=\dfrac{25-25i}{25}$. (1 mark)

Answer $1-i$. Check: $(1-i)(3+4i)=3+4i-3i-4i^2=7+i$ ✓. (½ mark)

E3. Board · 3 marks Let $z_1=2+3i$ and $z_2=1-i$. Find (i) $\dfrac{z_1}{z_2}$ in the form $a+ib$ (ii) $|z_1\bar z_2|$ (iii) $\operatorname{Re}(z_1\bar z_2)$.

💡 Show solution▶

(i) $\dfrac{(2+3i)(1+i)}{2}=\dfrac{2+2i+3i+3i^2}{2}$ $=-\dfrac12+\dfrac52i$. (1 mark)

(ii) $|z_1\bar z_2|=|z_1|\,|\bar z_2|=\sqrt{13}\cdot\sqrt2=\sqrt{26}$. (1 mark)

Answer (iii) $z_1\bar z_2=(2+3i)(1+i)=-1+5i$, so $\operatorname{Re}(z_1\bar z_2)=-1$. (Check (ii): $\sqrt{1+25}=\sqrt{26}$ ✓.) (1 mark)

E4. Board · Case study In a circuit, components with impedances $Z_1=4+3i\ \Omega$ and $Z_2=2-6i\ \Omega$ are connected in series (impedances add), and a voltage $V=30$ V is applied, so that the current is $I=\dfrac{V}{Z}$. (i) Find the total impedance $Z$. (1 mark) (ii) Find $|Z|$. (1 mark) (iii) Find the current $I$ in the form $a+ib$, and its size $|I|$. (2 marks)

💡 Show solution▶

(i) $Z=6-3i\ \Omega$. (1 mark)

(ii) $|Z|=\sqrt{36+9}=\sqrt{45}=3\sqrt5\approx6.71\ \Omega$. (1 mark)

Answer (iii) $I=\dfrac{30(6+3i)}{45}=4+2i$ A; $|I|=\sqrt{20}=2\sqrt5\approx4.47$ A (also $\dfrac{30}{3\sqrt5}=2\sqrt5$ ✓). (2 marks)

E5. JEE Main Let $|z|=1$ and $z\ne\pm1$. Then $w=\dfrac{z-1}{z+1}$ is (A) a non-zero real number (B) purely imaginary (C) of modulus $1$ (D) zero

💡 Show solution▶

Idea $|z|=1$ gives $z\bar z=1$, so $\bar z=\dfrac1z$. Then $\bar w=\dfrac{\bar z-1}{\bar z+1}=\dfrac{\frac1z-1}{\frac1z+1}=\dfrac{1-z}{1+z}=-w$.

Conclude $w+\bar w=0$ means $\operatorname{Re}w=0$; and $w\ne0$ because $z\ne1$. So $w$ is purely imaginary.

Trap (C) is tempting because $|z|=1$, but for $z=\tfrac35+\tfrac45i$: $w=\dfrac{-2+4i}{8+4i}=\dfrac i2$, of modulus $\tfrac12$.

Answer (B).

E6. JEE Advanced One or more options correct. For all complex numbers $z_1$, $z_2$: (A) $|z_1\bar z_2|=|z_1|\,|z_2|$ (B) $z_1\bar z_2+\bar z_1z_2$ is real (C) $|z_1+z_2|^2=|z_1|^2+|z_2|^2+2\operatorname{Re}(z_1\bar z_2)$ (D) $z_1\bar z_1=z_1^2$

💡 Show solution▶

(A) True: $|z_1\bar z_2|=|z_1|\,|\bar z_2|=|z_1|\,|z_2|$.

(B) True: $\bar z_1z_2=\overline{z_1\bar z_2}$, so the sum is $w+\bar w=2\operatorname{Re}w$ with $w=z_1\bar z_2$.

(C), (D) (C) True: $|z_1+z_2|^2=(z_1+z_2)(\bar z_1+\bar z_2)=|z_1|^2+|z_2|^2+(z_1\bar z_2+\bar z_1z_2)$, and the bracket is $2\operatorname{Re}(z_1\bar z_2)$ by (B). (D) False: for $z_1=i$, $z_1\bar z_1=1$ but $z_1^2=-1$.

Answer (A), (B), (C).

🔬 Research–Teaching NexusMini project · about 45 minutes

Research question. Why do cheap earphones lose the high notes? How much does a simple electronic filter weaken a signal at each frequency?

Why it matters. Every phone, speaker and hearing aid contains filters. The simplest one — a resistor $R$ and a capacitor $C$ — multiplies a signal of angular frequency $\omega$ by the complex number $H=\dfrac{1}{1+i\omega RC}$. The modulus $|H|$ is the fraction of the signal's size that survives, so the whole design question is a modulus calculation.

What to do.

Use Theorem 4.7 to show that $|H|=\dfrac{1}{\sqrt{1+(\omega RC)^2}}$.
Take $R=1000\ \Omega$ and $C=0.1\ \mu$F, so $RC=10^{-4}$ s. In a spreadsheet, compute $|H|$ for the frequencies $f$ = 100, 200, 500, 1000, 2000, 5000, 10 000 and 20 000 Hz, using $\omega=2\pi f$.
Plot $|H|$ against $f$ and find the cut-off frequency, where $|H|=\dfrac{1}{\sqrt2}\approx0.707$.
Listen to a tone generator app at a few of these frequencies and describe what changes when a signal is reduced to $0.3$ of its size.

Tools: A spreadsheet app; optionally a free tone-generator app

Hand in: Your table and graph, the cut-off frequency, and a short explanation of why high notes are weakened most.

Going further. Write $H$ in the form $a+ib$ by multiplying by the conjugate, and find the frequency at which its real and imaginary parts have equal size.

Instructor note: $|H|=\tfrac1{\sqrt2}$ when $\omega RC=1$, i.e. $\omega=10^4$ rad/s and $f=\tfrac{10^4}{2\pi}\approx1592$ Hz. In the form $a+ib$, $H=\dfrac{1-i\omega RC}{1+(\omega RC)^2}$, whose parts have equal size at the same frequency.

🧪 Self-check quiz

0/5 correct

Q1. The conjugate of $-2+5i$ is

Q2. $|-8+6i|$ equals

Q3. For $z=1-2i$, the product $z\bar z$ is

Q4. $\dfrac{2+i}{2-i}$ equals

Q5. If $|z_1|=2$ and $|z_2|=5$, then $\left|\dfrac{z_1^2}{z_2}\right|$ is

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