NCERT Solutions Class 11 Maths Chapter 4 Exercise 4.1

⚡ Complex Numbers & Quadratic Equations

Welcome to the world of imaginary numbers! This chapter introduces 'i' (iota), which allows us to solve equations that have no real solutions.

🧠 Cheat Sheet (Powers of i):

• The definition: \(i = \sqrt{-1}\)

• The cycle of powers:

- \(i^1 = i\)

- \(i^2 = -1\) (Most important!)

- \(i^3 = -i\)

- \(i^4 = 1\)

• Tip: For large powers like \(i^n\), divide \(n\) by 4. The remainder determines the value.

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🔥 Detailed Solutions

Question 1: Express in the form \(a + ib\): \((5i) (-\frac{3}{5}i)\).

Solution:

Multiply the real parts and imaginary parts.

\[ = 5 \times \left(-\frac{3}{5}\right) \times (i \times i) \]

\[ = -3 \times i^2 \]

Since \(i^2 = -1\):

\[ = -3(-1) = 3 \]

In \(a + ib\) form:

\[ \mathbf{3 + 0i} \]

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Question 2: Express in the form \(a + ib\): \(i^9 + i^{19}\).

Solution:

Reduce powers by dividing by 4.

- \(i^9 = i^{8+1} = (i^4)^2 \cdot i^1 = 1 \cdot i = i\)

- \(i^{19} = i^{16+3} = (i^4)^4 \cdot i^3 = 1 \cdot (-i) = -i\)

Now add them:

\[ i + (-i) = 0 \]

In \(a + ib\) form:

\[ \mathbf{0 + 0i} \]

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Question 3: Express in the form \(a + ib\): \(i^{-39}\).

Solution:

First, handle the negative exponent.

\[ i^{-39} = \frac{1}{i^{39}} \]

Divide 39 by 4: Remainder is 3 (\(39 = 4 \times 9 + 3\)).

\[ = \frac{1}{i^3} = \frac{1}{-i} \]

To remove \(i\) from the denominator, multiply and divide by \(i\) (Rationalize):

\[ = \frac{1}{-i} \times \frac{i}{i} = \frac{i}{-i^2} \]

Since \(-i^2 = -(-1) = 1\):

\[ = \frac{i}{1} = i \]

In \(a + ib\) form:

\[ \mathbf{0 + i} \]

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Question 4: Express in the form \(a + ib\): \(3(7 + i7) + i(7 + i7)\).

Solution:

Open the brackets:

\[ = 21 + 21i + 7i + 7i^2 \]

\[ = 21 + 28i + 7(-1) \]

\[ = 21 - 7 + 28i \]

\[ = \mathbf{14 + 28i} \]

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Question 5: Express in the form \(a + ib\): \((1 - i) - (-1 + i6)\).

Solution:

Open brackets and group like terms:

\[ = 1 - i + 1 - 6i \]

\[ = (1 + 1) + (-i - 6i) \]

\[ = \mathbf{2 - 7i} \]

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Question 6: Express in the form \(a + ib\): \((\frac{1}{5} + i\frac{2}{5}) - (4 + i\frac{5}{2})\).

Solution:

Group real parts and imaginary parts:

\[ = \left(\frac{1}{5} - 4\right) + i\left(\frac{2}{5} - \frac{5}{2}\right) \]

Take LCM for each part:

Real part:

\[ \frac{1 - 20}{5} = -\frac{19}{5} \]

Imaginary part:

\[ \frac{4 - 25}{10} = -\frac{21}{10} \]

Result:

\[ \mathbf{-\frac{19}{5} - i\frac{21}{10}} \]

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Question 7: Express in the form \(a + ib\): \([(\frac{1}{3} + i\frac{7}{3}) + (4 + i\frac{1}{3})] - (-\frac{4}{3} + i)\).

Solution:

First solve the terms inside the square bracket.

Real: \(\frac{1}{3} + 4 = \frac{13}{3}\)

Imaginary: \(\frac{7}{3} + \frac{1}{3} = \frac{8}{3}\)

Bracket result: \(\frac{13}{3} + i\frac{8}{3}\).

Now subtract the last term:

\[ (\frac{13}{3} + i\frac{8}{3}) - (-\frac{4}{3} + i) \]

\[ = (\frac{13}{3} + \frac{4}{3}) + i(\frac{8}{3} - 1) \]

\[ = \frac{17}{3} + i(\frac{5}{3}) \]

Result:

\[ \mathbf{\frac{17}{3} + i\frac{5}{3}} \]

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Question 8: Express in the form \(a + ib\): \((1 - i)^4\).

Solution:

We can write \((1-i)^4\) as \([(1-i)^2]^2\).

Step 1: Expand \((1-i)^2\):

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

\[ = 1 + (-1) - 2i = -2i \]

Step 2: Square the result:

\[ (-2i)^2 = 4i^2 \]

\[ = 4(-1) = -4 \]

In \(a + ib\) form:

\[ \mathbf{-4 + 0i} \]

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Question 9: Express in the form \(a + ib\): \((\frac{1}{3} + 3i)^3\).

Solution:

Use identity \((a+b)^3 = a^3 + b^3 + 3ab(a+b)\).

Here \(a = \frac{1}{3}\) and \(b = 3i\).

\[ = \left(\frac{1}{3}\right)^3 + (3i)^3 + 3\left(\frac{1}{3}\right)(3i)\left(\frac{1}{3} + 3i\right) \]

\[ = \frac{1}{27} + 27i^3 + 3i\left(\frac{1}{3} + 3i\right) \]

Remember \(i^3 = -i\):

\[ = \frac{1}{27} - 27i + i + 9i^2 \]

Substitute \(i^2 = -1\):

\[ = \frac{1}{27} - 26i - 9 \]

Group real parts:

\[ \left(\frac{1}{27} - 9\right) - 26i \]

\[ = \left(\frac{1 - 243}{27}\right) - 26i \]

\[ = \mathbf{-\frac{242}{27} - 26i} \]

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Question 10: Express in the form \(a + ib\): \((-2 - \frac{1}{3}i)^3\).

Solution:

Factor out -1 to make calculation easier: \([-1(2 + \frac{i}{3})]^3 = -(2 + \frac{i}{3})^3\).

\[ -(2^3 + (\frac{i}{3})^3 + 3(2)(\frac{i}{3})(2 + \frac{i}{3})) \]

\[ -(8 + \frac{i^3}{27} + 2i(2 + \frac{i}{3})) \]

\[ -(8 - \frac{i}{27} + 4i + \frac{2i^2}{3}) \]

\[ -(8 - \frac{2}{3} + i(4 - \frac{1}{27})) \]

\[ -(\frac{22}{3} + i\frac{107}{27}) \]

Result:

\[ \mathbf{-\frac{22}{3} - i\frac{107}{27}} \]

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Question 11: Find the multiplicative inverse of \(4 - 3i\).

Solution:

The multiplicative inverse of \(z\) is \(z^{-1} = \frac{1}{z}\).

Given \(z = 4 - 3i\).

\[ \frac{1}{4 - 3i} \]

Rationalize by multiplying numerator and denominator by the conjugate \((4 + 3i)\):

\[ = \frac{4 + 3i}{(4 - 3i)(4 + 3i)} \]

Denominator becomes \(a^2 + b^2\):

\[ = \frac{4 + 3i}{4^2 + 3^2} = \frac{4 + 3i}{16 + 9} = \frac{4 + 3i}{25} \]

Answer:

\[ \mathbf{\frac{4}{25} + \frac{3}{25}i} \]

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Question 12: Find the multiplicative inverse of \(\sqrt{5} + 3i\).

Solution:

Inverse \(z^{-1} = \frac{1}{\sqrt{5} + 3i}\). Rationalize.

\[ \frac{1}{\sqrt{5}+3i} \times \frac{\sqrt{5}-3i}{\sqrt{5}-3i} \]

\[ = \frac{\sqrt{5}-3i}{(\sqrt{5})^2 - (3i)^2} = \frac{\sqrt{5}-3i}{5 - 9(-1)} \]

\[ = \frac{\sqrt{5}-3i}{14} \]

Result:

\[ \mathbf{\frac{\sqrt{5}}{14} - i\frac{3}{14}} \]

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Question 13: Find the multiplicative inverse of \(-i\).

Solution:

Inverse \(z^{-1} = \frac{1}{-i}\). Rationalize.

\[ \frac{1}{-i} \times \frac{i}{i} = \frac{i}{-i^2} = \frac{i}{-(-1)} = \mathbf{i} \]

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Question 14: Express in the form \(a + ib\): \(\frac{(3 + i\sqrt{5})(3 - i\sqrt{5})}{(\sqrt{3} + \sqrt{2}i) - (\sqrt{3} - i\sqrt{2})}\).

Solution:

Numerator: It is in the form \((a+b)(a-b) = a^2 - b^2\).

\[ (3)^2 - (i\sqrt{5})^2 = 9 - 5i^2 = 9 - 5(-1) = 9 + 5 = 14 \]

Denominator: Open the brackets.

\[ \sqrt{3} + \sqrt{2}i - \sqrt{3} + \sqrt{2}i \]

\[ = 2\sqrt{2}i \]

Now divide Numerator by Denominator:

\[ \frac{14}{2\sqrt{2}i} = \frac{7}{\sqrt{2}i} \]

Rationalize (multiply/divide by \(\sqrt{2}i\)):

\[ = \frac{7\sqrt{2}i}{\sqrt{2}i \cdot \sqrt{2}i} = \frac{7\sqrt{2}i}{2i^2} \]

\[ = \frac{7\sqrt{2}i}{-2} \]

In \(a + ib\) form:

\[ \mathbf{0 - i\frac{7\sqrt{2}}{2}} \]

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🎓 Expert Tips for Complex Numbers:

🛑 iota in Denominator:

Never leave 'i' in the denominator. Always rationalize by multiplying with the conjugate.

Example: \(\frac{1}{a+ib} \times \frac{a-ib}{a-ib}\).

🛑 The "Sum of Squares" Trick:

Remember that \((a+ib)(a-ib) = a^2 + b^2\).

There is NO negative sign in the result!

Example: \((2+3i)(2-3i) = 4 + 9 = 13\).