ncert-solutions-class-11-maths-chapter-4-exercise-4 miscellaneous

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NCERT Solutions Class 11 Maths Chapter 4 Miscellaneous Exercise

🚀 Complex Numbers: The Final Challenge

This exercise tests your ability to mix concepts like conjugate, modulus, and algebraic operations. These 14 questions are critical for your final exams!

🧠 Cheat Sheet (Key Properties):

• Modulus:

- \(|z_1 z_2| = |z_1| |z_2|\)

- \(|\frac{z_1}{z_2}| = \frac{|z_1|}{|z_2|}\)

• Conjugate:

- \(\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}\)

- \(z \overline{z} = |z|^2\)

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🔥 Detailed Solutions (Q1 - Q14)

Question 1: Evaluate \([i^{18} + (\frac{1}{i})^{25}]^3\).

Solution:

Step 1: Simplify \(i^{18}\).

\(i^{18} = (i^4)^4 \cdot i^2 = 1 \cdot (-1) = -1\).

Step 2: Simplify \((\frac{1}{i})^{25}\).

\(\frac{1}{i} = -i\).

So, \((-i)^{25} = -(i^{25}) = -(i^{24} \cdot i) = -(1 \cdot i) = -i\).

Step 3: Substitute and cube.

\[ [-1 - i]^3 = -(1 + i)^3 \]

Expand using \((a+b)^3 = a^3 + b^3 + 3ab(a+b)\):

\[ -(1 + i^3 + 3i(1+i)) \]

\[ -(1 - i + 3i + 3i^2) \]

\[ -(1 + 2i - 3) \] (Since \(i^2 = -1\))

\[ -(-2 + 2i) = \mathbf{2 - 2i} \]

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Question 2: For any two complex numbers \(z_1\) and \(z_2\), prove that \(Re(z_1 z_2) = Re z_1 Re z_2 - Im z_1 Im z_2\).

Solution:

Let \(z_1 = x_1 + iy_1\) and \(z_2 = x_2 + iy_2\).

Multiply them:

\[ z_1 z_2 = (x_1 + iy_1)(x_2 + iy_2) \]

\[ = x_1 x_2 + i x_1 y_2 + i y_1 x_2 + i^2 y_1 y_2 \]

\[ = (x_1 x_2 - y_1 y_2) + i(x_1 y_2 + y_1 x_2) \]

The Real part is:

\[ Re(z_1 z_2) = x_1 x_2 - y_1 y_2 \]

Substituting values:

\[ Re(z_1 z_2) = Re z_1 Re z_2 - Im z_1 Im z_2 \]

Proved. ✅

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Question 3: Reduce \((\frac{1}{1-4i} - \frac{2}{1+i})(\frac{3-4i}{5+i})\) to standard form.

Solution:

Step 1: Simplify the first bracket.

\[ \frac{1(1+i) - 2(1-4i)}{(1-4i)(1+i)} = \frac{1+i - 2 + 8i}{1 + i - 4i - 4i^2} \]

\[ = \frac{-1 + 9i}{1 - 3i + 4} = \frac{-1 + 9i}{5 - 3i} \]

Step 2: Multiply with the second part.

\[ \left(\frac{-1 + 9i}{5 - 3i}\right) \left(\frac{3 - 4i}{5 + i}\right) \]

Numerator: \((-1+9i)(3-4i) = -3 + 4i + 27i - 36i^2 = 33 + 31i\)

Denominator: \((5-3i)(5+i) = 25 + 5i - 15i - 3i^2 = 28 - 10i\)

Step 3: Rationalize \(\frac{33 + 31i}{28 - 10i}\).

Multiply numerator and denominator by \(28+10i\).

\[ \frac{(33+31i)(28+10i)}{28^2 + 10^2} \]

\[ = \frac{924 + 330i + 868i + 310i^2}{784 + 100} \]

\[ = \frac{614 + 1198i}{884} \]

Divide by 2:

\[ \mathbf{\frac{307}{442} + i \frac{599}{442}} \]

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Question 4: If \(x - iy = \sqrt{\frac{a-ib}{c-id}}\), prove that \((x^2 + y^2)^2 = \frac{a^2 + b^2}{c^2 + d^2}\).

Solution:

Given: \(x - iy = \sqrt{\frac{a-ib}{c-id}}\).

Take conjugate on both sides:

\[ x + iy = \sqrt{\frac{a+ib}{c+id}} \]

Multiply both equations:

\[ (x-iy)(x+iy) = \sqrt{\frac{a-ib}{c-id}} \cdot \sqrt{\frac{a+ib}{c+id}} \]

\[ x^2 + y^2 = \sqrt{\frac{a^2+b^2}{c^2+d^2}} \]

Square both sides:

\[ (x^2 + y^2)^2 = \frac{a^2 + b^2}{c^2 + d^2} \]

Proved. ✅

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Question 5: If \(z_1 = 2 - i, z_2 = 1 + i\), find \(|\frac{z_1 + z_2 + 1}{z_1 - z_2 + 1}|\).

Solution:

Numerator: \(z_1 + z_2 + 1 = (2-i) + (1+i) + 1 = 4\).

Denominator: \(z_1 - z_2 + 1 = (2-i) - (1+i) + 1 = 2 - 2i\).

Expression becomes:

\[ |\frac{4}{2 - 2i}| = |\frac{2}{1 - i}| \]

Using property \(|\frac{z_1}{z_2}| = \frac{|z_1|}{|z_2|}\):

\[ \frac{|2|}{|1 - i|} = \frac{2}{\sqrt{1^2 + (-1)^2}} = \frac{2}{\sqrt{2}} = \mathbf{\sqrt{2}} \]

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Question 6: If \(a + ib = \frac{(x+i)^2}{2x^2+1}\), prove that \(a^2 + b^2 = \frac{(x^2+1)^2}{(2x^2+1)^2}\).

Solution:

Take modulus on both sides:

\[ |a + ib| = |\frac{(x+i)^2}{2x^2+1}| \]

\[ \sqrt{a^2+b^2} = \frac{|x+i|^2}{|2x^2+1|} \]

Since \(|x+i| = \sqrt{x^2+1}\) and denominator is real:

\[ \sqrt{a^2+b^2} = \frac{x^2+1}{2x^2+1} \]

Squaring both sides:

\[ a^2 + b^2 = \frac{(x^2+1)^2}{(2x^2+1)^2} \]

Proved. ✅

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Question 7: Let \(z_1 = 2 - i, z_2 = -2 + i\). Find (i) \(Re(\frac{z_1 z_2}{\overline{z_1}})\) (ii) \(Im(\frac{1}{z_1 \overline{z_1}})\).

Solution:

(i) \(z_1 z_2 = (2-i)(-2+i) = -4 + 2i + 2i - i^2 = -3 + 4i\).

\(\overline{z_1} = 2 + i\).

Now divide: \(\frac{-3+4i}{2+i} \times \frac{2-i}{2-i} = \frac{-6 + 3i + 8i - 4i^2}{4+1} = \frac{-2 + 11i}{5}\).

\(Re = \mathbf{-\frac{2}{5}}\).

(ii) \(z_1 \overline{z_1} = |z_1|^2 = 2^2 + (-1)^2 = 5\).

Expression is \(\frac{1}{5}\). This is purely real.

\(Im = \mathbf{0}\).

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Question 8: Find the real numbers \(x\) and \(y\) if \((x - iy)(3 + 5i)\) is the conjugate of \(-6 - 24i\).

Solution:

Conjugate of \(-6 - 24i\) is \(-6 + 24i\).

Equation:

\[ (x - iy)(3 + 5i) = -6 + 24i \]

Expand LHS:

\[ 3x + 5xi - 3yi - 5yi^2 = (3x + 5y) + i(5x - 3y) \]

Compare Real and Imaginary parts:

1. \(3x + 5y = -6\)

2. \(5x - 3y = 24\)

Solving these simultaneous equations:

Multiply (1) by 3 and (2) by 5:

\(9x + 15y = -18\)

\(25x - 15y = 120\)

Add them: \(34x = 102 \implies \mathbf{x = 3}\).

Substitute \(x=3\) in (1): \(9 + 5y = -6 \implies 5y = -15 \implies \mathbf{y = -3}\).

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Question 9: Find the modulus of \(\frac{1+i}{1-i} - \frac{1-i}{1+i}\).

Solution:

Simplify first:

\[ \frac{(1+i)^2 - (1-i)^2}{(1-i)(1+i)} \]

Numerator: \((1 + 2i - 1) - (1 - 2i - 1) = 2i - (-2i) = 4i\).

Denominator: \(1^2 - i^2 = 1 - (-1) = 2\).

Fraction reduces to: \(\frac{4i}{2} = 2i\).

Modulus \(|2i| = \sqrt{0^2 + 2^2} = \mathbf{2}\).

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Question 10: If \((x+iy)^3 = u+iv\), then show that \(\frac{u}{x} + \frac{v}{y} = 4(x^2 - y^2)\).

Solution:

Expand \((x+iy)^3 = x^3 + 3x^2(iy) + 3x(iy)^2 + (iy)^3\).

\[ u + iv = (x^3 - 3xy^2) + i(3x^2y - y^3) \]

Comparing parts:

\(u = x(x^2 - 3y^2) \implies \frac{u}{x} = x^2 - 3y^2\).

\(v = y(3x^2 - y^2) \implies \frac{v}{y} = 3x^2 - y^2\).

Adding them:

\[ (x^2 - 3y^2) + (3x^2 - y^2) = 4x^2 - 4y^2 = 4(x^2 - y^2) \]

Proved. ✅

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Question 11: If \(\alpha\) and \(\beta\) are different complex numbers with \(|\beta| = 1\), then find \(|\frac{\beta - \alpha}{1 - \overline{\alpha}\beta}|\).

Solution:

Using property \(|z|^2 = z \overline{z}\).

Let \(z = \frac{\beta - \alpha}{1 - \overline{\alpha}\beta}\).

Calculate \(|z|^2 = (\frac{\beta - \alpha}{1 - \overline{\alpha}\beta}) (\frac{\overline{\beta} - \overline{\alpha}}{1 - \alpha \overline{\beta}})\).

Expand numerator: \(\beta \overline{\beta} - \beta \overline{\alpha} - \alpha \overline{\beta} + \alpha \overline{\alpha}\).

Since \(|\beta|=1 \implies \beta \overline{\beta} = 1\).

Num = \(1 - \beta \overline{\alpha} - \alpha \overline{\beta} + |\alpha|^2\).

Expand denominator: \(1 - \alpha \overline{\beta} - \overline{\alpha} \beta + \alpha \overline{\alpha} \beta \overline{\beta}\).

Since \(\beta \overline{\beta} = 1\):

Denom = \(1 - \alpha \overline{\beta} - \overline{\alpha} \beta + |\alpha|^2\).

Since Numerator = Denominator, the ratio is 1.

So, \(|z|^2 = 1 \implies |z| = \mathbf{1}\).

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Question 12: Find the number of non-zero integral solutions of the equation \(|1 - i|^x = 2^x\).

Solution:

Modulus \(|1 - i| = \sqrt{1^2 + (-1)^2} = \sqrt{2}\).

Equation becomes:

\[ (\sqrt{2})^x = 2^x \]

\[ (2^{1/2})^x = 2^x \]

\[ 2^{x/2} = 2^x \]

Equating powers: \(\frac{x}{2} = x \implies x = 2x \implies x = 0\).

But we need non-zero integral solutions.

Since only \(x=0\) works, there are 0 non-zero solutions.

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Question 13: If \((a + ib)(c + id)(e + if)(g + ih) = A + iB\), then show that \((a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2) = A^2 + B^2\).

Solution:

Take modulus on both sides.

\[ |(a + ib)(c + id)(e + if)(g + ih)| = |A + iB| \]

Using property \(|z_1 z_2...| = |z_1| |z_2|...\):

\[ |a+ib| |c+id| |e+if| |g+ih| = |A+iB| \]

\[ \sqrt{a^2+b^2} \sqrt{c^2+d^2} \sqrt{e^2+f^2} \sqrt{g^2+h^2} = \sqrt{A^2+B^2} \]

Squaring both sides removes all square roots.

Proved. ✅

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Question 14: If \((\frac{1+i}{1-i})^m = 1\), then find the least positive integral value of \(m\).