📘 Chapter 3: Complex Number | NEB Mathematics Notes Class 12
Polar Form, Euler’s Form, De Moivre’s Theorem & Roots of Complex Numbers
🔢 Complex Number extends the real number system. This chapter covers polar form, Euler’s formula, De Moivre’s theorem, finding roots of complex numbers, and solving equations. Below are detailed, line-by-line solutions to typical NEB examination problems.
📖 1. Polar & Euler’s Form
Polar Form: For \(z = x + iy\), \(z = r(\cos\theta + i\sin\theta)\) where \(r = \sqrt{x^2 + y^2} = |z|\) and \(\theta = \tan^{-1}(y/x)\) (argument).
Euler’s Form: \(e^{i\theta} = \cos\theta + i\sin\theta\) ⇒ \(z = re^{i\theta}\)
De Moivre’s Theorem: \([r(\cos\theta + i\sin\theta)]^n = r^n(\cos n\theta + i\sin n\theta)\)
Euler’s Form: \(e^{i\theta} = \cos\theta + i\sin\theta\) ⇒ \(z = re^{i\theta}\)
De Moivre’s Theorem: \([r(\cos\theta + i\sin\theta)]^n = r^n(\cos n\theta + i\sin n\theta)\)
🎯 2. Express in Euler’s Form
1a \(2 + 2i\)
Step 1: Let \(2 + 2i = r\cos\theta + i r\sin\theta\)
\(r\cos\theta = 2\), \(r\sin\theta = 2\)
Step 2: \(r^2 = 2^2 + 2^2 = 8\) ⇒ \(r = 2\sqrt{2}\)
Step 3: \(\tan\theta = \frac{2}{2} = 1\) ⇒ \(\theta = \frac{\pi}{4}\)
Step 4: Euler’s form: \(2 + 2i = 2\sqrt{2} e^{i\pi/4}\)
\(r\cos\theta = 2\), \(r\sin\theta = 2\)
Step 2: \(r^2 = 2^2 + 2^2 = 8\) ⇒ \(r = 2\sqrt{2}\)
Step 3: \(\tan\theta = \frac{2}{2} = 1\) ⇒ \(\theta = \frac{\pi}{4}\)
Step 4: Euler’s form: \(2 + 2i = 2\sqrt{2} e^{i\pi/4}\)
✅ \(2\sqrt{2} e^{i\pi/4}\)
1b \(\sqrt{3} + i\)
\(r = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = 2\)
\(\tan\theta = \frac{1}{\sqrt{3}}\) ⇒ \(\theta = \frac{\pi}{6}\)
\(\sqrt{3} + i = 2e^{i\pi/6}\)
\(\tan\theta = \frac{1}{\sqrt{3}}\) ⇒ \(\theta = \frac{\pi}{6}\)
\(\sqrt{3} + i = 2e^{i\pi/6}\)
✅ \(2e^{i\pi/6}\)
1c \(2i\)
\(0 + 2i\) ⇒ \(r = \sqrt{0^2 + 2^2} = 2\)
\(\theta = \frac{\pi}{2}\)
\(2i = 2e^{i\pi/2}\)
\(\theta = \frac{\pi}{2}\)
\(2i = 2e^{i\pi/2}\)
✅ \(2e^{i\pi/2}\)
1d \(i – \sqrt{3}\)
\(-\sqrt{3} + i\) ⇒ \(r = \sqrt{3 + 1} = 2\)
Point in 2nd quadrant: \(\tan\theta = \frac{1}{-\sqrt{3}}\) ⇒ \(\theta = \frac{5\pi}{6}\)
\(i – \sqrt{3} = 2e^{i5\pi/6}\)
Point in 2nd quadrant: \(\tan\theta = \frac{1}{-\sqrt{3}}\) ⇒ \(\theta = \frac{5\pi}{6}\)
\(i – \sqrt{3} = 2e^{i5\pi/6}\)
✅ \(2e^{i5\pi/6}\)
1e \(1 – i\)
\(r = \sqrt{1^2 + (-1)^2} = \sqrt{2}\)
\(\tan\theta = -1\) ⇒ \(\theta = -\frac{\pi}{4}\) or \(\frac{7\pi}{4}\)
\(1 – i = \sqrt{2} e^{-i\pi/4}\)
\(\tan\theta = -1\) ⇒ \(\theta = -\frac{\pi}{4}\) or \(\frac{7\pi}{4}\)
\(1 – i = \sqrt{2} e^{-i\pi/4}\)
✅ \(\sqrt{2} e^{-i\pi/4}\)
1f \(\frac{1+i}{1-i}\)
\(\frac{1+i}{1-i} \times \frac{1+i}{1+i} = \frac{1 + 2i + i^2}{1 – i^2} = \frac{1 + 2i – 1}{1 + 1} = \frac{2i}{2} = i\)
\(i = 0 + i\) ⇒ \(r = 1\), \(\theta = \frac{\pi}{2}\) ⇒ \(i = e^{i\pi/2}\)
\(i = 0 + i\) ⇒ \(r = 1\), \(\theta = \frac{\pi}{2}\) ⇒ \(i = e^{i\pi/2}\)
✅ \(e^{i\pi/2}\)
⚡ 3. De Moivre’s Theorem Applications
3a \((\cos 32^\circ + i\sin 32^\circ)(\cos 13^\circ + i\sin 13^\circ)\)
De Moivre: \(\cos A + i\sin A = e^{iA}\)
= \(e^{i32^\circ} \cdot e^{i13^\circ} = e^{i45^\circ} = \cos 45^\circ + i\sin 45^\circ = \frac{1}{\sqrt{2}}(1 + i)\)
= \(e^{i32^\circ} \cdot e^{i13^\circ} = e^{i45^\circ} = \cos 45^\circ + i\sin 45^\circ = \frac{1}{\sqrt{2}}(1 + i)\)
✅ \(\frac{1}{\sqrt{2}}(1 + i)\)
3b \((\sin 40^\circ + i\cos 40^\circ)(\cos 40^\circ + i\sin 40^\circ)\)
\(\sin 40^\circ + i\cos 40^\circ = \cos 50^\circ + i\sin 50^\circ = e^{i50^\circ}\)
\(\cos 40^\circ + i\sin 40^\circ = e^{i40^\circ}\)
Product = \(e^{i90^\circ} = \cos 90^\circ + i\sin 90^\circ = i\)
\(\cos 40^\circ + i\sin 40^\circ = e^{i40^\circ}\)
Product = \(e^{i90^\circ} = \cos 90^\circ + i\sin 90^\circ = i\)
✅ \(i\)
📐 4. Apply De Moivre’s Theorem
4a \([2(\cos 15^\circ + i\sin 15^\circ)]^6\)
= \(2^6 [\cos(6 \times 15^\circ) + i\sin(6 \times 15^\circ)] = 64(\cos 90^\circ + i\sin 90^\circ) = 64(0 + i) = 64i\)
✅ \(64i\)
4b \([3(\cos 12^\circ + i\sin 12^\circ)]^3\)
= \(27[\cos 36^\circ + i\sin 36^\circ]\) (Numerical value approximated)
Exact form: \(27(\cos 36^\circ + i\sin 36^\circ)\)
Exact form: \(27(\cos 36^\circ + i\sin 36^\circ)\)
✅ \(27e^{i\pi/5}\)
4c \((\cos 18^\circ + i\sin 18^\circ)^5\)
= \(\cos(5 \times 18^\circ) + i\sin(5 \times 18^\circ) = \cos 90^\circ + i\sin 90^\circ = 0 + i = i\)
✅ \(i\)
4d \([\cos 9^\circ + i\sin 9^\circ]^{40}\)
= \(\cos(360^\circ) + i\sin(360^\circ) = 1 + 0i = 1\)
✅ \(1\)
🌱 5. Square Roots of Complex Numbers
5a Square root of \(1 + i\)
Let \(1 + i = r(\cos\theta + i\sin\theta)\)
\(r = \sqrt{1^2 + 1^2} = \sqrt{2}\), \(\tan\theta = 1\) ⇒ \(\theta = 45^\circ\)
\(\sqrt{1+i} = (\sqrt{2})^{1/2}[\cos(45^\circ/2 + n180^\circ) + i\sin(45^\circ/2 + n180^\circ)]\)
\(n=0\): \(\sqrt[4]{2}(\cos 22.5^\circ + i\sin 22.5^\circ)\)
\(n=1\): \(\sqrt[4]{2}(\cos 202.5^\circ + i\sin 202.5^\circ)\)
\(r = \sqrt{1^2 + 1^2} = \sqrt{2}\), \(\tan\theta = 1\) ⇒ \(\theta = 45^\circ\)
\(\sqrt{1+i} = (\sqrt{2})^{1/2}[\cos(45^\circ/2 + n180^\circ) + i\sin(45^\circ/2 + n180^\circ)]\)
\(n=0\): \(\sqrt[4]{2}(\cos 22.5^\circ + i\sin 22.5^\circ)\)
\(n=1\): \(\sqrt[4]{2}(\cos 202.5^\circ + i\sin 202.5^\circ)\)
✅ \(\pm \sqrt[4]{2}(\cos 22.5^\circ + i\sin 22.5^\circ)\)
5b Square root of \(-2 – 2\sqrt{3}i\)
\(r = \sqrt{(-2)^2 + (-2\sqrt{3})^2} = \sqrt{4 + 12} = 4\)
Point in 3rd quadrant: \(\tan\theta = \frac{-2\sqrt{3}}{-2} = \sqrt{3}\) ⇒ \(\theta = 240^\circ\)
\(\sqrt{z} = \sqrt{4}[\cos(240^\circ/2 + n180^\circ) + i\sin(240^\circ/2 + n180^\circ)]\)
\(n=0\): \(2(\cos 120^\circ + i\sin 120^\circ) = 2(-1/2 + i\sqrt{3}/2) = -1 + i\sqrt{3}\)
\(n=1\): \(2(\cos 300^\circ + i\sin 300^\circ) = 2(1/2 – i\sqrt{3}/2) = 1 – i\sqrt{3}\)
Point in 3rd quadrant: \(\tan\theta = \frac{-2\sqrt{3}}{-2} = \sqrt{3}\) ⇒ \(\theta = 240^\circ\)
\(\sqrt{z} = \sqrt{4}[\cos(240^\circ/2 + n180^\circ) + i\sin(240^\circ/2 + n180^\circ)]\)
\(n=0\): \(2(\cos 120^\circ + i\sin 120^\circ) = 2(-1/2 + i\sqrt{3}/2) = -1 + i\sqrt{3}\)
\(n=1\): \(2(\cos 300^\circ + i\sin 300^\circ) = 2(1/2 – i\sqrt{3}/2) = 1 – i\sqrt{3}\)
✅ \(\pm(-1 + i\sqrt{3})\)
5c Square root of \(2i\)
\(2i = 2(\cos 90^\circ + i\sin 90^\circ)\)
\(\sqrt{2i} = \sqrt{2}[\cos(45^\circ + n180^\circ) + i\sin(45^\circ + n180^\circ)]\)
\(n=0\): \(\sqrt{2}(1/\sqrt{2} + i/\sqrt{2}) = 1 + i\)
\(n=1\): \(\sqrt{2}(-1/\sqrt{2} – i/\sqrt{2}) = -1 – i\)
\(\sqrt{2i} = \sqrt{2}[\cos(45^\circ + n180^\circ) + i\sin(45^\circ + n180^\circ)]\)
\(n=0\): \(\sqrt{2}(1/\sqrt{2} + i/\sqrt{2}) = 1 + i\)
\(n=1\): \(\sqrt{2}(-1/\sqrt{2} – i/\sqrt{2}) = -1 – i\)
✅ \(\pm(1 + i)\)
🧊 6. Cube Roots
6 Cube roots of \(-1\)
\(-1 = \cos 180^\circ + i\sin 180^\circ = \cos(180^\circ + n360^\circ) + i\sin(180^\circ + n360^\circ)\)
\(\sqrt[3]{-1} = \cos(60^\circ + n120^\circ) + i\sin(60^\circ + n120^\circ)\), \(n = 0,1,2\)
\(n=0\): \(\cos 60^\circ + i\sin 60^\circ = \frac{1}{2} + i\frac{\sqrt{3}}{2}\)
\(n=1\): \(\cos 180^\circ + i\sin 180^\circ = -1\)
\(n=2\): \(\cos 300^\circ + i\sin 300^\circ = \frac{1}{2} – i\frac{\sqrt{3}}{2}\)
\(\sqrt[3]{-1} = \cos(60^\circ + n120^\circ) + i\sin(60^\circ + n120^\circ)\), \(n = 0,1,2\)
\(n=0\): \(\cos 60^\circ + i\sin 60^\circ = \frac{1}{2} + i\frac{\sqrt{3}}{2}\)
\(n=1\): \(\cos 180^\circ + i\sin 180^\circ = -1\)
\(n=2\): \(\cos 300^\circ + i\sin 300^\circ = \frac{1}{2} – i\frac{\sqrt{3}}{2}\)
✅ \(-1, \frac{1}{2}(1 \pm i\sqrt{3})\)
📝 7. Solving Equations
7a \(z^4 = 1\)
\(1 = \cos 0^\circ + i\sin 0^\circ = \cos(n360^\circ) + i\sin(n360^\circ)\)
\(z = \cos(n90^\circ) + i\sin(n90^\circ)\), \(n = 0,1,2,3\)
\(n=0\): \(1\), \(n=1\): \(i\), \(n=2\): \(-1\), \(n=3\): \(-i\)
\(z = \cos(n90^\circ) + i\sin(n90^\circ)\), \(n = 0,1,2,3\)
\(n=0\): \(1\), \(n=1\): \(i\), \(n=2\): \(-1\), \(n=3\): \(-i\)
✅ \(\pm 1, \pm i\)
7b \(z^6 = 1\)
\(z = \cos(n60^\circ) + i\sin(n60^\circ)\), \(n = 0,1,2,3,4,5\)
Roots: \(1, \frac{1}{2} \pm i\frac{\sqrt{3}}{2}, -\frac{1}{2} \pm i\frac{\sqrt{3}}{2}, -1\)
Roots: \(1, \frac{1}{2} \pm i\frac{\sqrt{3}}{2}, -\frac{1}{2} \pm i\frac{\sqrt{3}}{2}, -1\)
✅ \(\pm 1, \frac{1}{2}(1 \pm i\sqrt{3}), \frac{1}{2}(-1 \pm i\sqrt{3})\)
7c \(z^4 + 1 = 0\)
\(z^4 = -1 = \cos 180^\circ + i\sin 180^\circ = \cos(180^\circ + n360^\circ) + i\sin(180^\circ + n360^\circ)\)
\(z = \cos(45^\circ + n90^\circ) + i\sin(45^\circ + n90^\circ)\), \(n = 0,1,2,3\)
\(n=0\): \(\frac{1}{\sqrt{2}}(1 + i)\), \(n=1\): \(\frac{1}{\sqrt{2}}(-1 + i)\)
\(n=2\): \(-\frac{1}{\sqrt{2}}(1 + i)\), \(n=3\): \(-\frac{1}{\sqrt{2}}(-1 + i)\)
\(z = \cos(45^\circ + n90^\circ) + i\sin(45^\circ + n90^\circ)\), \(n = 0,1,2,3\)
\(n=0\): \(\frac{1}{\sqrt{2}}(1 + i)\), \(n=1\): \(\frac{1}{\sqrt{2}}(-1 + i)\)
\(n=2\): \(-\frac{1}{\sqrt{2}}(1 + i)\), \(n=3\): \(-\frac{1}{\sqrt{2}}(-1 + i)\)
✅ \(\pm \frac{1}{\sqrt{2}}(1 + i), \pm \frac{1}{\sqrt{2}}(-1 + i)\)
7d \(z^3 = 8i\)
\(8i = 8(\cos 90^\circ + i\sin 90^\circ) = 8[\cos(90^\circ + n360^\circ) + i\sin(90^\circ + n360^\circ)]\)
\(z = 2[\cos(30^\circ + n120^\circ) + i\sin(30^\circ + n120^\circ)]\), \(n = 0,1,2\)
\(n=0\): \(2(\cos 30^\circ + i\sin 30^\circ) = \sqrt{3} + i\)
\(n=1\): \(2(\cos 150^\circ + i\sin 150^\circ) = -\sqrt{3} + i\)
\(n=2\): \(2(\cos 270^\circ + i\sin 270^\circ) = -2i\)
\(z = 2[\cos(30^\circ + n120^\circ) + i\sin(30^\circ + n120^\circ)]\), \(n = 0,1,2\)
\(n=0\): \(2(\cos 30^\circ + i\sin 30^\circ) = \sqrt{3} + i\)
\(n=1\): \(2(\cos 150^\circ + i\sin 150^\circ) = -\sqrt{3} + i\)
\(n=2\): \(2(\cos 270^\circ + i\sin 270^\circ) = -2i\)
✅ \(\sqrt{3} + i, -\sqrt{3} + i, -2i\)
📐 8. Important Theorems
8 \(\text{Arg}(\overline{z}) = 2\pi – \text{Arg}(z)\)
Let \(z = r(\cos\theta + i\sin\theta)\) ⇒ \(\overline{z} = r(\cos\theta – i\sin\theta) = r[\cos(2\pi – \theta) + i\sin(2\pi – \theta)]\)
∴ \(\text{Arg}(\overline{z}) = 2\pi – \theta = 2\pi – \text{Arg}(z)\)
∴ \(\text{Arg}(\overline{z}) = 2\pi – \theta = 2\pi – \text{Arg}(z)\)
✅ Proved
9 If \(z = \cos\theta + i\sin\theta\), prove \(z^n – \frac{1}{z^n} = 2i\sin n\theta\)
By De Moivre: \(z^n = \cos n\theta + i\sin n\theta\)
\(z^{-n} = \cos n\theta – i\sin n\theta\)
\(z^n – z^{-n} = (\cos n\theta + i\sin n\theta) – (\cos n\theta – i\sin n\theta) = 2i\sin n\theta\)
\(z^{-n} = \cos n\theta – i\sin n\theta\)
\(z^n – z^{-n} = (\cos n\theta + i\sin n\theta) – (\cos n\theta – i\sin n\theta) = 2i\sin n\theta\)
✅ Proved
🔍 9. Fourth Roots
10 Fourth roots of \(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\)
Modulus: \(r = \sqrt{(-\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = 1\)
Argument: \(\tan\theta = \frac{\sqrt{3}/2}{-1/2} = -\sqrt{3}\) ⇒ \(\theta = 120^\circ\) (2nd quadrant)
\(z = \cos 120^\circ + i\sin 120^\circ = \cos(120^\circ + n360^\circ) + i\sin(120^\circ + n360^\circ)\)
\(z^{1/4} = \cos(30^\circ + n90^\circ) + i\sin(30^\circ + n90^\circ)\), \(n = 0,1,2,3\)
Roots: \(\frac{1}{2}(\sqrt{3} + i), \frac{1}{2}(-1 + i\sqrt{3}), -\frac{1}{2}(\sqrt{3} + i), -\frac{1}{2}(-1 + i\sqrt{3})\)
Argument: \(\tan\theta = \frac{\sqrt{3}/2}{-1/2} = -\sqrt{3}\) ⇒ \(\theta = 120^\circ\) (2nd quadrant)
\(z = \cos 120^\circ + i\sin 120^\circ = \cos(120^\circ + n360^\circ) + i\sin(120^\circ + n360^\circ)\)
\(z^{1/4} = \cos(30^\circ + n90^\circ) + i\sin(30^\circ + n90^\circ)\), \(n = 0,1,2,3\)
Roots: \(\frac{1}{2}(\sqrt{3} + i), \frac{1}{2}(-1 + i\sqrt{3}), -\frac{1}{2}(\sqrt{3} + i), -\frac{1}{2}(-1 + i\sqrt{3})\)
✅ \(\pm \frac{1}{2}(\sqrt{3} + i), \pm \frac{1}{2}(-1 + i\sqrt{3})\)
🧠 Complex Number Key Formulas
📌 Polar form: \(z = r(\cos\theta + i\sin\theta)\)
📌 Euler’s form: \(z = re^{i\theta}\)
📌 De Moivre’s Theorem: \([r(\cos\theta + i\sin\theta)]^n = r^n(\cos n\theta + i\sin n\theta)\)
📌 nth roots: \(z^{1/n} = r^{1/n}[\cos(\frac{\theta + 2k\pi}{n}) + i\sin(\frac{\theta + 2k\pi}{n})]\), \(k = 0,1,…,n-1\)