NCERT Class 11 Maths Complex Numbers and Quadratic Equations Miscellaneous Exercise

Overview of the Miscellaneous Exercise

The Miscellaneous Exercise of Chapter 5 acts as a bridge between foundational algebra and advanced complex number theory. It demands a flawless command over the properties of the imaginary unit $i$, the rationalization of complex expressions, and the utilization of algebraic identities. To excel here, you must be comfortable transitioning a complex number $z = x + iy$ to its polar form $r(\cos\theta + i\sin\theta)$ and applying complex algebraic identities such as $(a+b)^3$ directly to complex variables. The problems in this exercise generally fall into three high-yield categories: multi-step algebraic simplification, solving simultaneous equations with complex conjugates, and proving complex identities using the properties of modulus ($|z_1 z_2| = |z_1||z_2|$).

Core Formulae & Properties to Memorize

Properties of Conjugate

For any complex numbers $z_1$ and $z_2$, the conjugate distributed properties are crucial: $\overline{z_1 \pm z_2} = \overline{z_1} \pm \overline{z_2}$, $\overline{z_1 \cdot z_2} = \overline{z_1} \cdot \overline{z_2}$, and $\overline{\left(\frac{z_1}{z_2}\right)} = \frac{\overline{z_1}}{\overline{z_2}}$ (where $z_2 \neq 0$).

Properties of Modulus

Modulus properties simplify calculations without requiring full expansions: $|z|^2 = z \cdot \overline{z}$, $|z_1 z_2| = |z_1||z_2|$, and $\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}$.

Polar Form and Argument

Any complex number $z = x+iy$ can be expressed as $r(\cos \theta + i \sin \theta)$, where the modulus $r = \sqrt{x^2+y^2}$ and the principal argument $\theta \in (-\pi, \pi]$ is determined by checking the quadrant of the point $(x, y)$.

Step-by-Step Method: Simplifying Complex Fractions

1. Identify the Denominator — Examine the given complex fraction and locate the denominator of the form $c + id$.
2. Multiply by the Conjugate — Multiply both the numerator and the denominator by the complex conjugate of the denominator, which is $c - id$.
3. Apply algebraic identities — Simplify the denominator using the identity $(c+id)(c-id) = c^2 + d^2$. Expand the numerator using standard algebraic distribution.
4. Separate Real and Imaginary Parts — Group the terms of the simplified numerator to write the final result in standard form $A + iB$, where $A = \text{Re}(z)$ and $B = \text{Im}(z)$.

Worked Examples from the Miscellaneous Concepts

• Example 1: Find the real values of $x$ and $y$ if $(x - iy)(3 + 5i)$ is the conjugate of $-6 - 24i$. Step 1: Write down the given condition. The conjugate of $-6 - 24i$ is $-6 + 24i$. So, we set up the equation: $(x - iy)(3 + 5i) = -6 + 24i$. Step 2: Expand the left-hand side. $x(3+5i) - iy(3+5i) = 3x + 5xi - 3yi - 5y(i^2)$ Since $i^2 = -1$, this simplifies to: $(3x + 5y) + i(5x - 3y) = -6 + 24i$. Step 3: Equate the real and imaginary parts from both sides to form a system of linear equations: Real parts: $3x + 5y = -6$ ... (Equation 1) Imaginary parts: $5x - 3y = 24$ ... (Equation 2) Step 4: Solve the simultaneous equations. Multiply Equation 1 by 3 and Equation 2 by 5: $9x + 15y = -18$ $25x - 15y = 120$ Add both equations: $34x = 102 \implies x = 3$. Step 5: Substitute $x = 3$ back into Equation 1: $3(3) + 5y = -6 \implies 9 + 5y = -6 \implies 5y = -15 \implies y = -3$. Final Answer: The real values are $x = 3$ and $y = -3$.
• Example 2: Find the modulus of the complex number $z = \frac{1+i}{1-i} - \frac{1-i}{1+i}$. Step 1: Simplify the expression inside $z$ by finding a common denominator. $z = \frac{(1+i)^2 - (1-i)^2}{(1-i)(1+i)}$ Step 2: Expand the numerator terms. $(1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i$ $(1-i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i$ Therefore, the numerator is $2i - (-2i) = 4i$. Step 3: Simplify the denominator. $(1-i)(1+i) = 1^2 - i^2 = 1 - (-1) = 2$. Step 4: Write $z$ in standard form. $z = \frac{4i}{2} = 2i = 0 + 2i$. Step 5: Calculate the modulus $|z|$. $|z| = \sqrt{0^2 + 2^2} = \sqrt{4} = 2$. Final Answer: The modulus of the given complex number is 2.

Exam Traps & Crucial Study Tips

1. The Argument Quadrant Trap: When converting a complex number to polar form, do not blindly calculate $\theta = \tan^{-1}|y/x|$. Always locate which quadrant the point $(x, y)$ belongs to first. For example, for $z = -1 - i$ (Quadrant III), the principal argument is $-\frac{3\pi}{4}$, not $\frac{\pi}{4}$.
2. Avoid Tedious Expansions: If an exam question asks you to find the modulus of a complex multiplication or division (e.g., find $|z|$ where $z = \frac{(2+3i)(1-i)}{(3+4i)}$), do not multiply out the expression first! Use the property $|z| = \frac{|2+3i|\cdot|1-i|}{|3+4i|} = \frac{\sqrt{13}\cdot\sqrt{2}}{\sqrt{25}} = \frac{\sqrt{26}}{5}$ directly. This saves precious minutes and prevents calculation errors.

Practice Questions with Solutions

• Q: Evaluate: $\left[ i^{18} + \left(\frac{1}{i}\right)^{25} \right]^3$. A: Step 1: Simplify the individual power terms using the fact that $i^4 = 1$. $i^{18} = (i^4)^4 \cdot i^2 = 1 \cdot (-1) = -1$. Step 2: Simplify the second term. $\left(\frac{1}{i}\right)^{25} = \frac{1}{i^{25}} = \frac{1}{(i^4)^6 \cdot i} = \frac{1}{i}$. To remove $i$ from the denominator, multiply numerator and denominator by $i$: $\frac{1}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i$. Step 3: Substitute these values back into the expression: $[-1 + (-i)]^3 = [-(1 + i)]^3 = -(1 + i)^3$. Step 4: Use the algebraic identity $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ to expand $(1+i)^3$: $(1+i)^3 = 1^3 + 3(1)^2(i) + 3(1)(i^2) + i^3$ $= 1 + 3i - 3 - i$ $= -2 + 2i$. Step 5: Apply the negative sign from Step 3: $-(-2 + 2i) = 2 - 2i$. Final answer: $2 - 2i$
• Q: If $z_1 = 2 - i$ and $z_2 = 1 + i$, find the value of $\left| \frac{z_1 + z_2 + 1}{z_1 - z_2 + 1} \right|$. A: Step 1: Calculate the numerator complex number: Numerator $= z_1 + z_2 + 1 = (2-i) + (1+i) + 1 = 4$. Step 2: Calculate the denominator complex number: Denominator $= z_1 - z_2 + 1 = (2-i) - (1+i) + 1 = 2 - i - 1 - i + 1 = 2 - 2i$. Step 3: Write down the complete fraction: $\frac{z_1 + z_2 + 1}{z_1 - z_2 + 1} = \frac{4}{2-2i} = \frac{4}{2(1-i)} = \frac{2}{1-i}$. Step 4: Find the modulus of this expression. Use the modulus quotient property $\left|\frac{z_a}{z_b}\right| = \frac{|z_a|}{|z_b|}$: $\left| \frac{2}{1-i} \right| = \frac{|2|}{|1-i|} = \frac{2}{\sqrt{1^2 + (-1)^2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$. Final answer: $\sqrt{2}$
• Q: Find the modulus and argument of the complex number $z = \frac{1+2i}{1-3i}$. A: Step 1: Convert $z$ to standard form $x+iy$ by multiplying numerator and denominator by the conjugate of the denominator ($1+3i$): $z = \frac{(1+2i)(1+3i)}{(1-3i)(1+3i)} = \frac{1 + 3i + 2i + 6i^2}{1^2 - (3i)^2}$ $= \frac{1 + 5i - 6}{1 + 9} = \frac{-5 + 5i}{10} = -\frac{1}{2} + \frac{1}{2}i$. Step 2: Calculate the modulus $r$: $r = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{2}{4}} = \frac{1}{\sqrt{2}}$. Step 3: Determine the argument $\theta$. The real part is negative ($x = -1/2$) and the imaginary part is positive ($y = 1/2$), placing the point in the second quadrant. Find the acute angle $\alpha = \tan^{-1}\left|\frac{y}{x}\right| = \tan^{-1}\left|\frac{1/2}{-1/2}\right| = \tan^{-1}(1) = \frac{\pi}{4}$. Step 4: Since $z$ lies in Quadrant II, $\theta = \pi - \alpha = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$. Final answer: Modulus $= \frac{1}{\sqrt{2}}$, Argument $= \frac{3\pi}{4}$
• Q: If $\alpha$ and $\beta$ are different complex numbers with $|\beta| = 1$, then find the value of $\left| \frac{\beta - \alpha}{1 - \overline{\alpha}\beta} \right|$. A: Step 1: Let $E = \left| \frac{\beta - \alpha}{1 - \overline{\alpha}\beta} \right|$. To evaluate $E$, let us square it and use the property $|z|^2 = z \overline{z}$: $E^2 = \left( \frac{\beta - \alpha}{1 - \overline{\alpha}\beta} \right) \overline{\left( \frac{\beta - \alpha}{1 - \overline{\alpha}\beta} \right)}$ Step 2: Distribute the conjugate over the fraction: $E^2 = \left( \frac{\beta - \alpha}{1 - \overline{\alpha}\beta} \right) \left( \frac{\overline{\beta} - \overline{\alpha}}{1 - \alpha \overline{\beta}} \right)$ Step 3: Expand the numerator and the denominator product: Numerator $= (\beta - \alpha)(\overline{\beta} - \overline{\alpha}) = \beta\overline{\beta} - \beta\overline{\alpha} - \alpha\overline{\beta} + \alpha\overline{\alpha}$ Since $\beta\overline{\beta} = |\beta|^2 = 1^2 = 1$, and $\alpha\overline{\alpha} = |\alpha|^2$, this becomes: $= 1 - \beta\overline{\alpha} - \alpha\overline{\beta} + |\alpha|^2$. Step 4: Expand the denominator product: Denominator $= (1 - \overline{\alpha}\beta)(1 - α\overline{\beta}) = 1 - \alpha\overline{\beta} - \overline{\alpha}\beta + \alpha\overline{\alpha}\beta\overline{\beta}$ Since $\beta\overline{\beta} = 1$, the last term is simply $\alpha\overline{\alpha} = |\alpha|^2$: $= 1 - \alpha\overline{\beta} - \overline{\alpha}\beta + |\alpha|^2$. Step 5: Notice that the numerator is exactly equal to the denominator! Thus: $E^2 = \frac{1 - \beta\overline{\alpha} - \alpha\overline{\beta} + |\alpha|^2}{1 - \alpha\overline{\beta} - \overline{\alpha}\beta + |\alpha|^2} = 1$. Taking the positive square root (since modulus is always non-negative), we get $E = 1$. Final answer: 1

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