Complex Numbers – Study Notes (Class 12 Maths, Chapter 2)
1. The imaginary unit and powers of \(i\)
No real number squares to a negative value, so to solve equations such as \(x^{2}+1=0\) we introduce the imaginary unit \(i\), defined by \(i^{2}=-1\). Every higher power of \(i\) reduces to one of four values, because the powers repeat with period four:
\[ i^{1}=i,\qquad i^{2}=-1,\qquad i^{3}=-i,\qquad i^{4}=1. \]
To simplify \(i^{n}\), divide \(n\) by \(4\) and read the remainder: remainders \(0,1,2,3\) give \(1,i,-1,-i\) respectively. A handy consequence is that any four consecutive powers of \(i\) add to zero, i.e. \(i^{n}+i^{n+1}+i^{n+2}+i^{n+3}=0\).
2. Rectangular form and equality
A complex number is written \(z=x+iy\), where \(x=\operatorname{Re}(z)\) is the real part and \(y=\operatorname{Im}(z)\) is the imaginary part. If \(x=0\) the number is purely imaginary; if \(y=0\) it is real. Two complex numbers are equal precisely when their real parts agree and their imaginary parts agree, so a single complex equation secretly carries two real equations. That observation is the standard engine behind every “find \(x\) and \(y\)” problem.
3. The Argand plane
Each \(z=x+iy\) corresponds to the point \((x,y)\), which lets us picture complex numbers geometrically. The horizontal axis is the real axis and the vertical axis the imaginary axis; this picture is the Argand plane. A complex number can be read either as a point or as the position vector from the origin to that point, which is exactly why addition behaves like adding vectors.
4. Algebraic operations
Addition and subtraction act componentwise, \((a+ib)\pm(c+id)=(a\pm c)+i(b\pm d)\), mirroring the parallelogram law for vectors. Multiplication uses \(i^{2}=-1\):
\[ (a+ib)(c+id)=(ac-bd)+i(ad+bc). \]
One special case deserves attention: multiplying by \(i\) rotates a number through \(90^{\circ}\) anticlockwise about the origin while leaving its length unchanged. Repeated multiplication by \(i\) therefore walks a point around a circle in right-angle steps.
5. Conjugate of a complex number
The conjugate of \(z=x+iy\) is \(\bar z=x-iy\); geometrically it is the reflection of \(z\) in the real axis. Conjugation respects every arithmetic operation, and the product \(z\bar z=x^{2}+y^{2}\) is always a non-negative real number. This is precisely what makes division work: to compute a quotient you multiply numerator and denominator by the conjugate of the denominator, turning the denominator real. Key identities:
\[ \overline{z_{1}+z_{2}}=\bar z_{1}+\bar z_{2},\qquad \overline{z_{1}z_{2}}=\bar z_{1}\,\bar z_{2},\qquad \overline{\left(\tfrac{z_{1}}{z_{2}}\right)}=\dfrac{\bar z_{1}}{\bar z_{2}}. \]
Also, \(z\) is real \(\iff z=\bar z\), and \(z\) is purely imaginary \(\iff z=-\bar z\). These two tests convert “is real / is purely imaginary” questions into a couple of lines of algebra.
6. Modulus of a complex number
The modulus \(|z|=\sqrt{x^{2}+y^{2}}\) is the distance of \(z\) from the origin, and \(|z|^{2}=z\bar z\). The properties that earn the most marks are multiplicativity, \(|z_{1}z_{2}|=|z_{1}||z_{2}|\) and \(\left|\dfrac{z_{1}}{z_{2}}\right|=\dfrac{|z_{1}|}{|z_{2}|}\), together with the triangle inequality \(|z_{1}+z_{2}|\le|z_{1}|+|z_{2}|\) and its reverse form \(|z_{1}-z_{2}|\ge\big||z_{1}|-|z_{2}|\big|\). Taking the modulus of both sides of an equation is very often the quickest route to \(|z|\).
7. Square roots of a complex number
To find \(\sqrt{a+ib}\), set \(\sqrt{a+ib}=x+iy\) and square, then match real and imaginary parts. This leads to
\[ \sqrt{a+ib}=\pm\left(\sqrt{\dfrac{|z|+a}{2}}+i\,\dfrac{b}{|b|}\sqrt{\dfrac{|z|-a}{2}}\right),\qquad b\neq 0, \]
where the factor \(\dfrac{b}{|b|}\) records the sign of \(b\): \(x\) and \(y\) share a sign when \(b>0\) and take opposite signs when \(b<0\).
8. Geometry and locus
Conditions on \(z\) translate into curves. \(|z-z_{0}|=r\) is the circle of radius \(r\) centred at \(z_{0}\); \(|z-z_{0}|
9. Polar form, argument and principal argument
Any non-zero \(z\) can be written \(z=r(\cos\theta+i\sin\theta)\), where \(r=|z|\) and \(\theta\) is the argument — the angle the radius vector makes with the positive real axis. Adding \(2\pi\) returns to the same direction, so the argument is many-valued; the single value in \((-\pi,\pi]\) is the principal argument \(\operatorname{Arg}(z)\). To find it, first compute the reference angle \(\tan^{-1}\left|\dfrac{y}{x}\right|\), then fix its sign and size from the quadrant of \(z\). This quadrant check is where most marks slip away.
10. Euler form
Euler's formula \(e^{i\theta}=\cos\theta+i\sin\theta\) compresses the polar form to \(z=re^{i\theta}\). It makes multiplication and powers especially clean, since \(e^{i\theta_{1}}e^{i\theta_{2}}=e^{i(\theta_{1}+\theta_{2})}\).
11. De Moivre's theorem
For any integer \(n\),
\[ (\cos\theta+i\sin\theta)^{n}=\cos n\theta+i\sin n\theta. \]
This is the workhorse for raising complex numbers to powers and for deriving trigonometric identities. Always put the number in polar form first; raising the modulus to the power and multiplying the angle by \(n\) is far faster than multiplying everything out.
12. \(n\)th roots and \(n\)th roots of unity
The \(n\) distinct \(n\)th roots of \(z=r(\cos\theta+i\sin\theta)\) are
\[ z^{1/n}=r^{1/n}\left(\cos\dfrac{\theta+2k\pi}{n}+i\sin\dfrac{\theta+2k\pi}{n}\right),\qquad k=0,1,\dots,n-1. \]
They all share the modulus \(r^{1/n}\), so they lie on one circle, equally spaced, forming a regular \(n\)-gon. The case \(z=1\) gives the \(n\)th roots of unity: they sum to zero, their product is \((-1)^{\,n+1}\), and each is a power of a single primitive root. Remembering \(1+\omega+\omega^{2}=0\) for cube roots (with \(\omega^{3}=1\)) shortcuts a whole family of problems.
Common exam traps
- Quadrant of the argument. \(\tan^{-1}\dfrac{y}{x}\) on its own is not the argument — always settle the quadrant before quoting \(\operatorname{Arg}(z)\).
- The rule \(\sqrt{a}\,\sqrt{b}=\sqrt{ab}\) breaks when both \(a\) and \(b\) are negative; treating \(\sqrt{-4}\,\sqrt{-9}\) as \(\sqrt{36}\) flips the sign.
- Forgetting \(\bar z=\dfrac{1}{z}\) when \(|z|=1\). This single substitution dissolves most unit-modulus problems.
- Squaring a modulus equation can introduce extra solutions — always test your answer in the original equation.
- Counting roots. An \(n\)th-root problem has exactly \(n\) solutions; stopping early is a classic slip.