NCERT Class 11 Mathematics — Complex Numbers and Quadratic Equations, Miscellaneous Exercise on Chapter 4. All 14 questions solved.

The computational questions here are Exercise 4.1’s technique applied to messier expressions. The proofs all rest on one family of facts about the modulus and the conjugate:

$$z\bar z = |z|^2, \qquad |z_1 z_2| = |z_1||z_2|, \qquad \left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}$$

$$\overline{z_1 z_2} = \bar z_1 \bar z_2, \qquad \overline{z_1 + z_2} = \bar z_1 + \bar z_2$$

If $z = x + iy$ then $\overline{z} = x – iy$, and $x^2 + y^2 = |z|^2 = z\bar z$.

Key insight. When a question asks you to prove something about $a^2 + b^2$ or $x^2 + y^2$, it is asking about a modulus squared in disguise. Questions 4, 6 and 13 all become one line once you notice that: take the modulus of both sides and square. Trying to expand the real and imaginary parts by hand is doing the same calculation the hard way.

Question 1

Evaluate: $\left[i^{18} + \left(\dfrac{1}{i}\right)^{25}\right]^3$

Solution. Reduce each power separately.

$18 = 4(4) + 2$, so $i^{18} = i^2 = -1$.

For the second term, $\dfrac{1}{i} = \dfrac{i}{i^2} = -i$, so

$$\left(\frac1i\right)^{25} = (-i)^{25} = (-1)^{25}i^{25} = -i^{25}$$

and $25 = 4(6) + 1$, so $i^{25} = i$. Hence $\left(\frac1i\right)^{25} = -i$.

Adding inside the bracket:

$$i^{18} + \left(\frac1i\right)^{25} = -1 – i$$

Now cube it. First square:

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

Then

$$(-1-i)^3 = (-1-i)^2(-1-i) = 2i(-1-i) = -2i – 2i^2 = 2 – 2i$$

Question 2

For any two complex numbers $z_1$ and $z_2$, prove that

$$\operatorname{Re}(z_1 z_2) = \operatorname{Re}z_1 \operatorname{Re}z_2 – \operatorname{Im}z_1 \operatorname{Im}z_2$$

Solution. Write the two numbers in standard form:

$$z_1 = x_1 + iy_1, \qquad z_2 = x_2 + iy_2$$

so that $\operatorname{Re}z_1 = x_1$, $\operatorname{Im}z_1 = y_1$, and similarly for $z_2$. Multiplying:

$$z_1 z_2 = (x_1 + iy_1)(x_2 + iy_2) = x_1x_2 + i x_1y_2 + i y_1x_2 + i^2 y_1y_2$$

Using $i^2 = -1$ and grouping:

$$z_1 z_2 = (x_1x_2 – y_1y_2) + i(x_1y_2 + y_1x_2)$$

The real part is the first bracket:

$$\operatorname{Re}(z_1z_2) = x_1x_2 – y_1y_2 = \operatorname{Re}z_1\operatorname{Re}z_2 – \operatorname{Im}z_1\operatorname{Im}z_2 \qquad \blacksquare$$

The minus sign is the whole content of the result — it is exactly the $i^2 = -1$ appearing in the product.

Question 3

Reduce $\left(\dfrac{1}{1 – 4i} – \dfrac{2}{1 + i}\right)\left(\dfrac{3 – 4i}{5 + i}\right)$ to the standard form.

Solution. Handle the bracket first, over a common denominator:

$$\frac{1}{1-4i} – \frac{2}{1+i} = \frac{(1+i) – 2(1-4i)}{(1-4i)(1+i)}$$

Numerator: $1 + i – 2 + 8i = -1 + 9i$.

Denominator: $(1-4i)(1+i) = 1 + i – 4i – 4i^2 = 1 – 3i + 4 = 5 – 3i$.

So the bracket is $\dfrac{-1+9i}{5-3i}$, and the whole expression becomes

$$\frac{(-1+9i)(3-4i)}{(5-3i)(5+i)}$$

Numerator: $-3 + 4i + 27i – 36i^2 = -3 + 31i + 36 = 33 + 31i$.

Denominator: $25 + 5i – 15i – 3i^2 = 25 – 10i + 3 = 28 – 10i$.

Now multiply above and below by the conjugate $28 + 10i$:

$$\frac{(33+31i)(28+10i)}{28^2 + 10^2} = \frac{924 + 330i + 868i + 310i^2}{884} = \frac{614 + 1198i}{884}$$

Dividing numerator and denominator by $2$:

$$= \frac{307 + 599i}{442} = \frac{307}{442} + \frac{599}{442}i$$

Question 4

If $x – iy = \sqrt{\dfrac{a – ib}{c – id}}$, prove that $(x^2 + y^2)^2 = \dfrac{a^2 + b^2}{c^2 + d^2}$.

Solution. Take the modulus of both sides. Since $|x – iy| = \sqrt{x^2+y^2}$,

$$\sqrt{x^2 + y^2} = \left|\sqrt{\frac{a-ib}{c-id}}\right| = \sqrt{\left|\frac{a-ib}{c-id}\right|} = \sqrt{\frac{|a-ib|}{|c-id|}} = \sqrt{\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}}}$$

Squaring both sides:

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

Squaring once more:

$$(x^2 + y^2)^2 = \frac{a^2 + b^2}{c^2 + d^2} \qquad \blacksquare$$

The alternative route — squaring the original equation to get $(x-iy)^2 = \frac{a-ib}{c-id}$, cross-multiplying and comparing real and imaginary parts — works, but takes a page. Modulus is the right tool because the target involves only $x^2 + y^2$, $a^2+b^2$ and $c^2+d^2$.

Question 5

If $z_1 = 2 – i$, $z_2 = 1 + i$, find $\left|\dfrac{z_1 + z_2 + 1}{z_1 – z_2 + 1}\right|$.

Solution. Compute the numerator and denominator separately.

$$z_1 + z_2 + 1 = (2 – i) + (1 + i) + 1 = 4 + 0i = 4$$

$$z_1 – z_2 + 1 = (2 – i) – (1 + i) + 1 = 2 – 2i$$

So

$$\left|\frac{4}{2 – 2i}\right| = \frac{|4|}{|2-2i|} = \frac{4}{\sqrt{4 + 4}} = \frac{4}{2\sqrt2} = \frac{2}{\sqrt2} = \sqrt2$$

Using $\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}$ avoids rationalising the fraction at all.

Question 6

If $a + ib = \dfrac{(x+i)^2}{2x^2 + 1}$, prove that $a^2 + b^2 = \dfrac{(x^2+1)^2}{(2x^2+1)^2}$.

Solution. Note $a^2 + b^2 = |a + ib|^2$, so take the modulus of both sides and square. The denominator $2x^2 + 1$ is real and positive, so it comes out of the modulus unchanged:

$$|a + ib| = \frac{|x + i|^2}{2x^2 + 1} = \frac{\left(\sqrt{x^2 + 1}\right)^2}{2x^2+1} = \frac{x^2+1}{2x^2+1}$$

Squaring:

$$a^2 + b^2 = \frac{(x^2+1)^2}{(2x^2+1)^2} \qquad \blacksquare$$

(If you prefer, expand: $(x+i)^2 = x^2 – 1 + 2xi$, so $a = \frac{x^2-1}{2x^2+1}$ and $b = \frac{2x}{2x^2+1}$, and $(x^2-1)^2 + 4x^2 = x^4 + 2x^2 + 1 = (x^2+1)^2$ — the same answer, three lines longer.)

Question 7

Let $z_1 = 2 – i$, $z_2 = -2 + i$. Find

(i) $\operatorname{Re}\left(\dfrac{z_1 z_2}{\bar z_1}\right)$    (ii) $\operatorname{Im}\left(\dfrac{1}{z_1 \bar z_1}\right)$

Solution.

(i) First the product:

$$z_1 z_2 = (2-i)(-2+i) = -4 + 2i + 2i – i^2 = -4 + 4i + 1 = -3 + 4i$$

And $\bar z_1 = 2 + i$. So

$$\frac{z_1z_2}{\bar z_1} = \frac{-3 + 4i}{2 + i}$$

Multiply above and below by $2 – i$:

$$= \frac{(-3+4i)(2-i)}{4 + 1} = \frac{-6 + 3i + 8i – 4i^2}{5} = \frac{-6 + 11i + 4}{5} = \frac{-2 + 11i}{5}$$

$$\operatorname{Re}\left(\frac{z_1z_2}{\bar z_1}\right) = -\frac{2}{5}$$

(ii) Here $z_1\bar z_1 = |z_1|^2$, which is always real:

$$z_1 \bar z_1 = (2-i)(2+i) = 4 + 1 = 5$$

So $\dfrac{1}{z_1\bar z_1} = \dfrac15$, a real number, and its imaginary part is

$$\operatorname{Im}\left(\frac{1}{z_1\bar z_1}\right) = 0$$

Recognising $z\bar z = |z|^2$ turns part (ii) into no work at all.

Question 8

Find the real numbers $x$ and $y$ if $(x – iy)(3 + 5i)$ is the conjugate of $-6 – 24i$.

Solution. The conjugate of $-6 – 24i$ is $-6 + 24i$, so the condition is

$$(x – iy)(3 + 5i) = -6 + 24i$$

Expand the left side:

$$3x + 5xi – 3yi – 5yi^2 = (3x + 5y) + i(5x – 3y)$$

Two complex numbers are equal when their real and imaginary parts agree separately:

$$3x + 5y = -6 \qquad \text{(1)}$$ $$5x – 3y = 24 \qquad \text{(2)}$$

Multiply (1) by $3$ and (2) by $5$, then add to eliminate $y$:

$$9x + 15y = -18, \qquad 25x – 15y = 120$$

$$34x = 102 \quad\Longrightarrow\quad x = 3$$

Substituting into (1): $9 + 5y = -6$, so $5y = -15$ and $y = -3$.

Check: $(3 + 3i)(3+5i) = 9 + 15i + 9i + 15i^2 = 9 + 24i – 15 = -6 + 24i$ ✓

Question 9

Find the modulus of $\dfrac{1+i}{1-i} – \dfrac{1-i}{1+i}$.

Solution. Simplify each fraction first. Multiplying the first above and below by $1 + i$:

$$\frac{1+i}{1-i} = \frac{(1+i)^2}{(1-i)(1+i)} = \frac{1 + 2i – 1}{1 + 1} = \frac{2i}{2} = i$$

Similarly

$$\frac{1-i}{1+i} = \frac{(1-i)^2}{2} = \frac{1 – 2i – 1}{2} = -i$$

The two are conjugates, so subtracting doubles the imaginary part:

$$i – (-i) = 2i$$

$$|2i| = 2$$

Question 10

If $(x + iy)^3 = u + iv$, then show that $\dfrac{u}{x} + \dfrac{v}{y} = 4(x^2 – y^2)$.

Solution. Expand the cube using $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ with $a = x$, $b = iy$:

$$(x+iy)^3 = x^3 + 3x^2(iy) + 3x(iy)^2 + (iy)^3 = x^3 + 3x^2yi – 3xy^2 – y^3 i$$

Grouping real and imaginary parts:

$$u + iv = (x^3 – 3xy^2) + i(3x^2y – y^3)$$

so

$$u = x^3 – 3xy^2 = x(x^2 – 3y^2), \qquad v = 3x^2y – y^3 = y(3x^2 – y^2)$$

The factorisation is the point — dividing by $x$ and $y$ now costs nothing:

$$\frac{u}{x} + \frac{v}{y} = (x^2 – 3y^2) + (3x^2 – y^2) = 4x^2 – 4y^2 = 4(x^2 – y^2) \qquad \blacksquare$$

Question 11

If $\alpha$ and $\beta$ are different complex numbers with $|\beta| = 1$, then find $\left|\dfrac{\beta – \alpha}{1 – \bar\alpha\beta}\right|$.

Solution. The hypothesis $|\beta| = 1$ is equivalent to $\beta\bar\beta = 1$, and that is the substitution that makes the whole thing collapse. Replace the $1$ in the denominator:

$$1 – \bar\alpha\beta = \beta\bar\beta – \bar\alpha\beta = \beta(\bar\beta – \bar\alpha) = \beta\,\overline{(\beta – \alpha)}$$

using $\bar\beta – \bar\alpha = \overline{\beta – \alpha}$.

Now take moduli, remembering that a number and its conjugate have the same modulus:

$$|1 – \bar\alpha\beta| = |\beta|\,\left|\overline{\beta – \alpha}\right| = 1 \cdot |\beta – \alpha| = |\beta – \alpha|$$

Therefore

$$\left|\frac{\beta – \alpha}{1 – \bar\alpha\beta}\right| = \frac{|\beta – \alpha|}{|\beta – \alpha|} = 1$$

(The condition $\alpha \ne \beta$ is what keeps the denominator non-zero.)

Question 12

Find the number of non-zero integral solutions of the equation $|1 – i|^x = 2^x$.

Solution. Compute the modulus first:

$$|1 – i| = \sqrt{1^2 + (-1)^2} = \sqrt2$$

So the equation is

$$(\sqrt2)^x = 2^x \quad\Longrightarrow\quad 2^{x/2} = 2^x$$

Since $2^p = 2^q$ forces $p = q$ for real exponents:

$$\frac{x}{2} = x \quad\Longrightarrow\quad x = 2x \quad\Longrightarrow\quad x = 0$$

The only solution is $x = 0$, which the question explicitly excludes. Hence there are no non-zero integral solutions.

Question 13

If $(a + ib)(c + id)(e + if)(g + ih) = \mathrm{A} + i\mathrm{B}$, then show that

$$(a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2) = \mathrm{A}^2 + \mathrm{B}^2$$

Solution. Take the modulus of both sides of the given equation. The modulus of a product is the product of the moduli:

$$|a+ib|\,|c+id|\,|e+if|\,|g+ih| = |\mathrm A + i\mathrm B|$$

$$\sqrt{a^2+b^2}\,\sqrt{c^2+d^2}\,\sqrt{e^2+f^2}\,\sqrt{g^2+h^2} = \sqrt{\mathrm A^2 + \mathrm B^2}$$

Squaring both sides removes every radical at once:

$$(a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2) = \mathrm A^2 + \mathrm B^2 \qquad \blacksquare$$

Multiplying out four complex factors by hand would take a page and prove nothing extra; the modulus property does the entire job.

Question 14

If $\left(\dfrac{1+i}{1-i}\right)^m = 1$, then find the least positive integral value of $m$.

Solution. From question 9 we already know

$$\frac{1+i}{1-i} = i$$

So the equation becomes

$$i^m = 1$$

Powers of $i$ cycle with period $4$: $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$. The first positive exponent giving $1$ is

$$m = 4$$

Common mistakes

• Question 1, taking $\frac1i = i$. It is $-i$, because $\frac1i = \frac{i}{i^2} = \frac{i}{-1}$. Getting this backwards changes the whole bracket from $-1-i$ to $-1+i$.
• Question 3, dividing before combining. The bracket must be reduced to a single fraction first; distributing the second factor over both terms doubles the work.
• Questions 4, 6 and 13, expanding instead of taking moduli. Any target written purely in terms of $x^2 + y^2$, $a^2 + b^2$ and so on is a modulus statement. The proof is two lines if you see that and a page if you do not.
• Question 4, forgetting there are two squarings. One removes the square root from the definition, one raises $x^2+y^2$ to $(x^2+y^2)^2$. Stopping after the first gives $x^2 + y^2 = \sqrt{\frac{a^2+b^2}{c^2+d^2}}$, which is true but is not what was asked.
• Question 7(ii), computing $\frac{1}{z_1\bar z_1}$ the long way. $z\bar z$ is always real and equal to $|z|^2$, so the imaginary part is $0$ without any arithmetic.
• Question 8, forgetting to conjugate. The equation is $(x – iy)(3+5i) = -6 + 24i$, not $-6 – 24i$. Using the un-conjugated value gives $x = -3$, $y = 3$ — plausible-looking and wrong.
• Question 11, trying to write $\alpha$ and $\beta$ in $x + iy$ form. The substitution $1 = \beta\bar\beta$ is the entire proof; component form leads nowhere.
• Question 12, answering “$1$ solution”. $x = 0$ is the only solution and the question asks for non-zero ones, so the answer is $0$.
• Question 14, giving $m = 2$. $i^2 = -1$, not $1$. The cycle closes at $4$.

Practise next

• Exercise 5.1 — the next chapter, Linear Inequalities, which returns to real numbers and interval notation.
• Exercise 4.1 — worth working again alongside questions 1, 3 and 9, since every one of them is a conjugate-multiplication at heart.