Complex Numbers
Where the number line becomes a plane — and rotation, algebra and geometry turn out to be the same thing
- Why \(i=\sqrt{-1}\) had to be invented, and how its powers cycle with period four.
- To add, multiply, divide and conjugate complex numbers fluently, and to compute the modulus.
- To picture a complex number as a point — or an arrow — in the Argand plane.
- Polar and Euler form, and how multiplication becomes "scale and rotate."
- De Moivre's theorem, the nth roots of unity, and the special powers of \(\omega\).
- To use complex numbers as a geometry engine — rotations, distances, loci and inequalities.
Why Complex Numbers?
The real line, for all its richness, has a blind spot: the equation \(x^{2}+1=0\) has no solution, because no real number squares to a negative. For centuries this looked like a dead end. The breakthrough was to stop asking "does \(\sqrt{-1}\) exist?" and instead ask "what if we let it?"
Give that quantity a name — i — and insist that \(i^2=-1\). Astonishingly, nothing breaks. Every polynomial of degree \(n\) now has exactly \(n\) roots (the Fundamental Theorem of Algebra), and a whole new geometry of rotation falls into our lap. Complex numbers are not "imaginary" in any loose sense — they are as concrete as points on a page.
The Imaginary Unit & Powers of i
The imaginary unit is defined by a single rule:
The powers of \(i\) march in a four-step cycle and then repeat forever:
The rule \(\sqrt{a}\,\sqrt{b}=\sqrt{ab}\) only holds when at least one of \(a,b\) is non-negative. Writing \(\sqrt{-4}\,\sqrt{-9}=\sqrt{36}=6\) is wrong. Convert first: \(\sqrt{-4}\,\sqrt{-9}=(2i)(3i)=6i^2=-6\).
Complex Numbers Defined
A complex number is any expression of the form
If \(b=0\) the number is purely real; if \(a=0\) (and \(b\neq0\)) it is purely imaginary. So \(\mathbb{R}\subset\mathbb{C}\) — the reals are simply the complex numbers lying on the horizontal axis.
One complex equation is really two real equations — match real parts to real, imaginary to imaginary. This "comparing parts" trick solves a huge fraction of JEE problems.
Algebra of Complex Numbers
Treat \(i\) as an ordinary symbol, simplify, and replace \(i^2\) by \(-1\) at the end.
| Operation | Result |
|---|---|
| Addition | \((a+c)+i(b+d)\) |
| Subtraction | \((a-c)+i(b-d)\) |
| Multiplication | \((ac-bd)+i(ad+bc)\) |
| Division \((z_2\neq0)\) | \(\dfrac{(ac+bd)+i(bc-ad)}{c^{2}+d^{2}}\) |
The Conjugate
The conjugate of \(z=a+ib\) is \(\bar{z}=a-ib\) — its mirror image across the real axis. The conjugate is the single most useful tool in the chapter.
| Identity | Meaning |
|---|---|
| \(\overline{\bar{z}}=z\) | conjugating twice does nothing |
| \(z+\bar{z}=2\operatorname{Re}(z)\) | sum picks out the real part |
| \(z-\bar{z}=2i\operatorname{Im}(z)\) | difference picks out the imaginary part |
| \(z\bar{z}=|z|^{2}\) | product is real and non-negative |
| \(\overline{z_1\pm z_2}=\bar{z_1}\pm\bar{z_2}\) | distributes over \(\pm\) |
| \(\overline{z_1 z_2}=\bar{z_1}\,\bar{z_2}\) | distributes over product |
| \(z=\bar{z}\iff z\in\mathbb{R}\) | a number equals its conjugate only if real |
Modulus & Its Properties
The modulus \(|z|\) is the distance of \(z\) from the origin in the plane:
| Identity | In words |
|---|---|
| \(|z_1 z_2|=|z_1||z_2|\) | modulus of a product is the product of moduli |
| \(\left|\dfrac{z_1}{z_2}\right|=\dfrac{|z_1|}{|z_2|}\) | and likewise for quotients |
| \(|z^{n}|=|z|^{n}\) | powers scale the modulus |
| \(|\bar{z}|=|z|=|-z|\) | reflection and negation preserve length |
The length of a sum can never exceed the sum of lengths — equality holds only when \(z_1\) and \(z_2\) point the same way. It is the algebraic echo of "any side of a triangle is shorter than the other two combined."
The Argand Plane
Plot \(\operatorname{Re}(z)\) horizontally and \(\operatorname{Im}(z)\) vertically, and every complex number becomes a point — or, just as usefully, an arrow from the origin. This picture is the Argand plane. Addition of complex numbers is then exactly the parallelogram law for adding vectors.
Polar & Euler Form
Instead of "how far right, how far up," describe \(z\) by "how far out, in which direction" — its modulus \(r\) and argument \(\theta\). From the diagram, \(a=r\cos\theta\) and \(b=r\sin\theta\), so:
\(\tan\theta=b/a\) alone cannot tell which quadrant you are in. The principal argument lies in \((-\pi,\pi]\); always check the signs of \(a\) and \(b\) before fixing \(\theta\). For example, \(z=-1-i\) sits in the third quadrant, so \(\arg z=-\tfrac{3\pi}{4}\), not \(\tfrac{\pi}{4}\).
De Moivre's Theorem
If multiplying adds angles, then raising to a power multiplies the angle. That is De Moivre's theorem — the workhorse for powers and roots:
Equivalently \((e^{i\theta})^{n}=e^{in\theta}\). Combined with polar form, \(z^{n}=r^{n}\big(\cos n\theta+i\sin n\theta\big)\) — a horrible binomial expansion collapses to one line.
The nth Roots of Unity
The equation \(z^{n}=1\) has exactly \(n\) solutions, the nth roots of unity. By De Moivre:
The Cube Roots of Unity (ω)
The case \(n=3\) is so common it gets its own letter. The cube roots of unity are \(1,\ \omega,\ \omega^{2}\), where
So any power of \(\omega\) reduces using \(\omega^{3}=1\), and any sum of all three roots vanishes. These two facts factor sums of cubes too: \(a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a+\omega b+\omega^{2}c)(a+\omega^{2}b+\omega c)\).
Square Root of a Complex Number
To find \(\sqrt{a+ib}\), set \(\sqrt{a+ib}=x+iy\), square, and compare real and imaginary parts. That gives a tidy closed form:
Complex Numbers as a Geometry Engine
Because every complex number is a point, statements about \(z\) become statements about geometry. This is the part examiners love most.
| Complex statement | Geometric meaning |
|---|---|
| \(|z_1-z_2|\) | distance between the points \(z_1\) and \(z_2\) |
| \(|z-z_0|=r\) | circle, centre \(z_0\), radius \(r\) |
| \(|z-z_1|=|z-z_2|\) | perpendicular bisector of \(z_1 z_2\) |
| \(\dfrac{m z_2+n z_1}{m+n}\) | point dividing \(z_1 z_2\) in ratio \(m:n\) |
| \(\arg\!\left(\dfrac{z-z_1}{z-z_2}\right)=\alpha\) | arc on which \(z_1 z_2\) subtends angle \(\alpha\) |
Multiplying by \(e^{i\theta}\) turns an arrow through angle \(\theta\) without changing its length. This one line replaces pages of coordinate geometry for problems about squares, equilateral triangles and regular polygons.
Putting It to Work
Problem. Express \(\dfrac{3+2i}{2-i}\) in the form \(a+ib\).
Solution. Multiply top and bottom by the conjugate \(2+i\):
Problem. Write \(z=-1+i\sqrt{3}\) in polar form.
Solution. \(|z|=\sqrt{(-1)^2+(\sqrt3)^2}=\sqrt{4}=2\). The point is in the second quadrant (\(a<0,\ b>0\)), and the reference angle has \(\tan=\sqrt3\), so \(\arg z=\pi-\tfrac{\pi}{3}=\tfrac{2\pi}{3}\). Hence
Problem. Evaluate \((1+i)^{8}\).
Solution. Convert to polar: \(|1+i|=\sqrt2\), \(\arg=\tfrac{\pi}{4}\), so \(1+i=\sqrt2\,e^{i\pi/4}\). Then
Problem. Simplify \((1+\omega)(1+\omega^{2})(1+\omega^{4})(1+\omega^{8})\).
Solution. Since \(\omega^{3}=1\): \(\omega^{4}=\omega\) and \(\omega^{8}=\omega^{2}\). Also \(1+\omega=-\omega^{2}\) and \(1+\omega^{2}=-\omega\) (from \(1+\omega+\omega^2=0\)). The product becomes
Problem. Find \(\sqrt{3+4i}\).
Solution. Let \(\sqrt{3+4i}=x+iy\). Squaring: \(x^{2}-y^{2}=3\) and \(2xy=4\). Also \(x^{2}+y^{2}=|3+4i|=5\). Adding the first and last: \(2x^{2}=8\Rightarrow x=\pm2\); then \(y=\pm1\) with \(xy>0\). So
Problem. Describe the set of points \(z\) with \(|z-1|=|z+1|\).
Solution. The left side is the distance from \(z\) to \(1\); the right side, the distance to \(-1\). Points equidistant from \(1\) and \(-1\) form the perpendicular bisector of that segment — the imaginary axis, \(\operatorname{Re}(z)=0\).
Chapter Summary
\(i^2=-1\); powers cycle every four. A complex number is \(z=a+ib\) with real and imaginary parts; equate parts to solve.
\(\bar z=a-ib\), \(z\bar z=|z|^2\). Divide by multiplying with the conjugate. Triangle inequality bounds \(|z_1+z_2|\).
Plot in the plane; \(z=r\,e^{i\theta}\). Multiplication adds angles and multiplies moduli; \(\times i\) = 90° turn.
\((e^{i\theta})^n=e^{in\theta}\). The \(n\) roots of unity sit on a regular \(n\)-gon and sum to zero.
\(\omega^3=1\), \(1+\omega+\omega^2=0\) — the two identities that crush cube-root problems.
Distances, circles, bisectors and rotations all read straight off complex equations; rotate with \(z'-z_0=(z-z_0)e^{i\theta}\).
Problems
Work these by converting to the most convenient form — rectangular for algebra, polar for powers and roots. Difficulty rises down the list.
- Simplify \(i^{37}+\dfrac{1}{i^{37}}\) and \(i^{-1}+i^{-2}+i^{-3}+i^{-4}\).
- Express \(\dfrac{(2+i)}{(1-i)(1+2i)}\) in the form \(a+ib\).
- For \(z=4-3i\), find \(\bar z\), \(|z|\) and the multiplicative inverse \(z^{-1}\).
- Find the modulus and principal argument of \(z=-\sqrt{3}-i\) and write it in polar form.
- Use De Moivre's theorem to evaluate \(\left(\dfrac{1+i}{1-i}\right)^{6}\).
- If \(\omega\) is a non-real cube root of unity, find the value of \((1-\omega+\omega^{2})(1+\omega-\omega^{2})\).
- Find both square roots of \(-8-6i\).
- Solve \(z^{4}=1\) and plot the four roots on the Argand plane.
- Prove that if \(|z_1|=|z_2|=1\) and \(z_1 z_2\neq -1\), then \(\dfrac{z_1+z_2}{1+z_1 z_2}\) is real.
- Identify the curve given by \(|z-3|+|z+3|=10\).
- The points \(z_1\) and \(z_2\) are two vertices of an equilateral triangle. Using the rotation formula, find the third vertex.
- If \(z+\dfrac{1}{z}=2\cos\theta\), show that \(z^{n}+\dfrac{1}{z^{n}}=2\cos n\theta\).