By Vivek Sharma·MATHEMATICS·Ch. 2 · Complex NumbersFriday, October 23, 2026· 20 min read

Complex Numbers: Start with the Basics

Understand i, calculate with a + bi, and see numbers as points on a plane.

You can begin with ordinary algebra, fractions, and Pythagoras’ theorem. We will build the new ideas one at a time. Complex numbers extend the real numbers so that equations such as x² + 1 = 0 have solutions. They also give engineers a convenient language for rotations, waves, and alternating current.

1. Why introduce a new number?

The square of any real number is nonnegative, so no real number can satisfy $x^2=-1$ . We introduce the imaginary unit $i$ , defined by $i^2=-1$ . The equation now has two solutions: $i$ and $-i$ .

A complex number has the form $z=a+bi$ , where $a$ and $b$ are real numbers. The real part is $a$ and the imaginary part is $b$ . Despite its name, ‘imaginary’ is simply a mathematical label; these numbers are useful in real physical models.

Every real number is also complex: $5=5+0i$ . A purely imaginary number has zero real part, such as $-3i=0-3i$ . Two complex numbers are equal exactly when their real parts match and their imaginary parts match.

z=a+bi,\qquad \operatorname{Re}(z)=a,\qquad \operatorname{Im}(z)=b

a+bi=c+di\quad\Longleftrightarrow\quad a=c\text{ and }b=d

Worked example · Read the parts

For $z=4-7i$ , the real part is $4$ and the imaginary part is $-7$ .
If $x+2+(y-1)i=5+3i$ with real $x,y$ , compare parts: $x+2=5$ and $y-1=3$ .
Solve the ordinary equations: $x=3$ and $y=4$ .

Remember: A complex number has two real components, linked by the rule $i^2=-1$ .

2. Powers of i repeat in a cycle

Multiply by $i$ repeatedly. After four multiplications we return to $1$ , so the pattern repeats every four powers. Also $i^0=1$ .

For a nonnegative integer exponent, divide it by $4$ and use the remainder: $0$ gives $1$ , $1$ gives $i$ , $2$ gives $-1$ , and $3$ gives $-i$ .

i^1=i,\qquad i^2=-1,\qquad i^3=-i,\qquad i^4=1

Worked example · A large exponent

To find $i^{27}$ , write $27=4\times6+3$ .
Then $i^{27}=(i^4)^6i^3=1^6(-i)=-i$ .
For a negative real number, the principal square root is $\sqrt{-9}=3i$ . Solving $x^2=-9$ gives both $x=3i$ and $x=-3i$ .

Remember: Reduce powers of $i$ using the four-step cycle. Distinguish a principal square root from both solutions of an equation.

3. Add, subtract, and multiply

For addition and subtraction, collect real parts together and imaginary parts together. When subtracting a bracket, change the sign of every term inside it.

For multiplication, expand the brackets just as in algebra, then replace $i^2$ by $-1$ and collect terms.

(a+bi)+(c+di)=(a+c)+(b+d)i

(a+bi)(c+di)=(ac-bd)+(ad+bc)i

Worked example · Three operations

Add: $(3+2i)+(1-4i)=4-2i$ .
Subtract: $(3+2i)-(1-4i)=3+2i-1+4i=2+6i$ .
Multiply: $(3+2i)(1-4i)=3-12i+2i-8i^2=11-10i$ .

Remember: Use ordinary algebra and finish by replacing every $i^2$ with $-1$ .

4. Conjugates make division manageable

The conjugate of $z=a+bi$ is $\overline z=a-bi$ . It changes the sign of the imaginary part. Multiplying a number by its conjugate gives the real number $a^2+b^2$ because the imaginary terms cancel.

To divide by a nonzero complex number, multiply the numerator and denominator by the denominator’s conjugate. This is multiplication by $1$ , so it keeps the value unchanged.

z\overline z=(a+bi)(a-bi)=a^2+b^2

\frac{a+bi}{c+di}=\frac{(ac+bd)+(bc-ad)i}{c^2+d^2}\qquad(c+di\ne0)

Worked example · Divide step by step

For $(3+2i)/(1-i)$ , the denominator’s conjugate is $1+i$ .
Multiply top and bottom: $(3+2i)(1+i)/[(1-i)(1+i)]$ .
The numerator is $3+5i+2i^2=1+5i$ . The denominator is $1-i^2=2$ .
The answer in standard form is $1/2+(5/2)i$ . As a check, multiplying this answer by $1-i$ returns $3+2i$ .

Remember: Conjugation turns a nonzero complex denominator into a positive real denominator.

5. Draw a complex number

The complex plane, also called the Argand plane, has a horizontal real axis and a vertical imaginary axis. Plot $a+bi$ at the point $(a,b)$ . For $3+4i$ , move three units right and four units up.

The modulus $|z|$ is the distance from the origin. Pythagoras’ theorem gives $|a+bi|=\sqrt{a^2+b^2}$ . The distance between two complex numbers $z_1$ and $z_2$ is $|z_1-z_2|$ .

Conjugation reflects a point across the real axis: $3+4i$ and $3-4i$ have the same modulus. Adding complex numbers adds their horizontal and vertical components, just like adding displacement vectors.

|a+bi|=\sqrt{a^2+b^2},\qquad z\overline z=|z|^2

\frac1z=\frac{\overline z}{|z|^2}\qquad(z\ne0)

The point 3 + 4i has modulus 5. Conjugation reflects it across the real axis.

Worked example · Distance and a locus

For $z=3+4i$ , the modulus is $\sqrt{9+16}=5$ .
The distance between $3+4i$ and $1+i$ is $|(3+4i)-(1+i)|=|2+3i|=\sqrt{13}$ .
A locus is a set of points satisfying a condition. The equation $|z-(1+i)|=2$ describes a circle with centre $(1,1)$ and radius $2$ .

Remember: Modulus measures distance; the complex plane makes the algebra visible.

6. Describe a point by length and angle

For $z\ne0$ , an argument $\theta$ is the angle from the positive real axis to the line from the origin to $z$ , measured anticlockwise. We use radians: $\pi$ radians equals $180^\circ$ , so one full turn is $2\pi$ .

If $r=|z|$ , right-triangle trigonometry gives $a=r\cos\theta$ and $b=r\sin\theta$ . This is the polar form. Angles differing by a whole turn describe the same point.

We choose the principal argument $\operatorname{Arg}(z)$ in $(-\pi,\pi]$ . All arguments are $\operatorname{Arg}(z)+2k\pi$ , where $k$ is any integer. Zero has modulus zero but no defined argument.

To find the angle, first locate the quadrant using the signs of $a$ and $b$ . The ratio $b/a$ alone cannot distinguish opposite quadrants. On a calculator or in code, atan2(b,a) accounts for the quadrant; use radian mode.

z=r(\cos\theta+i\sin\theta),\qquad r=\sqrt{a^2+b^2}

\tan\theta=\frac ba\qquad(a\ne0)

Worked example · Get the quadrant right

For $z=-1+i$ , the modulus is $\sqrt2$ . The point is in quadrant II.
The reference angle is $\pi/4$ , so the principal argument is $3\pi/4$ , not $-\pi/4$ .
Thus $-1+i=\sqrt2[\cos(3\pi/4)+i\sin(3\pi/4)]$ . For $z=-2i$ , the principal argument is $-\pi/2$ .

Remember: Rectangular form gives horizontal and vertical components; polar form gives length and direction.

7. Solve an equation with complex roots

The quadratic formula still works when its discriminant is negative. For real coefficients, nonreal roots occur as conjugate pairs. You can check a proposed root by substitution.

Ax^2+Bx+C=0\quad\Longrightarrow\quad x=\frac{-B\pm\sqrt{B^2-4AC}}{2A}\qquad(A\ne0)

Worked example · Solve x² − 4x + 13 = 0

Here $A=1$ , $B=-4$ , and $C=13$ . The discriminant is $16-52=-36$ .
Use $\sqrt{-36}=6i$ : $x=(4\pm6i)/2=2\pm3i$ .
Check $x=2+3i$ : $(2+3i)^2-4(2+3i)+13=(-5+12i)-(8+12i)+13=0$ . The other root is its conjugate $2-3i$ .

Remember: A negative discriminant gives two nonreal conjugate roots when the coefficients are real.

8. Your quick revision sheet

Start in rectangular form for addition and subtraction. Use a conjugate for division. Draw the point before finding its angle. In the next lessons, polar form will make multiplication, powers, roots, and oscillations much easier.

i^2=-1,\quad\overline{a+bi}=a-bi,\quad|a+bi|=\sqrt{a^2+b^2}

z=r(\cos\theta+i\sin\theta),\quad a=r\cos\theta,\quad b=r\sin\theta

Remember: Try the objective questions without looking back, then read the explanation for each answer.