NCERT Class 11 Mathematics — Complex Numbers and Quadratic Equations, Exercise 4.1. All 14 questions solved.

Everything in this exercise follows from a single definition,

$$i^2 = -1$$

and the rule that complex numbers add, subtract and multiply exactly like ordinary algebraic expressions in $i$, with $i^2$ replaced by $-1$ wherever it appears.

Two tools are used repeatedly:

Powers of $i$ repeat with period four.

$$i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1, \quad i^5 = i, \ldots$$

So $i^n$ depends only on the remainder of $n$ on division by $4$.

The multiplicative inverse of $z = a + ib$ (with $z \ne 0$) is

$$z^{-1} = \frac{\bar z}{|z|^2} = \frac{a – ib}{a^2 + b^2}$$

Key insight. Every division by a complex number is really a multiplication by its conjugate. $\dfrac{1}{a+ib}$ becomes real in the denominator the moment you multiply above and below by $a – ib$, because $(a+ib)(a-ib) = a^2 + b^2$ has no $i$ in it. Questions 11 to 14 are that one manoeuvre, four times.

Express each of the complex numbers given in questions 1 to 10 in the form $a + ib$.

Question 1

$$(5i)\left(-\frac{3}{5}i\right)$$

Solution. Multiply the numerical coefficients and the $i$ terms separately:

$$(5i)\left(-\frac35 i\right) = 5 \times \left(-\frac35\right) \times i^2 = -3 \times (-1) = 3$$

Written in the required form, the imaginary part is zero:

Question 2

$$i^9 + i^{19}$$

Solution. Reduce each exponent modulo $4$.

$$9 = 4(2) + 1 \quad\Longrightarrow\quad i^9 = (i^4)^2 \cdot i = 1 \cdot i = i$$

$$19 = 4(4) + 3 \quad\Longrightarrow\quad i^{19} = (i^4)^4 \cdot i^3 = -i$$

Adding:

$$i^9 + i^{19} = i + (-i) = 0$$

Question 3

$$i^{-39}$$

Solution. A negative exponent is a reciprocal:

$$i^{-39} = \frac{1}{i^{39}}$$

Now $39 = 4(9) + 3$, so $i^{39} = i^3 = -i$. Hence

$$i^{-39} = \frac{1}{-i}$$

Multiply above and below by $i$ to clear the imaginary denominator:

$$= \frac{1}{-i} \times \frac{i}{i} = \frac{i}{-i^2} = \frac{i}{1} = i$$

Question 4

$$3(7 + i7) + i(7 + i7)$$

Solution. Notice the common factor $(7 + i7)$ — factoring it out saves an expansion:

$$= (3 + i)(7 + i7) = 7(3 + i)(1 + i)$$

$$(3+i)(1+i) = 3 + 3i + i + i^2 = 3 + 4i – 1 = 2 + 4i$$

$$7(2 + 4i) = 14 + 28i$$

Question 5

$$(1 – i) – (-1 + i6)$$

Solution. Subtraction acts separately on the real and imaginary parts, but the minus sign must reach both terms of the second bracket:

$$= 1 – i + 1 – 6i = (1 + 1) + (-1 – 6)i = 2 – 7i$$

Question 6

$$\left(\frac{1}{5} + i\frac{2}{5}\right) – \left(4 + i\frac{5}{2}\right)$$

Solution. Real parts:

$$\frac15 – 4 = \frac{1 – 20}{5} = -\frac{19}{5}$$

Imaginary parts:

$$\frac25 – \frac52 = \frac{4 – 25}{10} = -\frac{21}{10}$$

Question 7

$$\left[\left(\frac{1}{3} + i\frac{7}{3}\right) + \left(4 + i\frac{1}{3}\right)\right] – \left(-\frac{4}{3} + i\right)$$

Solution. Work the inner bracket first.

Real parts: $\dfrac13 + 4 = \dfrac{13}{3}$. Imaginary parts: $\dfrac73 + \dfrac13 = \dfrac83$. So the bracket is $\dfrac{13}{3} + i\dfrac83$.

Now subtract $-\dfrac43 + i$:

$$\text{real: } \frac{13}{3} + \frac43 = \frac{17}{3}, \qquad \text{imaginary: } \frac83 – 1 = \frac53$$

Question 8

$$(1 – i)^4$$

Solution. Rather than expanding a fourth power directly, square twice.

$$(1 – i)^2 = 1 – 2i + i^2 = 1 – 2i – 1 = -2i$$

Then

$$(1-i)^4 = \left[(1-i)^2\right]^2 = (-2i)^2 = 4i^2 = -4$$

Question 9

$$\left(\frac{1}{3} + 3i\right)^3$$

Solution. Use $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ with $a = \frac13$, $b = 3i$.

$$a^3 = \frac{1}{27}$$ $$3a^2b = 3 \cdot \frac19 \cdot 3i = i$$ $$3ab^2 = 3 \cdot \frac13 \cdot 9i^2 = 9(-1) = -9$$ $$b^3 = 27i^3 = -27i$$

Collecting real and imaginary parts:

$$\text{real: } \frac{1}{27} – 9 = \frac{1 – 243}{27} = -\frac{242}{27}$$

$$\text{imaginary: } 1 – 27 = -26$$

Question 10

$$\left(-2 – \frac{1}{3}i\right)^3$$

Solution. Factor out the minus sign first — it makes the cube cleaner:

$$\left(-2 – \frac13 i\right)^3 = -\left(2 + \frac13 i\right)^3$$

Expanding $\left(2 + \frac13 i\right)^3$ with $a = 2$, $b = \frac13 i$:

$$a^3 = 8$$ $$3a^2b = 3(4)\left(\frac13 i\right) = 4i$$ $$3ab^2 = 3(2)\left(\frac19 i^2\right) = -\frac{2}{3}$$ $$b^3 = \frac{1}{27}i^3 = -\frac{1}{27}i$$

$$\text{real: } 8 – \frac23 = \frac{22}{3}, \qquad \text{imaginary: } 4 – \frac{1}{27} = \frac{107}{27}$$

So $\left(2 + \frac13 i\right)^3 = \frac{22}{3} + \frac{107}{27}i$, and negating:

Find the multiplicative inverse of each of the complex numbers given in questions 11 to 13.

Question 11

$$4 – 3i$$

Solution. With $z = 4 – 3i$, the conjugate is $\bar z = 4 + 3i$ and

$$|z|^2 = 4^2 + (-3)^2 = 16 + 9 = 25$$

$$z^{-1} = \frac{\bar z}{|z|^2} = \frac{4 + 3i}{25} = \frac{4}{25} + \frac{3}{25}i$$

Check: $(4-3i)\left(\frac{4+3i}{25}\right) = \frac{16 + 9}{25} = 1$ ✓

Question 12

$$\sqrt5 + 3i$$

Solution. Here $\bar z = \sqrt5 – 3i$ and

$$|z|^2 = (\sqrt5)^2 + 3^2 = 5 + 9 = 14$$

$$z^{-1} = \frac{\sqrt5 – 3i}{14} = \frac{\sqrt5}{14} – \frac{3}{14}i$$

The irrational real part causes no difficulty — $|z|^2$ is still an ordinary integer, because squaring removes the surd.

Question 13

$$-i$$

Solution. Here $z = 0 – i$, so $\bar z = i$ and $|z|^2 = 0^2 + (-1)^2 = 1$:

$$z^{-1} = \frac{i}{1} = i$$

Directly: $\dfrac{1}{-i} \times \dfrac{i}{i} = \dfrac{i}{-i^2} = i$, and the check $(-i)(i) = -i^2 = 1$ confirms it.

Question 14

Express the following expression in the form $a + ib$:

$$\frac{(3 + i\sqrt5)(3 – i\sqrt5)}{(\sqrt3 + \sqrt2 i) – (\sqrt3 – i\sqrt2)}$$

Solution. Both parts simplify dramatically before any division is needed.

Numerator. It is a product of conjugates, so it is $a^2 + b^2$:

$$(3 + i\sqrt5)(3 – i\sqrt5) = 3^2 – (i\sqrt5)^2 = 9 – (-5) = 14$$

Denominator. The real parts cancel and the imaginary parts add:

$$(\sqrt3 + \sqrt2 i) – (\sqrt3 – \sqrt2 i) = 2\sqrt2\,i$$

So the expression is

$$\frac{14}{2\sqrt2\,i} = \frac{7}{\sqrt2\,i}$$

Clear the $i$ from the denominator by multiplying above and below by $i$:

$$= \frac{7}{\sqrt2\, i} \times \frac{i}{i} = \frac{7i}{\sqrt2\,i^2} = \frac{7i}{-\sqrt2} = -\frac{7}{\sqrt2}i$$

Rationalising the surd:

$$= -\frac{7\sqrt2}{2}i$$

Common mistakes

• Writing $i^2 = 1$. It is $-1$, and every sign in the exercise depends on it. In question 1 the answer is $+3$ precisely because $i^2 = -1$ turns $-3$ into $+3$.
• Question 3, treating $i^{-39}$ as $-i^{39}$. A negative exponent is a reciprocal, not a negation.
• Questions 2 and 3, reducing the exponent wrongly. $19 = 4(4) + 3$, so $i^{19} = i^3 = -i$; taking the remainder as $1$ flips the sign of the answer.
• Question 5, not distributing the minus. $-(-1 + 6i)$ is $+1 – 6i$; changing only the first term gives $2 + 5i$ instead of $2 – 7i$.
• Question 8, expanding $(1-i)^4$ term by term. Squaring twice is quicker and far less error-prone, and $(1-i)^2 = -2i$ is worth remembering in its own right.
• Questions 9 and 10, forgetting $i^3 = -i$. The $b^3$ term carries $i^3$, not $i$, so its sign flips. This is where most of the arithmetic marks are lost.
• Question 10, cubing without factoring out the minus. $(-2 – \frac13 i)^3$ is the negative of $(2 + \frac13 i)^3$ because an odd power preserves the sign — but only if you take it out first rather than tracking eight signs.
• Questions 11 to 13, forgetting to square the modulus. The inverse is $\dfrac{\bar z}{|z|^2}$, and $|z|^2 = a^2 + b^2$ — not $\sqrt{a^2+b^2}$.
• Question 14, expanding the numerator in full. It is a conjugate pair, so it collapses to $9 + 5 = 14$ in one step. Multiplying it out gives the same answer with four times the opportunity to slip.

Practise next

• Miscellaneous Exercise on Chapter 4 — modulus, conjugate identities, and the standard-form reductions that build directly on questions 11 to 14 here.
• Exercise 3.2 — worth revising alongside the polar form that follows this exercise in the textbook, since the argument of a complex number is read off the same standard angles.