The Problem that Real Numbers Couldn't Solve
Consider the elementary equation: $x^2 + 1 = 0 \implies x^2 = -1$.
Since the square of any real number is always non-negative ($x^2 \ge 0$), no real number can satisfy this equation. To overcome this limitation, mathematicians introduced the imaginary unit , denoted by $i$ (iota), defined by:
$$i = \sqrt{-1} \quad \implies \quad i^2 = -1$$
Definition: Complex Number
A number of the form $z = a + ib$, where $a, b \in \mathbb{R}$ and $i = \sqrt{-1}$, is called a complex number .
$a$ is the Real part of $z$, written as $\text{Re}(z) = a$.
$b$ is the Imaginary part of $z$, written as $\text{Im}(z) = b$ (Note: $b$ is a real number, not including $i$!).
If $b = 0$, $z = a$ is purely real .
If $a = 0$ and $b \ne 0$, $z = ib$ is purely imaginary .
The Cyclic Pattern
The powers of $i$ repeat in a cycle of 4:
$i^0 = 1$
$i^1 = i$
$i^2 = -1$
$i^3 = i^2 \cdot i = (-1)i = -i$
$i^4 = (i^2)^2 = (-1)^2 = 1$
Master Formula for Any Power $n$: Divide $n$ by $4$. If $n = 4k + r$ (where $r \in \{0, 1, 2, 3\}$ is the remainder):
$$i^n = i^{4k + r} = (i^4)^k \cdot i^r = (1)^k \cdot i^r = i^r$$
Golden Property: The sum of any four consecutive powers of $i$ is ALWAYS ZERO:
$$i^n + i^{n+1} + i^{n+2} + i^{n+3} = i^n(1 + i - 1 - i) = 0$$
1. Equality of Two Complex Numbers
$$a_1 + ib_1 = a_2 + ib_2 \iff a_1 = a_2 \quad \text{and} \quad b_1 = b_2$$
Real parts must equal real parts, and imaginary parts must equal imaginary parts!
Addition & Subtraction
$$(a + ib) \pm (c + id) = (a \pm c) + i(b \pm d)$$
Add/subtract real parts and imaginary parts separately.
Multiplication (FOIL with $i^2 = -1$)
$$(a + ib)(c + id) = ac + iad + ibc + i^2 bd$$
$$= (ac - bd) + i(ad + bc)$$
Division (Conjugate Multiplication)
To compute $\frac{a + ib}{c + id}$, multiply numerator and denominator by the conjugate of the denominator $(c - id)$:
$$\frac{a + ib}{c + id} \times \frac{c - id}{c - id} = \frac{(ac + bd) + i(bc - ad)}{c^2 + d^2}$$
Identities
All algebraic identities hold: $(z_1 \pm z_2)^2 = z_1^2 \pm 2z_1 z_2 + z_2^2$, $z_1^2 - z_2^2 = (z_1 - z_2)(z_1 + z_2)$, etc.
Conjugate of a Complex Number
If $z = a + ib$, its conjugate is $\bar{z} = a - ib$ (simply flip the sign of the imaginary part!).
Properties: $\overline{z_1 \pm z_2} = \bar{z}_1 \pm \bar{z}_2$, $\overline{z_1 z_2} = \bar{z}_1 \bar{z}_2$, $\overline{(\bar{z})} = z$, and $\mathbf{z \bar{z} = a^2 + b^2 = |z|^2}$.
Modulus (Absolute Value) of a Complex Number
The modulus of $z = a + ib$ is the distance of point $(a, b)$ from the origin in the Argand plane:
$$|z| = \sqrt{a^2 + b^2} \ge 0$$
Properties: $|z_1 z_2| = |z_1||z_2|$, $\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}$, $|z| = |\bar{z}| = |-z|$.
Multiplicative Inverse ($z^{-1}$)
The multiplicative inverse of non-zero $z$ is the number $z^{-1}$ such that $z \cdot z^{-1} = 1$:
$$z^{-1} = \frac{1}{z} = \frac{\bar{z}}{z \bar{z}} = \frac{\bar{z}}{|z|^2} = \frac{a - ib}{a^2 + b^2}$$
The Argand Plane
Every complex number $z = a + ib$ uniquely corresponds to an ordered pair $(a, b)$ on a Cartesian coordinate plane:
The horizontal axis is the Real axis .
The vertical axis is the Imaginary axis .
Polar Form of a Complex Number
Let $r = |z| = \sqrt{a^2 + b^2}$ and $\theta = \text{arg}(z)$. The polar representation is:
$$z = r(\cos \theta + i\sin \theta)$$
where $\theta$ is the Principal Argument , restricted to the interval: $\mathbf{-\pi < \theta \le \pi}$.
Step 1: Compute Acute Reference Angle $\alpha$
$$\tan \alpha = \left|\frac{b}{a}\right| \implies \alpha = \tan^{-1}\left|\frac{b}{a}\right|$$
Step 2: Determine Quadrant & Assign $\theta$
Quadrant I ($a>0, b>0$): $\theta = \alpha$Quadrant II ($a<0, b>0$): $\theta = \pi - \alpha$Quadrant III ($a<0, b<0$): $\theta = -(\pi - \alpha) = -\pi + \alpha$Quadrant IV ($a>0, b<0$): $\theta = -\alpha$
Solving $ax^2 + bx + c = 0$ when $D < 0$
When the discriminant $D = b^2 - 4ac < 0$, the roots are not real, but complex conjugates:
$$\sqrt{D} = \sqrt{-(4ac - b^2)} = i\sqrt{4ac - b^2}$$
$$x = \frac{-b \pm i\sqrt{4ac - b^2}}{2a}$$
Evaluate: (i) $i^{35}$, (ii) $i^{-39}$, (iii) $i^9 + i^{19}$.
(i) $i^{35}$: $35 = 4(8) + 3$. Thus $i^{35} = (i^4)^8 \cdot i^3 = (1)^8 \cdot (-i) = -i$.
(ii) $i^{-39}$: $\frac{1}{i^{39}}$. Since $39 = 4(9) + 3$, $i^{39} = -i$. So $\frac{1}{-i} = \frac{i}{-i^2} = \frac{i}{-(-1)} = i$.
(iii) $i^9 + i^{19}$: $i^9 = i^{4(2)+1} = i$. $i^{19} = i^{4(4)+3} = -i$. Sum $= i + (-i) = 0$.
Express $\frac{5 + \sqrt{2}i}{1 - \sqrt{2}i}$ in the form $a + ib$.
Step 1: Multiply numerator and denominator by conjugate of denominator $(1 + \sqrt{2}i)$:
$$\frac{5 + \sqrt{2}i}{1 - \sqrt{2}i} \times \frac{1 + \sqrt{2}i}{1 + \sqrt{2}i} = \frac{5 + 5\sqrt{2}i + \sqrt{2}i + (\sqrt{2}i)^2}{1^2 - (\sqrt{2}i)^2}$$
Step 2: Simplify: $(\sqrt{2}i)^2 = 2i^2 = -2$, and denominator $= 1 - (-2) = 3$.
$$\frac{(5 - 2) + (5\sqrt{2} + \sqrt{2})i}{3} = \frac{3 + 6\sqrt{2}i}{3} = 1 + 2\sqrt{2}i$$
Final Answer: $a = 1,\; b = 2\sqrt{2}$.
Find the multiplicative inverse of $z = 4 - 3i$.
Step 1: Find conjugate $\bar{z} = 4 + 3i$.
Step 2: Find $|z|^2 = 4^2 + (-3)^2 = 16 + 9 = 25$.
Step 3: Compute inverse:
$$z^{-1} = \frac{\bar{z}}{|z|^2} = \frac{4 + 3i}{25} = \frac{4}{25} + \frac{3}{25}i$$
Convert $z = -1 + i\sqrt{3}$ into polar form.
Step 1: Modulus $r$:
$$r = |z| = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2$$
Step 2: Acute angle $\alpha$:
$$\tan \alpha = \left|\frac{b}{a}\right| = \left|\frac{\sqrt{3}}{-1}\right| = \sqrt{3} \implies \alpha = \frac{\pi}{3}$$
Step 3: Principal Argument $\theta$:
Since $a = -1 < 0$ and $b = \sqrt{3} > 0$, $z$ lies in Quadrant II .
$$\theta = \pi - \alpha = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$$
Step 4: Polar form:
$$z = 2\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right)$$
If $x + iy = \frac{a + ib}{a - ib}$, prove that $x^2 + y^2 = 1$.
Method using Modulus:
Take the modulus on both sides:
$$|x + iy| = \left|\frac{a + ib}{a - ib}\right|$$
Property of Modulus: $\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}$.
$$\sqrt{x^2 + y^2} = \frac{|a + ib|}{|a - ib|} = \frac{\sqrt{a^2 + b^2}}{\sqrt{a^2 + (-b)^2}} = \frac{\sqrt{a^2 + b^2}}{\sqrt{a^2 + b^2}} = 1$$
Squaring both sides:
$$x^2 + y^2 = 1^2 = 1$$
Hence proved! (Much faster than expanding algebraically!)
Solve the quadratic equation: $\sqrt{2}x^2 + x + \sqrt{2} = 0$.
Step 1: Compute discriminant:
Here $a = \sqrt{2}$, $b = 1$, $c = \sqrt{2}$.
$$D = b^2 - 4ac = 1^2 - 4(\sqrt{2})(\sqrt{2}) = 1 - 4(2) = 1 - 8 = -7$$
Step 2: Roots:
Since $D = -7 < 0$, $\sqrt{D} = \sqrt{-7} = i\sqrt{7}$.
$$x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-1 \pm i\sqrt{7}}{2\sqrt{2}} = -\frac{1}{2\sqrt{2}} \pm i\frac{\sqrt{7}}{2\sqrt{2}}$$
NCERT Exercise 4.1 — Question 11
Q: Find the multiplicative inverse of $-i$.
View Full Solution
Let $z = -i = 0 - 1i$.
Conjugate $\bar{z} = 0 + 1i = i$.
Modulus squared $|z|^2 = 0^2 + (-1)^2 = 1$.
$$z^{-1} = \frac{\bar{z}}{|z|^2} = \frac{i}{1} = i$$
NCERT Exercise 4.2 / Polar Form Problem
Q: Convert $z = -1 - i$ into polar form.
View Full Solution
$r = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2}$.
$\tan \alpha = |-1 / -1| = 1 \implies \alpha = \frac{\pi}{4}$.
Since both $a < 0$ and $b < 0$, $z$ is in Quadrant III :
$$\theta = -(\pi - \alpha) = -\left(\pi - \frac{\pi}{4}\right) = -\frac{3\pi}{4}$$
Polar form: $z = \sqrt{2}\left(\cos\left(-\frac{3\pi}{4}\right) + i\sin\left(-\frac{3\pi}{4}\right)\right)$.
NCERT Miscellaneous Exercise — Question 5
Q: Convert $\frac{1 + 7i}{(2 - i)^2}$ into polar form.
View Full Solution
First expand denominator: $(2 - i)^2 = 4 - 4i + i^2 = 3 - 4i$.
Simplify fraction: $\frac{1 + 7i}{3 - 4i} \times \frac{3 + 4i}{3 + 4i} = \frac{3 + 4i + 21i + 28i^2}{9 - 16i^2} = \frac{-25 + 25i}{25} = -1 + i$.
$r = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$.
$\tan \alpha = 1 \implies \alpha = \frac{\pi}{4}$. In Q2: $\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.
Polar form: $\sqrt{2}\left(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\right)$.
The rule $\sqrt{a} \sqrt{b} = \sqrt{ab}$ is ONLY valid when at least one of $a$ or $b$ is non-negative! If both are negative: $\sqrt{-1} \sqrt{-1} = i \cdot i = -1$. But $\sqrt{(-1)(-1)} = \sqrt{1} = 1$. Concluding that $-1 = 1$ is a catastrophic logical fallacy!
Many students compute $\theta = \tan^{-1}(b/a)$ blindly. For $-1 - i$, $\frac{-1}{-1} = 1$, giving $\theta = \pi/4$. But $-1 - i$ is in Quadrant III, where $\pi/4$ is in Quadrant I! Always sketch the point first to identify the quadrant.
For $z = 3 + 4i$, $\text{Im}(z) = 4$, NOT $4i$. Writing $\text{Im}(z) = 4i$ will lose marks.
Q1: What is the value of $i^{100} + i^{101} + i^{102} + i^{103}$?
Answer
0. The sum of four consecutive powers of $i$ is always zero ($1 + i - 1 - i = 0$).
Q2: What is the conjugate of the complex number $5$?
Answer
5. Since $5 = 5 + 0i$, its conjugate is $5 - 0i = 5$. A real number is its own conjugate!
Q3: What is the modulus of $\frac{1 + i}{1 - i}$?
Answer
1. $|(1+i)/(1-i)| = |1+i| / |1-i| = \sqrt{2}/\sqrt{2} = 1$. (Also, $(1+i)/(1-i) = i$, and $|i| = 1$).
Q4: In which quadrant does the principal argument of $-3 - 4i$ lie?
Answer
Quadrant III (both real and imaginary parts are negative). The argument is in $(-\pi, -\pi/2)$.
Q5: Solve for real $x, y$: $(x + 2y) + i(2x - 3y) = 5 - 4i$.
Answer
Equating parts: $x + 2y = 5$ and $2x - 3y = -4$. Multiply first eq by 2: $2x + 4y = 10$. Subtract: $7y = 14 \implies y = 2$. Then $x = 5 - 2(2) = 1$. So $x = 1, y = 2$.