NCERT Exemplar Solutions Class 11 Maths Complex Numbers and Quadratic Equations — Short and Long Answer Type Q1–24

NCERT Exemplar Problems, Class 11 Mathematics — Complex Numbers and Quadratic Equations, Short and Long Answer questions 1 to 24. All 24 questions solved.

These questions run on three tools: powers of $i$ (which repeat with period 4), the identity $z\bar z = |z|^2$ together with “multiply top and bottom by the conjugate”, and writing $z = x + iy$ to turn a condition on $z$ into an equation in the real numbers $x$ and $y$. Questions 7 to 10 and 13 to 24 are mostly the third kind, and the loci in them (circles, lines) are exactly what the same equation gives in coordinate geometry.

One printed answer is wrong: Question 23. It is explained where it occurs. Also, argument, polar form and Argand-plane loci (Questions 10, 13, 16, 18, 20, 23, 24) are no longer in the current NCERT Class 11 chapter, but they remain good practice and are solved here.

Key insight. Never fight a complex condition in the abstract. Put $z = x + iy$, split the equation into its real part and its imaginary part, and you have ordinary simultaneous equations in $x$ and $y$. When a modulus is involved, remember it is a real number, so the imaginary part of the other side must vanish first.

Question 1

For a positive integer $n$, find the value of $(1 – i)^n\left(1 – \dfrac1i\right)^n$.

Solution. Since $\dfrac1i = -i$, we get $1 – \dfrac1i = 1 + i$. Both factors carry the same power $n$, so combine them before expanding:

$$(1-i)^n(1+i)^n = \big[(1-i)(1+i)\big]^n = (1 – i^2)^n = 2^n.$$

Question 2

Evaluate $\displaystyle\sum_{n=1}^{13}\left(i^n + i^{n+1}\right)$, where $n \in \mathbf N$.

Solution. Any four consecutive powers of $i$ add to $0$, because $i + i^2 + i^3 + i^4 = i – 1 – i + 1 = 0$. So only the leftover terms matter.

$\sum_{n=1}^{13} i^n = i + i^2 + \dots + i^{13}$: the first twelve terms cancel in threes of four, leaving $i^{13} = i$.

$\sum_{n=1}^{13} i^{n+1} = i^2 + i^3 + \dots + i^{14}$: the first twelve terms ($i^2$ to $i^{13}$) cancel, leaving $i^{14} = i^2 = -1$.

The total is $i + (-1)$.

Question 3

If $\left(\dfrac{1+i}{1-i}\right)^3 – \left(\dfrac{1-i}{1+i}\right)^3 = x + iy$, then find $(x, y)$.

Solution. $\dfrac{1+i}{1-i} = \dfrac{(1+i)^2}{2} = \dfrac{2i}{2} = i$, and its reciprocal $\dfrac{1-i}{1+i} = -i$. Then

$$i^3 – (-i)^3 = -i – i = -2i.$$

So $x = 0$, $y = -2$.

Question 4

If $\dfrac{(1+i)^2}{2-i} = x + iy$, then find the value of $x + y$.

Solution. $(1+i)^2 = 2i$. To clear the complex denominator, multiply top and bottom by the conjugate $2 + i$:

$$\frac{2i(2+i)}{(2-i)(2+i)} = \frac{4i + 2i^2}{5} = \frac{-2 + 4i}{5}.$$

So $x = -\dfrac25$, $y = \dfrac45$, and $x + y = \dfrac25$.

Question 5

If $\left(\dfrac{1-i}{1+i}\right)^{100} = a + ib$, then find $(a, b)$.

Solution. As in Question 3, $\dfrac{1-i}{1+i} = -i$. Then $(-i)^{100} = \big((-i)^4\big)^{25} = 1$, because $(-i)^4 = 1$. So $a + ib = 1 + 0i$.

Question 6

If $a = \cos\theta + i\sin\theta$, find the value of $\dfrac{1+a}{1-a}$.

Solution. Use the half-angle forms $1 + \cos\theta = 2\cos^2\frac\theta2$, $1 – \cos\theta = 2\sin^2\frac\theta2$, $\sin\theta = 2\sin\frac\theta2\cos\frac\theta2$:

$$\frac{1+a}{1-a} = \frac{2\cos\frac\theta2\left(\cos\frac\theta2 + i\sin\frac\theta2\right)}{2\sin\frac\theta2\left(\sin\frac\theta2 – i\cos\frac\theta2\right)}.$$

Now $\sin\frac\theta2 – i\cos\frac\theta2 = -i\left(\cos\frac\theta2 + i\sin\frac\theta2\right)$, so the bracketed factors cancel and

$$\frac{1+a}{1-a} = \frac{\cos\frac\theta2}{-i\sin\frac\theta2} = i\cot\frac\theta2.$$

Question 7

If $(1+i)z = (1-i)\bar z$, then show that $z = -i\bar z$.

Solution. Divide both sides by $1 + i$ and then simplify the ratio, which we already know from Question 3:

$$z = \frac{1-i}{1+i}\,\bar z = (-i)\,\bar z.$$

Question 8

If $z = x + iy$, then show that $z\bar z + 2(z + \bar z) + b = 0$, where $b \in \mathbf R$, represents a circle.

Solution. Here $z\bar z = x^2 + y^2$ and $z + \bar z = 2x$, so the equation becomes

$$x^2 + y^2 + 4x + b = 0 \implies (x+2)^2 + y^2 = 4 – b.$$

This is a circle with centre $(-2, 0)$ and radius $\sqrt{4-b}$. The book does not say so, but it needs $b < 4$ for a genuine circle: at $b = 4$ it shrinks to the single point $(-2, 0)$, and for $b > 4$ there is no point at all.

Question 9

If the real part of $\dfrac{\bar z + 2}{\bar z – 1}$ is $4$, then show that the locus of the point representing $z$ in the complex plane is a circle.

Solution. Write $z = x + iy$, so $\bar z = x – iy$. Multiply top and bottom by the conjugate of the denominator, $(x-1) + iy$:

$$\frac{(x+2) – iy}{(x-1) – iy}\cdot\frac{(x-1)+iy}{(x-1)+iy} = \frac{\big[(x+2)(x-1) + y^2\big] + i\big[\,y(x+2) – y(x-1)\,\big]}{(x-1)^2 + y^2}.$$

The real part is $\dfrac{x^2 + x – 2 + y^2}{(x-1)^2 + y^2} = 4$. Cross-multiply:

$$x^2 + y^2 + x – 2 = 4x^2 + 4y^2 – 8x + 4 \implies 3x^2 + 3y^2 – 9x + 6 = 0 \implies x^2 + y^2 – 3x + 2 = 0.$$

Completing the square, $\left(x – \frac32\right)^2 + y^2 = \frac14$: a circle with centre $\left(\frac32, 0\right)$ and radius $\frac12$.

Question 10

Show that the complex number $z$, satisfying the condition $\arg\left(\dfrac{z-1}{z+1}\right) = \dfrac\pi4$, lies on a circle.

Solution. Argument and polar form are no longer in the current NCERT syllabus, but the method is the same. With $z = x + iy$, multiply by the conjugate of the denominator:

$$\frac{z-1}{z+1} = \frac{\big[(x-1) + iy\big]\big[(x+1) – iy\big]}{(x+1)^2 + y^2} = \frac{(x^2 + y^2 – 1) + 2iy}{(x+1)^2 + y^2}.$$

An argument of $\frac\pi4$ means the real and imaginary parts are equal and positive (first quadrant, on the line at $45^\circ$):

$$x^2 + y^2 – 1 = 2y > 0 \implies x^2 + (y-1)^2 = 2, \quad y > 0.$$

This is a circle with centre $(0, 1)$ and radius $\sqrt2$. Strictly, $z$ lies on the part of it above the $x$-axis (the major arc), because the condition $y > 0$ excludes the lower portion, where the argument would be $-\frac{3\pi}{4}$.

Question 11

Solve the equation $|z| = z + 1 + 2i$.

Solution. The left side $|z|$ is a real number, so the right side must have zero imaginary part. With $z = x + iy$, the imaginary part of $z + 1 + 2i$ is $y + 2$, so $y = -2$. The real parts give

$$\sqrt{x^2 + 4} = x + 1.$$

Squaring (which needs $x + 1 \ge 0$): $x^2 + 4 = x^2 + 2x + 1$, so $x = \dfrac32$. That satisfies $x + 1 \ge 0$, so it is not an extraneous root. Check: $|z| = \sqrt{\frac94 + 4} = \frac52$ and $z + 1 + 2i = \frac32 + 1 = \frac52$.

Question 12

If $|z + 1| = z + 2(1 + i)$, then find $z$.

Solution. Again the left side is real, so the imaginary part of the right side vanishes: $z + 2 + 2i$ has imaginary part $y + 2$, so $y = -2$. The real parts give

$$\sqrt{(x+1)^2 + 4} = x + 2 \implies x^2 + 2x + 5 = x^2 + 4x + 4 \implies x = \frac12.$$

The condition $x + 2 \ge 0$ holds. Check: $|z+1| = \left|\frac32 – 2i\right| = \frac52 = \frac12 + 2$.

Question 13

If $\arg(z – 1) = \arg(z + 3i)$, then find $x – 1 : y$, where $z = x + iy$.

Solution. Two complex numbers have the same argument when the points $z – 1 = (x-1) + iy$ and $z + 3i = x + i(y+3)$ lie on the same ray from the origin, so their slopes are equal:

$$\frac{y}{x-1} = \frac{y+3}{x} \implies xy = (x-1)(y+3) = xy + 3x – y – 3 \implies y = 3(x – 1).$$

Hence $\dfrac{x-1}{y} = \dfrac13$.

Question 14

Show that $\left|\dfrac{z-2}{z-3}\right| = 2$ represents a circle. Find its centre and radius.

Solution. Clear the denominator and square: $|z-2|^2 = 4|z-3|^2$. With $z = x + iy$,

$$(x-2)^2 + y^2 = 4\big[(x-3)^2 + y^2\big] \implies 3x^2 + 3y^2 – 20x + 32 = 0 \implies x^2 + y^2 – \frac{20}{3}x + \frac{32}{3} = 0.$$

Complete the square: $\left(x – \frac{10}{3}\right)^2 + y^2 = \frac{100}{9} – \frac{32}{3} = \frac49$.

Question 15

If $\dfrac{z-1}{z+1}$ is a purely imaginary number ($z \neq -1$), then find the value of $|z|$.

Solution. From Question 10, $\dfrac{z-1}{z+1} = \dfrac{(x^2 + y^2 – 1) + 2iy}{(x+1)^2 + y^2}$. Purely imaginary means the real part is zero:

$$x^2 + y^2 – 1 = 0 \implies |z|^2 = 1.$$

(The imaginary part $2y$ must be non-zero, so $y \ne 0$, but that does not change $|z|$.)

Question 16

$z_1$ and $z_2$ are two complex numbers such that $|z_1| = |z_2|$ and $\arg(z_1) + \arg(z_2) = \pi$. Then show that $z_1 = -\bar z_2$.

Solution. Argument and polar form are outside the current NCERT chapter. Let the common modulus be $r$ and $\arg z_1 = \alpha$, so $\arg z_2 = \pi – \alpha$. Then $z_2 = r\big(\cos(\pi-\alpha) + i\sin(\pi-\alpha)\big) = r(-\cos\alpha + i\sin\alpha)$, so

$$-\bar z_2 = -r(-\cos\alpha – i\sin\alpha) = r(\cos\alpha + i\sin\alpha) = z_1.$$

Question 17

If $|z_1| = 1$ ($z_1 \ne -1$) and $z_2 = \dfrac{z_1 – 1}{z_1 + 1}$, then show that the real part of $z_2$ is zero.

Solution. $|z_1| = 1$ means $z_1\bar z_1 = 1$, so $\bar z_1 = \dfrac{1}{z_1}$. Take the conjugate of $z_2$:

$$\bar z_2 = \frac{\bar z_1 – 1}{\bar z_1 + 1} = \frac{\frac1{z_1} – 1}{\frac1{z_1} + 1} = \frac{1 – z_1}{1 + z_1} = -z_2.$$

So $\bar z_2 = -z_2$, i.e. $z_2 + \bar z_2 = 0$, and $\operatorname{Re}(z_2) = \frac{z_2 + \bar z_2}{2} = 0$.

Question 18

If $z_1, z_2$ and $z_3, z_4$ are two pairs of conjugate complex numbers, then find $\arg\left(\dfrac{z_1}{z_4}\right) + \arg\left(\dfrac{z_2}{z_3}\right)$.

Solution. Argument is outside the current NCERT chapter. Conjugate pairs mean $z_2 = \bar z_1$ and $z_4 = \bar z_3$. Arguments of a product add, so the sum of the two arguments is the argument of the product:

$$\frac{z_1}{z_4}\cdot\frac{z_2}{z_3} = \frac{z_1\bar z_1}{\bar z_3 z_3} = \frac{|z_1|^2}{|z_3|^2},$$

a positive real number, whose argument is $0$.

Question 19

If $|z_1| = |z_2| = \dots = |z_n| = 1$, then show that $|z_1 + z_2 + z_3 + \dots + z_n| = \left|\dfrac1{z_1} + \dfrac1{z_2} + \dfrac1{z_3} + \dots + \dfrac1{z_n}\right|$.

Solution. For each $k$, $|z_k| = 1$ gives $z_k\bar z_k = 1$, so $\dfrac1{z_k} = \bar z_k$. The right side is therefore

$$|\bar z_1 + \bar z_2 + \dots + \bar z_n| = \left|\overline{z_1 + z_2 + \dots + z_n}\right| = |z_1 + z_2 + \dots + z_n|,$$

because a complex number and its conjugate have the same modulus.

Question 20

If for complex numbers $z_1$ and $z_2$, $\arg(z_1) – \arg(z_2) = 0$, then show that $|z_1 – z_2| = |z_1| – |z_2|$.

Solution. Argument is outside the current NCERT chapter. Equal arguments mean both numbers lie on the same ray from the origin: $z_1 = r_1(\cos\theta + i\sin\theta)$ and $z_2 = r_2(\cos\theta + i\sin\theta)$ with $r_1 = |z_1|$, $r_2 = |z_2|$. Then

$$z_1 – z_2 = (r_1 – r_2)(\cos\theta + i\sin\theta), \qquad |z_1 – z_2| = |r_1 – r_2|.$$

The statement as printed needs a condition. $|r_1 – r_2|$ equals $r_1 – r_2$ only when $|z_1| \ge |z_2|$. Without it, what is true is $|z_1 – z_2| = \big||z_1| – |z_2|\big|$. The book’s statement is the case $|z_1| \ge |z_2|$.

Question 21

Solve the system of equations $\operatorname{Re}(z^2) = 0$, $|z| = 2$.

Solution. With $z = x + iy$: $z^2 = x^2 – y^2 + 2ixy$, so $\operatorname{Re}(z^2) = x^2 – y^2 = 0$ and $|z| = 2$ gives $x^2 + y^2 = 4$. Adding, $2x^2 = 4$, so $x^2 = 2$ and then $y^2 = 2$. Every sign combination works, since only the squares appear.

Question 22

Find the complex number satisfying the equation $z + \sqrt2\,|z + 1| + i = 0$.

Solution. $|z+1|$ is real, so the imaginary part of $z + i$ must vanish: $y + 1 = 0$, so $y = -1$. The real parts give

$$x + \sqrt2\sqrt{(x+1)^2 + 1} = 0 \implies \sqrt2\sqrt{(x+1)^2 + 1} = -x.$$

This needs $x \le 0$. Squaring: $2\big[(x+1)^2 + 1\big] = x^2 \implies x^2 + 4x + 4 = 0 \implies x = -2$, which satisfies $x \le 0$. Check: $-2 + \sqrt2\,|{-1} – i| = -2 + \sqrt2\cdot\sqrt2 = 0$.

Question 23

Write the complex number $z = \dfrac{1-i}{\cos\frac\pi3 + i\sin\frac\pi3}$ in polar form.

Solution. Polar form is outside the current NCERT chapter. Write the numerator in polar form: $|1 – i| = \sqrt2$ and its argument is $-\frac\pi4$ (fourth quadrant), so $1 – i = \sqrt2\left(\cos\frac\pi4 – i\sin\frac\pi4\right)$. Dividing by $\cos\frac\pi3 + i\sin\frac\pi3$ means the moduli divide (the denominator has modulus $1$) and the arguments subtract:

$$\arg z = -\frac\pi4 – \frac\pi3 = -\frac{7\pi}{12}.$$

So $z = \sqrt2\left[\cos\left(-\dfrac{7\pi}{12}\right) + i\sin\left(-\dfrac{7\pi}{12}\right)\right]$. Check by direct division: $(1-i)\left(\frac12 – \frac{\sqrt3}{2}i\right) = \frac{1-\sqrt3}{2} – \frac{1+\sqrt3}{2}i \approx -0.366 – 1.366i$, and $\sqrt2\cos\frac{7\pi}{12} \approx -0.366$, $-\sqrt2\sin\frac{7\pi}{12} \approx -1.366$. The principal argument is $-\frac{7\pi}{12}$; the same angle written positively is $\frac{17\pi}{12}$.

The printed key is wrong here. It gives $\sqrt2\left(\cos\frac{5\pi}{12} + i\sin\frac{5\pi}{12}\right)$, which is in the first quadrant, while $z$ is in the third.

Question 24

If $z$ and $w$ are two complex numbers such that $|zw| = 1$ and $\arg(z) – \arg(w) = \dfrac\pi2$, then show that $\bar z w = -i$.

Solution. Argument is outside the current NCERT chapter. Let $|z| = r$, so $|w| = \dfrac1r$ from $|zw| = 1$, and let $\arg z = \alpha$, so $\arg w = \alpha – \frac\pi2$. Then

$$\bar z\,w = r(\cos\alpha – i\sin\alpha)\cdot\frac1r\left[\cos\left(\alpha – \tfrac\pi2\right) + i\sin\left(\alpha – \tfrac\pi2\right)\right].$$

The first factor has argument $-\alpha$ and the second $\alpha – \frac\pi2$, so the product has modulus $1$ and argument $-\frac\pi2$:

$$\bar z w = \cos\left(-\tfrac\pi2\right) + i\sin\left(-\tfrac\pi2\right) = -i.$$

Common mistakes

• Question 2: adding the two sums as if each had $13$ cancelling terms. Thirteen is not a multiple of $4$, so one leftover term ($i^{13}$, then $i^{14}$) survives from each sum.
• Question 4: writing $\dfrac{(1+i)^2}{2-i} = \dfrac{2i}{2-i}$ and stopping, or “rationalising” with $2 – i$ instead of its conjugate $2 + i$. Only the conjugate makes the denominator real.
• Questions 11 and 12: not noticing that $|z|$ is real. Students square the whole equation first and end up with a mess; setting the imaginary part to zero gives $y$ immediately, and the squared equation then gives $x$. Forgetting to check $x + 1 \ge 0$ (or $x + 2 \ge 0$) risks an extraneous root.
• Question 13: equating only the tangents is fine here, but students then forget the answer is a ratio $x – 1 : y$ and leave it as $y = 3x – 3$ without converting to $1 : 3$.
• Question 14: squaring $\left|\dfrac{z-2}{z-3}\right| = 2$ as $|z-2|^2 = 2|z-3|^2$, forgetting that the $2$ must also be squared to $4$. That gives the wrong radius.
• Question 23: adding the arguments instead of subtracting for a quotient. Dividing by $\cos\frac\pi3 + i\sin\frac\pi3$ rotates the point back by $\frac\pi3$.

Practise next

• The other pages of this unit: Fill in the Blanks, True or False and Matching, Q25–34 and Objective Type, Q35–50.
• NCERT Class 11 Maths Exercise 4.1 — Complex Numbers: conjugates, modulus and standard form, the base for Questions 1 to 7.
• NCERT Class 11 Maths Miscellaneous Exercise — Complex Numbers: modulus and conjugate identities like Questions 17 and 19.