Complex Numbers and Quadratic Equations Ex 5.1 Class 11 NCERT

Algebra of Complex Numbers & Cyclic Powers of i

To operate on complex numbers, we treat them like standard real binomials while applying the fundamental rule: $i^2 = -1$. When adding or subtracting two complex numbers, we combine their real and imaginary components separately: $(a+ib) \pm (c+id) = (a \pm c) + i(b \pm d)$. Multiplication follows the distributive law: $(a+ib)(c+id) = (ac - bd) + i(ad + bc)$. Division requires rationalizing the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator. One of the primary focal points of Exercise 5.1 is managing powers of the imaginary unit $i$. Its values cycle every four steps: $i^1 = i$, $i^2 = -1$, $i^3 = -i$, and $i^4 = 1$. Consequently, any integer power $i^n$ can be quickly simplified by finding the remainder $r$ when $n$ is divided by 4: $i^n = i^r$.

Key Terminology and Formulas

Complex Number

Any number of the form z = a + ib, where a and b are real numbers and i is the imaginary unit satisfying i² = -1.

Conjugate of a Complex Number

For a given complex number z = a + ib, its conjugate is denoted as z̅ = a - ib, obtained by reversing the sign of its imaginary component.

Modulus Squared

The square of the absolute value of z, defined as |z|² = a² + b², representing the distance squared from the origin in the Argand plane.

Multiplicative Inverse

The reciprocal of a non-zero complex number, denoted as z⁻¹ or 1/z, calculated using the mathematical formula z⁻¹ = z̅ / |z|².

Step-by-Step Method: Finding the Multiplicative Inverse

1. Express the complex number clearly — Identify the given complex number and write it in the standard algebraic format, $z = a + ib$.
2. Determine the conjugate of z — Obtain the complex conjugate $\bar{z} = a - ib$ by swapping the sign of the imaginary coefficient.
3. Calculate the squared modulus — Compute $|z|^2 = a^2 + b^2$. Ensure you square both coefficients and sum them up (do not include $i$ in this computation).
4. Divide conjugate by modulus squared — Evaluate the division $z^{-1} = \frac{\bar{z}}{|z|^2} = \frac{a - ib}{a^2 + b^2}$, and split it into real and imaginary parts: $\frac{a}{a^2+b^2} - i\frac{b}{a^2+b^2}$.

Exam Traps & Crucial Tips

To prevent silly errors and secure full marks in Ex 5.1, keep these strategic board tips in mind:

1. The Negative Sign Trap in Modulus: When calculating $|z|^2$ for a number like $z = 4 - 3i$, the imaginary part is $b = -3$. Therefore, $|z|^2 = 4^2 + (-3)^2 = 16 + 9 = 25$. Students often make the mistake of calculating $16 - 9$, which leads to incorrect solutions.
2. Strict standard form $a + ib$: CBSE marking schemes require the final answer to be explicitly in standard form. If your final simplified value is $5$, write it as $5 + 0i$. If it simplifies to $-i$, express it as $0 - i$.
3. Negative exponents of $i$: To solve problems like $i^{-39}$, rewrite them as $\frac{1}{i^{39}}$ first. Then, reduce the denominator power: $i^{39} = i^{36} \cdot i^3 = 1 \cdot (-i) = -i$. This leaves $\frac{1}{-i}$, which simplifies to $i$ after multiplying numerator and denominator by $i$.

Practice Questions with Solutions

• Q: Express the following complex number in standard $a + ib$ form: $(5i) \times \left(-\frac{3}{5}i\right)$ A: Step 1: Write down the product expression: $z = (5i) \times \left(-\frac{3}{5}i\right)$ Step 2: Multiply the real coefficients and the imaginary parts separately: $z = 5 \times \left(-\frac{3}{5}\right) \times (i \times i)$ $z = -3 \times i^2$ Step 3: Substitute $i^2 = -1$: $z = -3 \times (-1) = 3$ Step 4: Express the final real integer in standard form $a + ib$: $z = 3 + 0i$ Final answer: $3 + 0i$
• Q: Express the complex number $i^9 + i^{19}$ in standard $a + ib$ form. A: Step 1: Simplify $i^9$ by dividing the exponent 9 by 4: $9 = 4(2) + 1 \implies i^9 = (i^4)^2 \cdot i^1 = (1)^2 \cdot i = i$ Step 2: Simplify $i^{19}$ by dividing the exponent 19 by 4: $19 = 4(4) + 3 \implies i^{19} = (i^4)^4 \cdot i^3 = (1)^4 \cdot (-i) = -i$ Step 3: Add the simplified results: $i^9 + i^{19} = i + (-i) = 0$ Step 4: Express this in standard $a + ib$ form: $0 + 0i$ Final answer: $0 + 0i$
• Q: Find the multiplicative inverse of the complex number $4 - 3i$. A: Step 1: Set $z = 4 - 3i$. Here, the real part is $a = 4$ and the imaginary part is $b = -3$. Step 2: Find the complex conjugate $\bar{z}$: $\bar{z} = 4 - (-3i) = 4 + 3i$ Step 3: Calculate the modulus squared $|z|^2$: $|z|^2 = a^2 + b^2 = 4^2 + (-3)^2 = 16 + 9 = 25$ Step 4: Use the multiplicative inverse formula $z^{-1} = \frac{\bar{z}}{|z|^2}$: $z^{-1} = \frac{4 + 3i}{25} = \frac{4}{25} + \frac{3}{25}i$ Final answer: $\frac{4}{25} + \frac{3}{25}i$
• Q: Express the complex number $(1 - i)^4$ in standard $a + ib$ form. A: Step 1: Rewrite $(1 - i)^4$ as $[(1 - i)^2]^2$. Step 2: Expand the inner expression using $(x - y)^2 = x^2 - 2xy + y^2$: $(1 - i)^2 = 1^2 - 2(1)(i) + i^2$ $= 1 - 2i - 1$ (since $i^2 = -1$) $= -2i$ Step 3: Square the result obtained in Step 2: $[(1 - i)^2]^2 = (-2i)^2 = 4i^2$ Step 4: Substitute $i^2 = -1$: $4(-1) = -4$ Step 5: Write the answer in $a + ib$ form: $-4 + 0i$ Final answer: $-4 + 0i$

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