Chapter 4 · Complex Numbers and Quadratic Equations

4.1 Complex Numbers and Their Algebra

Add to the real numbers one new number $i$ with $i^2=-1$ and you get the complex numbers $a+ib$. They add, subtract and multiply by the familiar rules of algebra — with $i^2$ replaced by $-1$ wherever it appears.

Every time mathematics met an equation it could not solve, it grew a new kind of number. $x+5=2$ has no solution in $\mathbb{N}$, so we use the negative integers; $3x=2$ has none in $\mathbb{Z}$, so we use fractions; $x^2=2$ has none in $\mathbb{Q}$, so we use irrational numbers such as $\sqrt2$. The real numbers $\mathbb{R}$ seem to finish the job — every point of the number line is one. Yet the equation $x^2+1=0$ still has no solution, because the square of a real number is never negative.

In this section we take the same step once more. We introduce a number $i$ whose square is $-1$, build the numbers $a+ib$ from it, and agree that they obey all the usual rules of algebra. The surprise is that nothing breaks: we can add, subtract and multiply exactly as with binomials such as $a+b\sqrt2$, and the only new rule is $i^2=-1$. You will learn to write any expression in the standard form $a+ib$, to use equality of complex numbers to find unknowns, to simplify any power of $i$, and to use the algebraic identities you already know.

The Imaginary Unit and the Form a + ib

📘 Definition 4.1 — The imaginary unit $i$

The imaginary unit $i$ is a number with $$i^2=-1.$$ For a positive real number $a$ we define $\sqrt{-a}=i\sqrt a$. In particular $\sqrt{-1}=i$, $\sqrt{-9}=3i$ and $\sqrt{-5}=i\sqrt5$.

Both $i$ and $-i$ square to $-1$, since $(-i)^2=(-1)^2i^2=-1$. So the equation $x^2+1=0$ now has two solutions, $x=i$ and $x=-i$, and more generally $x^2=-a$ (with $a\gt0$) has the two solutions $x=\pm i\sqrt a$. The name "imaginary" is an accident of history — René Descartes used it as an insult — but $i$ is no less real a mathematical object than $\sqrt2$ or $-3$.

📘 Definition 4.2 — Complex number; real and imaginary parts

A complex number is a number of the form $$z=a+ib,\qquad a,b\in\mathbb{R}.$$ The real number $a$ is the real part of $z$, written $\operatorname{Re}z$, and the real number $b$ is the imaginary part, written $\operatorname{Im}z$. The set of all complex numbers is denoted $\mathbb{C}$.

If $b=0$, $z=a$ is a real number; so $\mathbb{R}\subset\mathbb{C}$. If $a=0$ and $b\ne0$, $z=ib$ is called purely imaginary.

For example, $z=3-4i$ has $\operatorname{Re}z=3$ and $\operatorname{Im}z=-4$ — note that the imaginary part is the real number $-4$, not $-4i$. The number $-7$ is complex with imaginary part $0$, and $2i$ is purely imaginary with real part $0$. We write $3-4i$ rather than $3+(-4)i$, and either $3+4i$ or $3+i4$ for the same number; this course uses $3+4i$.

📘 Definition 4.3 — Equality of complex numbers

Two complex numbers are equal when their real parts are equal and their imaginary parts are equal: $$a+ib=c+id\iff a=c\ \text{and}\ b=d,$$ where $a,b,c,d\in\mathbb{R}$.

One equation between complex numbers is therefore two equations between real numbers — a fact that makes complex equations a powerful tool for finding two real unknowns at once. In particular $a+ib=0$ means $a=0$ and $b=0$.

✏️ Example 1 — Writing numbers in the form $a+ib$
Write each number in the form $a+ib$ and state its real and imaginary parts: (a) $\sqrt{-36}$ (b) $4-\sqrt{-5}$ (c) $\dfrac{-1+\sqrt{-3}}{2}$ (d) $7$ (e) $\sqrt{-16}+3\sqrt{-25}-2$.
Rule Replace each $\sqrt{-a}$ (with $a\gt0$) by $i\sqrt a$, then collect the real terms and the terms containing $i$.
(a), (b) $\sqrt{-36}=6i=0+6i$: real part $0$, imaginary part $6$ (purely imaginary). $4-\sqrt{-5}=4-i\sqrt5$: real part $4$, imaginary part $-\sqrt5$.
(c), (d) $\dfrac{-1+i\sqrt3}{2}=-\tfrac12+\tfrac{\sqrt3}{2}i$: real part $-\tfrac12$, imaginary part $\tfrac{\sqrt3}{2}$. $7=7+0i$: real part $7$, imaginary part $0$.
(e) $4i+3(5i)-2=-2+19i$: real part $-2$, imaginary part $19$.
✏️ Example 2 — Finding unknowns from equality
(a) Find real numbers $x$ and $y$ such that $(2x-y)+i(x+3y)=7+7i$.
(b) Find the real number $x$ for which $(x^2-4)+i(x+2)=0$.
(a) Compare $x$ and $y$ are real, so $2x-y$ and $x+3y$ are the real and imaginary parts of the left side. Equating parts: $2x-y=7$ and $x+3y=7$.
(a) Solve From the first, $y=2x-7$. Then $x+3(2x-7)=7$, so $7x=28$, $x=4$ and $y=1$.
(a) Check $2(4)-1=7$ ✓ and $4+3(1)=7$ ✓.
(b) $0=0+0i$, so we need both $x^2-4=0$ (giving $x=\pm2$) and $x+2=0$ (giving $x=-2$). Only $x=-2$ satisfies both. ($x=2$ makes the real part zero but leaves $4i$.)

Addition, Subtraction and Multiplication

📘 Definition 4.4 — Sum and difference

For $z_1=a+ib$ and $z_2=c+id$: $$z_1+z_2=(a+c)+i(b+d),$$ $$z_1-z_2=(a-c)+i(b-d).$$ Real parts combine with real parts, imaginary parts with imaginary parts.

📐 Theorem 4.1 — Properties of addition

For all $z_1,z_2,z_3\in\mathbb{C}$:

  • Closure: $z_1+z_2\in\mathbb{C}$.
  • Commutative law: $z_1+z_2=z_2+z_1$.
  • Associative law: $(z_1+z_2)+z_3=z_1+(z_2+z_3)$.
  • Additive identity: $0=0+0i$ satisfies $z+0=z$ for every $z$.
  • Additive inverse: for $z=a+ib$, the number $-z=-a-ib$ satisfies $z+(-z)=0$. Subtraction is adding the inverse: $z_1-z_2=z_1+(-z_2)$.

Proof. Each statement is a pair of statements about real numbers, one for the real parts and one for the imaginary parts. For example, $z_1+z_2=(a+c)+i(b+d)$ and $z_2+z_1=(c+a)+i(d+b)$ are equal because $a+c=c+a$ and $b+d=d+b$ in $\mathbb{R}$. The other laws follow in the same way from the corresponding laws for real numbers. $\blacksquare$

Multiplication is where $i$ earns its keep. Multiply $a+ib$ by $c+id$ exactly as you would multiply two binomials, then use $i^2=-1$: $$\begin{aligned}(a+ib)(c+id)&=ac+iad+ibc+i^2bd\\&=(ac-bd)+i(ad+bc).\end{aligned}$$ We take this as the definition.

📘 Definition 4.5 — Product of complex numbers

For $z_1=a+ib$ and $z_2=c+id$, $$z_1z_2=(ac-bd)+i(ad+bc).$$ In particular, for a real number $k$, $k(a+ib)=ka+ikb$.

📐 Theorem 4.2 — Properties of multiplication

For all $z_1,z_2,z_3\in\mathbb{C}$:

  • Closure: $z_1z_2\in\mathbb{C}$.
  • Commutative law: $z_1z_2=z_2z_1$.
  • Associative law: $(z_1z_2)z_3=z_1(z_2z_3)$.
  • Multiplicative identity: $1=1+0i$ satisfies $z\cdot1=z$ for every $z$.
  • Distributive law: $z_1(z_2+z_3)=z_1z_2+z_1z_3$.

Proof of the commutative law. With $z_1=a+ib$, $z_2=c+id$: $z_1z_2=(ac-bd)+i(ad+bc)$ and $z_2z_1=(ca-db)+i(cb+da)$. These are equal because multiplication and addition of real numbers are commutative. The other laws are checked in the same way, by expanding both sides and comparing real and imaginary parts. $\blacksquare$

Every non-zero complex number also has a multiplicative inverse; we find it in Section 4.2.

✏️ Example 3 — Sums, differences and products
Let $z_1=2+3i$ and $z_2=1-4i$. Find (a) $z_1+z_2$ (b) $z_1-z_2$ (c) $3z_1-2z_2$ (d) $z_1z_2$ (e) $z_2^2$.
(a)–(c) $z_1+z_2=(2+1)+i(3-4)=3-i$. $z_1-z_2=(2-1)+i(3+4)=1+7i$. $3z_1-2z_2=(6+9i)-(2-8i)=4+17i$.
(d) Expand $(2+3i)(1-4i)=2-8i+3i-12i^2=2-5i+12=14-5i$.
(d) Check By the formula with $a=2$, $b=3$, $c=1$, $d=-4$: $ac-bd=2+12=14$ and $ad+bc=-8+3=-5$ ✓.
(e) $(1-4i)^2=1-8i+16i^2=1-8i-16=-15-8i$.
Interactive 1 — The Multiplication Box

Multiplying $(a+ib)(c+id)$ produces four partial products, one in each cell of the box. Watch them appear one at a time; in the last cell $i\cdot i=i^2=-1$ turns $+bd$ into $-bd$. Then the cells are collected: the two real cells give the real part, the two cells containing $i$ give the imaginary part. Change $a$, $b$, $c$, $d$ to try your own product.

product—
real part $ac-bd$—
imaginary part $ad+bc$—

Powers of i

Because $i^2=-1$, the powers of $i$ cannot grow: $$i^1=i,\quad i^2=-1,\quad i^3=i^2\cdot i=-i,$$ $$i^4=(i^2)^2=1,\quad i^5=i^4\cdot i=i,\ \dots$$ After four steps we are back at $1$, and the pattern $i,\,-1,\,-i,\,1$ repeats for ever. Negative powers follow the same cycle backwards: since $i\cdot(-i)=-i^2=1$, we have $i^{-1}=\dfrac1i=-i$.

📐 Theorem 4.3 — Powers of $i$

For every integer $k$ (positive, negative or zero), $$i^{4k}=1,\qquad i^{4k+1}=i,$$ $$i^{4k+2}=-1,\qquad i^{4k+3}=-i.$$ So to find $i^n$, write $n=4k+r$ with $r\in\{0,1,2,3\}$ (the remainder when $n$ is divided by $4$); then $i^n=i^r$.

Proof. $i^4=(i^2)^2=(-1)^2=1$, so $i^{4k}=(i^4)^k=1^k=1$ for every integer $k$ — for negative $k$ too, since $i^{-4}=\dfrac{1}{i^4}=1$. (The laws of indices hold for integer powers of complex numbers because multiplication in $\mathbb{C}$ is commutative and associative.) Then $i^{4k+r}=i^{4k}\cdot i^r=i^r$, and $i^0=1$, $i^1=i$, $i^2=-1$, $i^3=-i$. $\blacksquare$

✏️ Example 4 — Simplifying powers of $i$
Simplify (a) $i^{35}$ (b) $i^{-39}$ (c) $i^{4n+1}-i^{4n-1}$, where $n$ is an integer (d) $1+i+i^2+\cdots+i^{10}$.
(a) $35=4\cdot8+3$, so $i^{35}=i^3=-i$.
(b) $-39=4\cdot(-10)+1$ (remainder $1$, not $-3$), so $i^{-39}=i$. Check: $i^{39}=i^{36+3}=-i$, and $\dfrac{1}{-i}=\dfrac{i}{-i\cdot i}=\dfrac{i}{1}=i$ ✓.
(c) $i^{4n+1}=i$ and $4n-1=4(n-1)+3$, so $i^{4n-1}=-i$. The difference is $i-(-i)=2i$.
(d) Any four consecutive powers add to zero: $i^m(1+i+i^2+i^3)=i^m(1+i-1-i)=0$. The eleven terms $i^0,\dots,i^{10}$ contain two full blocks of four ($i^0$ to $i^7$), leaving $i^8+i^9+i^{10}=1+i-1=i$. So the sum is $i$.
Interactive 2 — The Four-Step Cycle of i

Picture $1$ to the east, $i$ to the north, $-1$ to the west and $-i$ to the south (Section 4.3 explains why this picture is natural). Each multiplication by $i$ moves the pointer one quarter turn on. On the right, the running sum $1+i+i^2+\cdots+i^n$ takes one unit step in the direction of each new power — east, north, west, south — and so walks round a square, returning to $0$ after every four terms.

7
$n=4k+r$—
$i^n$—
$1+i+\cdots+i^n$—

Algebraic Identities in ℂ

The identities you learnt for real numbers, such as $(a+b)^2=a^2+2ab+b^2$, are proved using only the commutative, associative and distributive laws. Theorems 4.1 and 4.2 say that these laws hold in $\mathbb{C}$ as well — so the identities hold for complex numbers too.

📐 Theorem 4.4 — Algebraic identities for complex numbers

For all $z_1,z_2\in\mathbb{C}$:

  • $(z_1+z_2)^2=z_1^2+2z_1z_2+z_2^2$ and $(z_1-z_2)^2=z_1^2-2z_1z_2+z_2^2$;
  • $(z_1+z_2)(z_1-z_2)=z_1^2-z_2^2$;
  • $(z_1+z_2)^3=z_1^3+3z_1^2z_2+3z_1z_2^2+z_2^3$ and $(z_1-z_2)^3=z_1^3-3z_1^2z_2+3z_1z_2^2-z_2^3$.

Proof of the first. $(z_1+z_2)^2=(z_1+z_2)(z_1+z_2)$ $=z_1(z_1+z_2)+z_2(z_1+z_2)$ by the distributive law, $=z_1^2+z_1z_2+z_2z_1+z_2^2$ $=z_1^2+2z_1z_2+z_2^2$ by the commutative law ($z_2z_1=z_1z_2$). The others are proved in the same way. $\blacksquare$

✏️ Example 5 — Squares and cubes using the identities
Express in the form $a+ib$: (a) $(2+3i)^2$ (b) $(1-2i)^3$ (c) $(3+2i)(3-2i)$ (d) $(1+i)^8$.
(a) $(2+3i)^2=4+2(2)(3i)+(3i)^2=4+12i-9=-5+12i$.
(b) With $z_1=1$, $z_2=2i$: $(1-2i)^3=1-3(2i)+3(2i)^2-(2i)^3=1-6i-12-8i^3=1-6i-12+8i=-11+2i$, using $i^3=-i$.
(b) Check $(1-2i)^2=1-4i-4=-3-4i$, and $(-3-4i)(1-2i)=-3+6i-4i+8i^2=-11+2i$ ✓.
(c) Difference of squares: $3^2-(2i)^2=9-4i^2=9+4=13$ — a real number. Pairs like $3+2i$ and $3-2i$ are studied in Section 4.2.
(d) $(1+i)^2=1+2i+i^2=2i$, so $(1+i)^8=\big((1+i)^2\big)^4=(2i)^4=16i^4=16$.
✏️ Example 6 — Bombelli's calculation
(a) Show that $(2+i)^3=2+11i$ and $(2-i)^3=2-11i$.
(b) The cubic-equation formula gives a root of $x^3=15x+4$ as $x=u+v$, where $u^3=2+11i$ and $v^3=2-11i$. Taking $u=2+i$ and $v=2-i$, find $x$ and check that it is a root.
(a) $(2+i)^3=8+3(4)(i)+3(2)(i^2)+i^3=8+12i-6-i=2+11i$. Replacing $i$ by $-i$ throughout (which changes only the signs of the odd powers of $i$): $(2-i)^3=8-12i-6+i=2-11i$.
(b) $x=(2+i)+(2-i)=4$: the imaginary parts cancel.
Check $4^3=64$ and $15(4)+4=64$ ✓. Complex numbers were needed on the way, yet the answer is real — exactly the surprise that made mathematicians take $i$ seriously.
⚠️ Common mistakes with complex numbers

1. $\sqrt{a}\,\sqrt{b}=\sqrt{ab}$ fails when $a$ and $b$ are both negative. $\sqrt{-4}\,\sqrt{-9}=(2i)(3i)=6i^2=-6$, not $\sqrt{36}=6$. Always convert $\sqrt{-a}$ to $i\sqrt a$ first, then multiply.

2. The imaginary part is real. $\operatorname{Im}(5-2i)=-2$, not $-2i$.

3. $i^{-1}=-i$, not $i$. And for $i^{-n}$ use the remainder in $\{0,1,2,3\}$: $-39=4(-10)+1$.

4. Forgetting $i^2=-1$ in a product. $(2+3i)(1-4i)$ contains $-12i^2=+12$; leaving it as $-12$ gives the wrong real part.

5. Complex numbers cannot be ordered. Statements like $3+2i\gt1+i$ or $i\gt0$ have no meaning. (If $i\gt0$, multiplying by $i$ would give $i^2\gt0$, i.e. $-1\gt0$; if $i\lt0$, multiplying by the negative number $i$ would reverse the sign and again give $-1\gt0$.) Only equality makes sense.

📘 In the CBSE syllabus — school assessment only (not in the year-end board exam)

Taught and tested in school, but not in the year-end board exam; JEE Main and JEE Advanced both list quadratic equations in the complex number system and expect you to solve any quadratic equation with real coefficients, including those with a negative discriminant.

Method. For $ax^2+bx+c=0$ with $a,b,c\in\mathbb{R}$, $a\ne0$ and discriminant $D=b^2-4ac\lt0$, the quadratic formula still works once $\sqrt D$ is read as $i\sqrt{-D}$: $$x=\dfrac{-b\pm i\sqrt{-D}}{2a}.$$ The two roots have the same real part and opposite imaginary parts.

Worked example. Solve $x^2-4x+13=0$.
Solution. $D=16-52=-36$, so $\sqrt D=\sqrt{-36}=6i$ and $x=\dfrac{4\pm6i}{2}=2\pm3i$. Check $x=2+3i$: $(2+3i)^2-4(2+3i)+13$ $=(-5+12i)-8-12i+13=0$ ✓.

Exercises

I
The form $a+ib$ and equality.
  • 1 Write $2+\sqrt{-49}$ in the form $a+ib$ and state its real and imaginary parts.
  • 2 Write $3\sqrt{-2}-\sqrt{-8}$ in the form $a+ib$.
  • 3 Write $\dfrac{-2-\sqrt{-12}}{4}$ in the form $a+ib$.
  • 4 Which of $5i$, $0$, $-3$, $\sqrt{-7}$ are purely imaginary?
  • 5 Find real $x$, $y$ with $(x+y)+i(x-y)=6+2i$.
  • 6 Find real $x$, $y$ with $(3x-7)+2iy=-5y+(5+x)i$.
  • 7 For which real $x$ is $(x^2-9)+i(x-3)$ (a) real (b) purely imaginary (c) zero?
  • 8 Find real $a$, $b$ with $(a+ib)+(3-2i)=2(1+4i)$.
1 $2+7i$: real part $2$, imaginary part $7$.
2 $3i\sqrt2-2i\sqrt2=i\sqrt2=0+\sqrt2\,i$ (using $\sqrt{-8}=i\sqrt8=2i\sqrt2$).
3 $\sqrt{-12}=2i\sqrt3$, so $\dfrac{-2-2i\sqrt3}{4}=-\tfrac12-\tfrac{\sqrt3}{2}i$.
4 $5i$ and $\sqrt{-7}=i\sqrt7$. ($0$ and $-3$ are real; a purely imaginary number needs a non-zero imaginary part.)
5 $x+y=6$, $x-y=2$: adding, $2x=8$, so $x=4$, $y=2$.
6 Real parts: $3x-7=-5y$; imaginary parts: $2y=5+x$, so $x=2y-5$. Then $3(2y-5)-7=-5y$, $11y=22$: $y=2$, $x=-1$. Check: both sides equal $-10+4i$ ✓.
7 (a) Imaginary part $0$: $x=3$. (b) Real part $0$ and imaginary part non-zero: $x^2=9$ and $x\ne3$, so $x=-3$. (c) Both parts $0$: $x=3$.
8 $(a+3)+i(b-2)=2+8i$: $a=-1$, $b=10$.
II
Arithmetic. Let $z_1=3-2i$, $z_2=-1+4i$, $z_3=2i$.
  • 1 $z_1+z_2$
  • 2 $z_1-z_2$
  • 3 $2z_1+3z_2$
  • 4 $z_1z_2$
  • 5 $z_1z_3$
  • 6 $z_2^2$
  • 7 Compute $z_1(z_2+z_3)$ and $z_1z_2+z_1z_3$, and check the distributive law.
  • 8 $(1+2i)(2-i)(1+i)$
  • 9 Find the real number $k$ for which $(2+ki)(1+3i)$ is (a) purely imaginary (b) real.
  • 10 Write down the additive inverse of $4-7i$.
Check: add, subtract or multiply any two complex numbers
+$i$+$i$
result—
real part—
imaginary part—
1 $2+2i$.
2 $4-6i$.
3 $(6-4i)+(-3+12i)=3+8i$.
4 $-3+12i+2i-8i^2=-3+14i+8=5+14i$.
5 $6i-4i^2=4+6i$.
6 $1-8i+16i^2=-15-8i$.
7 $z_2+z_3=-1+6i$ and $(3-2i)(-1+6i)=-3+18i+2i-12i^2=9+20i$. Also $z_1z_2+z_1z_3=(5+14i)+(4+6i)=9+20i$ — equal ✓.
8 $(1+2i)(2-i)=2-i+4i-2i^2=4+3i$; then $(4+3i)(1+i)=4+4i+3i+3i^2=1+7i$.
9 $(2+ki)(1+3i)=(2-3k)+i(6+k)$. (a) $2-3k=0$: $k=\tfrac23$ (and then the imaginary part $6\tfrac23\ne0$). (b) $6+k=0$: $k=-6$ (the product is then $20$).
10 $-4+7i$.
III
Powers of $i$. Simplify parts 1–7.
  • 1 $i^{18}$
  • 2 $i^{-35}$
  • 3 $i^{99}+i^{100}$
  • 4 $(-i)^6$
  • 5 $i^n+i^{n+1}+i^{n+2}+i^{n+3}$, $n\in\mathbb{Z}$
  • 6 $\big(i^{30}-i^{-15}\big)^2$
  • 7 $1+i^2+i^4+i^6+\cdots+i^{20}$
  • 8 Show that, for every integer $n$, $i^n+i^{-n}$ is one of $2$, $0$, $-2$, and say which for each remainder of $n$ on division by $4$.
1 $18=16+2$: $i^{18}=-1$.
2 $-35=4(-9)+1$: $i^{-35}=i$.
3 $i^{99}=i^3=-i$ and $i^{100}=1$: the sum is $1-i$.
4 $(-i)^6=(-1)^6i^6=i^6=i^2=-1$.
5 $i^n(1+i+i^2+i^3)=i^n(1+i-1-i)=0$.
6 $i^{30}=i^2=-1$; $-15=4(-4)+1$ so $i^{-15}=i$. Then $(-1-i)^2=1+2i+i^2=2i$.
7 $i^{2k}=(i^2)^k=(-1)^k$, so the sum is $1-1+1-\cdots+1$ with $11$ terms ($k=0$ to $10$): the answer is $1$.
8 $n=4k$: $1+1=2$. $n=4k+1$: $i^{-n}=1/i=-i$, sum $i-i=0$. $n=4k+2$: $-1+(-1)=-2$. $n=4k+3$: $-i+i=0$.
IV
Identities and applications.
  • 1 Express $(4-3i)^2$ in the form $a+ib$.
  • 2 Express $(-2+i)^3$ in the form $a+ib$.
  • 3 Simplify $(\sqrt3+i)(\sqrt3-i)$.
  • 4 Simplify $(1-i)^6$.
  • 5 Find the least positive integer $m$ for which $(1+i)^m$ is a positive real number.
  • 6 Show that $(x+1+i)(x+1-i)=x^2+2x+2$ for every complex number $x$.
  • 7 AC circuits. Electrical engineers describe a component by a complex impedance $Z$ (in ohms, $\Omega$); impedances in series add, and the voltage across a component is $V=IZ$. (a) Find the total impedance of $Z_1=3+4i\ \Omega$ and $Z_2=5-2i\ \Omega$ in series. (b) Find $V$ when the current is $I=2-i$ amperes and $Z=3+4i\ \Omega$.
  • 8 Show that $(a+ib)(c+id)$ is real exactly when $ad+bc=0$, and use this to find the real $k$ for which $(2+3i)(4+ki)$ is real. What is the product then?
1 $16-24i+9i^2=7-24i$.
2 $(-2)^3+3(-2)^2(i)+3(-2)(i^2)+i^3=-8+12i+6-i=-2+11i$.
3 $(\sqrt3)^2-i^2=3+1=4$.
4 $(1-i)^2=1-2i+i^2=-2i$, so $(1-i)^6=(-2i)^3=-8i^3=8i$.
5 $(1+i)^2=2i$, $(1+i)^4=(2i)^2=-4$, $(1+i)^6=-4\cdot2i=-8i$, $(1+i)^8=16$. The odd powers have both parts non-zero ($(1+i)^1=1+i$, $(1+i)^3=-2+2i$, $(1+i)^5=-4-4i$, $(1+i)^7=8-8i$). So $m=8$.
6 $(x+1)^2-i^2=x^2+2x+1+1=x^2+2x+2$, by the difference-of-squares identity with $z_1=x+1$, $z_2=i$.
7 (a) $Z_1+Z_2=8+2i\ \Omega$. (b) $V=(2-i)(3+4i)=6+8i-3i-4i^2=10+5i$ volts.
8 The product is $(ac-bd)+i(ad+bc)$, which is real exactly when $ad+bc=0$. Here $a=2$, $b=3$, $c=4$, $d=k$: $2k+12=0$, $k=-6$, and the product is $ac-bd=8+18=26$.
🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksJEE MainJEE Advanced

How this is examined: Board papers ask for powers of $i$ (1 mark), a product or power in the form $a+ib$ (2 marks), and real unknowns found by equating real and imaginary parts (3 marks). JEE uses the four-step cycle of $i$ in sums, and tests whether you know which real-number rules survive in $\mathbb{C}$.

E1. Board · 1 mark The value of $i^{57}+\dfrac{1}{i^{25}}$ is (A) $0$   (B) $2i$   (C) $-2i$   (D) $2$

Idea $57=56+1$ and $25=24+1$, so $i^{57}=i$ and $i^{25}=i$; then $\dfrac1i=-i$.
Trap (B) comes from taking $\dfrac1i=i$.
Answer (A) $i+(-i)=0$.

E2. Board · 2 marks Express $(5-3i)^3$ in the form $a+ib$.

Expand $(5-3i)^3=125-3(25)(3i)+3(5)(3i)^2-(3i)^3=125-225i-135+27i$, using $(3i)^2=-9$ and $(3i)^3=27i^3=-27i$. (1 mark)
Answer $-10-198i$. (1 mark)

E3. Board · 3 marks Find real numbers $x$ and $y$ such that $(x+iy)(2-3i)=4+i$.

Expand $(x+iy)(2-3i)=2x-3ix+2iy-3i^2y=(2x+3y)+i(2y-3x)$. (1 mark)
Equate $2x+3y=4$ and $-3x+2y=1$. (1 mark)
Answer Multiply the first by $2$ and the second by $3$ and subtract: $13x=5$, $x=\tfrac{5}{13}$; then $y=\tfrac{4-2x}{3}=\tfrac{14}{13}$. Check: $2x+3y=\tfrac{10+42}{13}=4$ ✓. (1 mark)

E4. JEE Main $\displaystyle\sum_{k=1}^{13}\big(i^k+i^{k+1}\big)$ equals (A) $i$   (B) $i-1$   (C) $-i$   (D) $0$

Idea Four consecutive powers of $i$ add to $0$. $\sum_{k=1}^{13}i^k$: the first $12$ terms cancel, leaving $i^{13}=i$. $\sum_{k=1}^{13}i^{k+1}=i^2+\cdots+i^{14}$: the first $12$ cancel, leaving $i^{14}=-1$.
Trap (A) counts only the first sum; (D) assumes every block cancels completely.
Answer (B) $i+(-1)=i-1$.
⏱ Fast route $i^k+i^{k+1}=i^k(1+i)$, so the sum is $(1+i)\sum_{k=1}^{13}i^k=(1+i)i=i-1$.

E5. JEE Main Numerical value. Find the least positive integer $n$ for which $(1+i)^{2n}=(1-i)^{2n}$.

Squares $(1+i)^2=2i$ and $(1-i)^2=-2i$, so the equation is $(2i)^n=(-2i)^n$, i.e. $(2i)^n=(-1)^n(2i)^n$.
Solve $(2i)^n\ne0$, so we need $(-1)^n=1$: $n$ even. ($n=1$ gives $2i\ne-2i$.)
Answer $n=2$: both sides equal $(2i)^2=-4$.

E6. JEE Advanced One or more options correct. Which statements are true? (A) $(z_1+z_2)^2=z_1^2+2z_1z_2+z_2^2$ for all $z_1,z_2\in\mathbb{C}$   (B) $i^{4k+3}=-i$ for every integer $k$, including negative $k$   (C) $\sqrt{-a}\,\sqrt{-b}=\sqrt{ab}$ for all real $a,b\gt0$   (D) if $z\in\mathbb{C}$ and $z^2=0$, then $z=0$

(A), (B) (A) is Theorem 4.4. (B) is Theorem 4.3: $i^{4k}=1$ for every integer $k$, so $i^{4k+3}=i^3=-i$.
(C) False: $\sqrt{-a}\sqrt{-b}=(i\sqrt a)(i\sqrt b)=i^2\sqrt{ab}=-\sqrt{ab}$, e.g. $\sqrt{-4}\sqrt{-9}=-6$.
(D) True: with $z=a+ib$, $z^2=(a^2-b^2)+2abi=0$ gives $ab=0$ and $a^2=b^2$; if $a=0$ then $b^2=0$, and if $b=0$ then $a^2=0$. Either way $a=b=0$.
Answer (A), (B), (D).
🔬 Research–Teaching NexusMini project · about 45 minutes

Research question. How do engineers use complex numbers to tune a radio circuit — and at what frequency does a resistor–inductor–capacitor circuit behave like a plain resistor?

Why it matters. Every radio, Wi-Fi router and wireless phone charger contains a tuned circuit. Engineers give each component a complex impedance: a resistor $R$ ohms has $Z=R$, an inductor of $L$ henries has $Z=i\omega L$, and a capacitor of $C$ farads has $Z=\dfrac{1}{i\omega C}=-\dfrac{i}{\omega C}$ (using $\tfrac1i=-i$), where $\omega$ is the angular frequency in radians per second. In series the impedances simply add — complex addition doing real engineering.

What to do.

  1. Take $R=10\ \Omega$, $L=0.1$ H and $C=100\ \mu$F $=10^{-4}$ F in series. Show that the total impedance is $Z=10+i\Big(0.1\,\omega-\dfrac{10^4}{\omega}\Big)$.
  2. In a spreadsheet, list $\omega=100,150,200,\dots,600$ and compute $\operatorname{Re}Z$ and $\operatorname{Im}Z$ for each.
  3. Find, from the table and then exactly, the value of $\omega$ for which $\operatorname{Im}Z=0$ — the resonant frequency, where the circuit behaves like a $10\ \Omega$ resistor.
  4. Repeat with $C$ halved and describe how the resonant frequency moves.

Tools: A spreadsheet app on a phone or computer

Hand in: Your table, a plot of $\operatorname{Im}Z$ against $\omega$, the resonant frequency in both cases, and two sentences explaining what "tuning a radio" changes in this model.

Going further. Show that in general $\operatorname{Im}Z=0$ exactly when $\omega=\dfrac{1}{\sqrt{LC}}$.

Instructor note: $0.1\,\omega=10^4/\omega$ gives $\omega^2=10^5$, $\omega\approx316.2$ rad/s; halving $C$ multiplies the resonant frequency by $\sqrt2$ (to about $447.2$ rad/s). The general result comes from $\omega L=\tfrac{1}{\omega C}$.

🧪 Self-check quiz

0/5 correct

Q1. $i^{-1}$ equals

Q2. The imaginary part of $4-7i$ is

Q3. $(1+i)^2$ equals

Q4. $\sqrt{-9}\times\sqrt{-16}$ equals

Q5. If $a$ and $b$ are real and $(a+2)+ib=5-3i$, then $(a,b)$ is

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