Complex Numbers And Quadratic Equations Class 11 NCERT

Introduction to Complex Numbers & Why They Matter

In our previous study of real numbers, we noticed a fundamental limitation: the square of any real number is always non-negative. Consequently, equations like $x^2 + 1 = 0$ have no solution in the set of real numbers. To resolve this, mathematicians defined the imaginary unit $i$ (iota) such that $i^2 = -1$. A number of the form $z = a + ib$, where $a$ and $b$ are real numbers, is called a complex number. Here, $a$ is the real part, denoted by $\text{Re}(z)$, and $b$ is the imaginary part, denoted by $\text{Im}(z)$. If $b = 0$, the number is purely real; if $a = 0$, it is purely imaginary. This demonstrates that the real number system is simply a subset of the larger complex number system.

Key Terminologies and Definitions

Complex Number

Any number that can be expressed in the form $z = a + ib$, where $a, b \in \mathbb{R}$ and $i = \sqrt{-1}$.

Conjugate of a Complex Number

For any complex number $z = a + ib$, its conjugate is denoted by $\bar{z}$ and defined as $a - ib$. Geometrically, it is the reflection of $z$ across the real axis.

Modulus

The modulus of $z = a + ib$, written as $|z|$, is the non-negative real number $\sqrt{a^2 + b^2}$. It represents the distance of the point $(a, b)$ from the origin in the Argand plane.

Multiplicative Inverse

For a non-zero complex number $z = a + ib$, the multiplicative inverse is $z^{-1} = \frac{1}{z} = \frac{\bar{z}}{|z|^2}$.

Algebraic Operations on Complex Numbers

1. Addition and Subtraction — Combine real parts and imaginary parts separately. For $z_1 = a + ib$ and $z_2 = c + id$, we get $z_1 \pm z_2 = (a \pm c) + i(b \pm d)$.
2. Multiplication — Multiply using the distributive law, keeping in mind that $i^2 = -1$. Thus, $(a + ib)(c + id) = (ac - bd) + i(ad + bc)$.
3. Division and Rationalization — To evaluate $\frac{z_1}{z_2}$, multiply both numerator and denominator by the conjugate of the denominator $\bar{z}_2$ to convert the denominator into a real value.

Exam Trap: Square Roots of Negative Numbers

A very common error in CBSE board exams involves the incorrect application of the identity $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$. This property is only valid when at least one of $a$ or $b$ is non-negative!

For negative real numbers, always express them in terms of $i$ first. For instance:
$\sqrt{-3} \times \sqrt{-12} = (i\sqrt{3}) \times (i\sqrt{12}) = i^2\sqrt{36} = -1 \times 6 = -6$
Never write: $\sqrt{-3} \times \sqrt{-12} = \sqrt{(-3) \times (-12)} = \sqrt{36} = 6$. This is a critical trap where students lose easy marks!

Practice Questions with Solutions

• Q: Express the complex number $\frac{5 + \sqrt{2}i}{1 - \sqrt{2}i}$ in the standard form $a + ib$. A: Step 1: Identify the conjugate of the denominator. The denominator is $1 - \sqrt{2}i$, so its conjugate is $1 + \sqrt{2}i$. Step 2: Multiply the numerator and the denominator by this conjugate: $\frac{(5 + \sqrt{2}i)(1 + \sqrt{2}i)}{(1 - \sqrt{2}i)(1 + \sqrt{2}i)}$ Step 3: Expand the numerator: $5(1) + 5\sqrt{2}i + \sqrt{2}i + (\sqrt{2}i)^2 = 5 + 6\sqrt{2}i - 2 = 3 + 6\sqrt{2}i$. Step 4: Expand the denominator: $1^2 - (\sqrt{2}i)^2 = 1 - (-2) = 3$. Step 5: Divide: $\frac{3 + 6\sqrt{2}i}{3} = 1 + 2\sqrt{2}i$. Final answer: $1 + 2\sqrt{2}i$
• Q: Find the multiplicative inverse of the complex number $z = 2 - 3i$. A: Step 1: Write down the formula for the multiplicative inverse: $z^{-1} = \frac{\bar{z}}{|z|^2}$. Step 2: Find the conjugate $\bar{z} = 2 + 3i$. Step 3: Calculate $|z|^2 = a^2 + b^2 = 2^2 + (-3)^2 = 4 + 9 = 13$. Step 4: Substitute these values into the formula: $z^{-1} = \frac{2 + 3i}{13} = \frac{2}{13} + \frac{3}{13}i$. Final answer: $\frac{2}{13} + \frac{3}{13}i$
• Q: Solve the quadratic equation $x^2 + 3x + 9 = 0$ using the quadratic formula. A: Step 1: Compare with the standard quadratic equation $ax^2 + bx + c = 0$ to get $a = 1, b = 3, c = 9$. Step 2: Calculate the discriminant: $D = b^2 - 4ac = 3^2 - 4(1)(9) = 9 - 36 = -27$. Step 3: Since $D < 0$, the roots are complex conjugate pairs. Use the formula $x = \frac{-b \pm \sqrt{D}}{2a}$. Step 4: Simplify the root: $\sqrt{-27} = i\sqrt{27} = 3\sqrt{3}i$. Step 5: Substitute: $x = \frac{-3 \pm 3\sqrt{3}i}{2}$. Final answer: $x = \frac{-3 \pm 3\sqrt{3}i}{2}$
• Q: Evaluate the value of the algebraic expression $i^{18} + \left(\frac{1}{i}\right)^{25}$. A: Step 1: Simplify $i^{18}$. Since $18 = 4(4) + 2$, we have $i^{18} = (i^4)^4 \cdot i^2 = (1)^4 \cdot (-1) = -1$. Step 2: Simplify $\frac{1}{i}$ by multiplying numerator and denominator by $i$: $\frac{1}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i$. Step 3: Evaluate $\left(\frac{1}{i}\right)^{25} = (-i)^{25} = (-1)^{25} \cdot i^{25} = - (i^{24} \cdot i) = - ((i^4)^6 \cdot i) = - (1 \cdot i) = -i$. Step 4: Add the two results: $-1 + (-i) = -1 - i$. Final answer: $-1 - i$

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