NCERT Exemplar Solutions Class 11 Maths Complex Numbers and Quadratic Equations — Fill in the Blanks, True or False, Matching and Very Short Answer Q25–34

NCERT Exemplar Problems, Class 11 Mathematics — Complex Numbers and Quadratic Equations, Fill in the Blanks, True or False, Matching and Very Short Answer questions 25 to 34. All 10 questions solved.

Questions 25, 26 and 27 are really 26 small problems packed under three numbers, so each part is answered separately. They test the facts of the chapter: $i^4 = 1$, $|z|^2 = z\bar z$, that $\sqrt{-a}\cdot\sqrt{-b}$ is not $\sqrt{ab}$, and that $|z – a|$ is a distance, so $|z – a| = |z – b|$ is a perpendicular bisector and $|z – a| \le r$ is a disc. Questions 28 to 34 are one-line computations.

Every printed answer agrees with mine. Two blanks, 25 (vi) and (vii), are only true in a special case, and that is said where they occur. Argument and polar form (25 (vii), (x), 27 (a), (b), 31, 33) are no longer in the current NCERT Class 11 chapter, but are solved here.

Key insight. Read $|z – a|$ as “the distance from $z$ to $a$”. Almost every locus question on this page then becomes coordinate geometry: equal distances give a perpendicular bisector, a fixed distance gives a circle, and “at most” or “at least” gives the inside or outside of it.

Question 25

Fill in the blanks of the following.

Question 25 (i)

For any two complex numbers $z_1, z_2$ and any real numbers $a, b$, $|az_1 – bz_2|^2 + |bz_1 + az_2|^2 = \_\_\_\_$.

Solution. Use $|u|^2 = u\bar u$ and expand. Because $a$ and $b$ are real, $\overline{az_1 – bz_2} = a\bar z_1 – b\bar z_2$:

$$|az_1 – bz_2|^2 = a^2|z_1|^2 + b^2|z_2|^2 – ab(z_1\bar z_2 + \bar z_1 z_2),$$ $$|bz_1 + az_2|^2 = b^2|z_1|^2 + a^2|z_2|^2 + ab(z_1\bar z_2 + \bar z_1 z_2).$$

The cross terms cancel when added.

Question 25 (ii)

The value of $\sqrt{-25}\times\sqrt{-9}$ is $\_\_\_\_$.

Solution. The rule $\sqrt a\sqrt b = \sqrt{ab}$ holds only when at least one of $a, b$ is non-negative. Here both are negative, so convert each root first: $\sqrt{-25} = 5i$ and $\sqrt{-9} = 3i$. Then $5i\times 3i = 15i^2 = -15$.

Question 25 (iii)

The number $\dfrac{(1-i)^3}{1-i^3}$ is equal to $\_\_\_\_$.

Solution. $i^3 = -i$, so $1 – i^3 = 1 + i$. And $(1-i)^3 = (1-i)(1-i)^2 = (1-i)(-2i) = -2i + 2i^2 = -2 – 2i = -2(1+i)$. Dividing by $1 + i$ gives $-2$.

Question 25 (iv)

The sum of the series $i + i^2 + i^3 + \dots$ up to $1000$ terms is $\_\_\_\_$.

Solution. Each block of four consecutive powers adds to $0$, and $1000 = 4 \times 250$, so there are exactly $250$ blocks and nothing left over.

Question 25 (v)

The multiplicative inverse of $1 + i$ is $\_\_\_\_$.

Solution. $\dfrac{1}{1+i} = \dfrac{1-i}{(1+i)(1-i)} = \dfrac{1-i}{2}$.

Question 25 (vi)

If $z_1$ and $z_2$ are complex numbers such that $z_1 + z_2$ is a real number, then $z_2 = \_\_\_\_$.

Solution. Write $z_1 = a + ib$, $z_2 = c + id$. The sum has imaginary part $b + d$, and that is zero exactly when $d = -b$. So $z_2 = c – ib = (c – a) + \overline{z_1}$, where $c$ is any real number. The intended answer, the one the key prints, is $\bar z_1$, which is the case $c = a$ (equal real parts). Strictly, though, $z_2 = \bar z_1$ is only a special case: $z_1 = i$, $z_2 = 2 – i$ has a real sum $2$ but $z_2 \ne \bar z_1 = -i$. The fully general blank is $z_2 = k – z_1$ with $k$ real.

Question 25 (vii)

$\arg(z) + \arg(\bar z)$ $(\bar z \ne 0)$ is $\_\_\_\_$.

Solution. The point $\bar z$ is the mirror image of $z$ in the real axis, so its argument is the negative of that of $z$, and the sum is $0$. The one exception is a negative real $z$: with principal arguments in $(-\pi, \pi]$, both $z$ and $\bar z = z$ have argument $\pi$, and the sum is $2\pi$. For every other $z$ the answer is $0$, which is what the key prints.

Question 25 (viii)

If $|z + 4| \le 3$, then the greatest and least values of $|z + 1|$ are $\_\_\_\_$ and $\_\_\_\_$.

Solution. $|z + 4| \le 3$ is the disc with centre $-4$ and radius $3$. $|z + 1|$ is the distance from $z$ to the point $-1$. That point is $3$ units from the centre, exactly on the boundary circle. The farthest point of the disc from $-1$ is the opposite end of the diameter, at distance $3 + 3 = 6$, and the nearest is $-1$ itself, at distance $0$. In symbols, $|z+1| = |(z+4) – 3| \le |z+4| + 3 \le 6$.

Question 25 (ix)

If $\left|\dfrac{z-2}{z+2}\right| = \dfrac\pi6$, then the locus of $z$ is $\_\_\_\_$.

Solution. As printed, this is a modulus (not an argument) equal to the constant $k = \frac\pi6$, which is positive and not $1$. So $|z – 2| = k|z + 2|$; with $z = x + iy$,

$$(x-2)^2 + y^2 = k^2\big[(x+2)^2 + y^2\big] \implies (1-k^2)(x^2 + y^2) – 4(1+k^2)x + 4(1-k^2) = 0.$$

Since $k \ne 1$ this is a circle, with centre $\left(\dfrac{2(1+k^2)}{1-k^2}, 0\right)$ and radius $\dfrac{4k}{1-k^2}$ (the circle of Apollonius). It becomes a straight line only when $k = 1$.

Question 25 (x)

If $|z| = 4$ and $\arg(z) = \dfrac{5\pi}{6}$, then $z = \_\_\_\_$.

Solution. $z = |z|(\cos\theta + i\sin\theta) = 4\left(\cos\frac{5\pi}6 + i\sin\frac{5\pi}6\right)$. Since $\cos\frac{5\pi}6 = -\frac{\sqrt3}2$ and $\sin\frac{5\pi}6 = \frac12$, $z = -2\sqrt3 + 2i$ (second quadrant, as expected).

Question 26

State True or False for the following.

Question 26 (i)

The order relation is defined on the set of complex numbers.

Solution. No ordering “$<$” compatible with addition and multiplication exists on the complex numbers (for example it would force $i^2 = -1 > 0$ or $-i^2 > 0$, both absurd). We can order the moduli, which are real, but not the numbers.

Question 26 (ii)

Multiplication of a non-zero complex number by $-i$ rotates the point about the origin through a right angle in the anti-clockwise direction.

Solution. Multiplying by $i$ turns the point $90^\circ$ anti-clockwise ($1 \to i$). Multiplying by $-i$ does the opposite: $1 \to -i$, a $90^\circ$ clockwise turn.

Question 26 (iii)

For any complex number $z$, the minimum value of $|z| + |z – 1|$ is $1$.

Solution. $|z| + |z – 1|$ is the sum of the distances from $z$ to $0$ and to $1$. By the triangle inequality $|z| + |z – 1| \ge |z – (z-1)| = 1$, and equality holds for every point on the segment from $0$ to $1$ (for example $z = 0$).

Question 26 (iv)

The locus represented by $|z – 1| = |z – i|$ is a line perpendicular to the join of $(1, 0)$ and $(0, 1)$.

Solution. The locus is the set of points equidistant from $(1, 0)$ and $(0, 1)$, the perpendicular bisector of the join of those points; in coordinates it is the line $y = x$. A perpendicular bisector is in particular a line perpendicular to the join, so the statement is true (it leaves out “bisector”, but nothing in it is wrong).

Question 26 (v)

If $z$ is a complex number such that $z \ne 0$ and $\operatorname{Re}(z) = 0$, then $\operatorname{Im}(z^2) = 0$.

Solution. $\operatorname{Re}(z) = 0$ means $z = iy$ with $y \ne 0$. Then $z^2 = -y^2$, a real number, so its imaginary part is $0$.

Question 26 (vi)

The inequality $|z – 4| < |z – 2|$ represents the region given by $x > 3$.

Solution. $z$ is closer to $4$ than to $2$. The boundary is the perpendicular bisector of $2$ and $4$, the line $x = 3$, and the side containing $4$ is $x > 3$. Algebraically, $(x-4)^2 + y^2 < (x-2)^2 + y^2$ gives $-8x + 16 < -4x + 4$, so $x > 3$.

Question 26 (vii)

Let $z_1$ and $z_2$ be two complex numbers such that $|z_1 + z_2| = |z_1| + |z_2|$, then $\arg(z_1 – z_2) = 0$.

Solution. Equality in the triangle inequality means $z_1$ and $z_2$ point the same way from the origin, so $\arg z_1 = \arg z_2$. That says nothing about $\arg(z_1 – z_2)$. Take $z_1 = 1$, $z_2 = 2$: then $|z_1 + z_2| = 3 = |z_1| + |z_2|$, but $z_1 – z_2 = -1$, whose argument is $\pi$, not $0$.

Question 26 (viii)

$2$ is not a complex number.

Solution. Every real number is a complex number: $2 = 2 + 0i$.

Question 27

Match the statements of Column A and Column B.

Solution. Work each item of Column A out and then find it in Column B.

• (a) The polar form of $i + \sqrt3$: modulus $\sqrt{3 + 1} = 2$, argument $\tan^{-1}\frac{1}{\sqrt3} = \frac\pi6$, so $2\left(\cos\frac\pi6 + i\sin\frac\pi6\right)$, which is (v).
• (b) The amplitude of $-1 + \sqrt{-3} = -1 + i\sqrt3$: second quadrant, $\tan\theta = -\sqrt3$, so $\theta = \pi – \frac\pi3 = \frac{2\pi}3$, which is (iii).
• (c) $|z + 2| = |z – 2|$: equidistant from $(-2, 0)$ and $(2, 0)$, the perpendicular bisector of that segment (the $y$-axis), which is (i).
• (d) $|z + 2i| = |z – 2i|$: equidistant from $(0, -2)$ and $(0, 2)$, the perpendicular bisector of that segment (the $x$-axis), which is (iv).
• (e) $|z + 4i| \ge 3$: distance from $(0, -4)$ at least $3$, so on or outside the circle with centre $(0, -4)$ and radius $3$, which is (ii).
• (f) $|z + 4| \le 3$: distance from $(-4, 0)$ at most $3$, so on or inside the circle with centre $(-4, 0)$ and radius $3$, which is (vi).
• (g) $\dfrac{1+2i}{1-i} = \dfrac{(1+2i)(1+i)}{2} = \dfrac{-1+3i}{2}$; its conjugate is $\dfrac{-1-3i}{2}$, with both coordinates negative, so the third quadrant, which is (viii).
• (h) The reciprocal of $1 – i$ is $\dfrac{1+i}{2}$, with both coordinates positive, so the first quadrant, which is (vii).

Question 28

What is the conjugate of $\dfrac{2-i}{(1-2i)^2}$?

Solution. First $(1-2i)^2 = 1 – 4i + 4i^2 = -3 – 4i$. Then multiply by the conjugate of the denominator:

$$\frac{2-i}{-3-4i}\cdot\frac{-3+4i}{-3+4i} = \frac{(2-i)(-3+4i)}{9+16} = \frac{-6 + 8i + 3i – 4i^2}{25} = \frac{-2 + 11i}{25}.$$

The conjugate flips the sign of the imaginary part.

Question 29

If $|z_1| = |z_2|$, is it necessary that $z_1 = z_2$?

Solution. No. Equal moduli only say the two points are the same distance from the origin; they can be anywhere on the same circle. For example $z_1 = 1$ and $z_2 = i$ both have modulus $1$ but are different.

Question 30

If $\dfrac{(a^2+1)^2}{2a – i} = x + iy$, what is the value of $x^2 + y^2$?

Solution. $x^2 + y^2$ is the square of the modulus of the left side, and a modulus of a quotient is the quotient of moduli (here $a$ is real):

$$x^2 + y^2 = \frac{(a^2+1)^4}{|2a – i|^2} = \frac{(a^2+1)^4}{4a^2 + 1}.$$

Question 31

Find $z$ if $|z| = 4$ and $\arg(z) = \dfrac{5\pi}{6}$.

Solution. This is the same as 25 (x). Argument is no longer in the current NCERT chapter. $z = 4\left(\cos\frac{5\pi}6 + i\sin\frac{5\pi}6\right) = 4\left(-\frac{\sqrt3}2 + \frac i2\right)$.

Question 32

Find $\left|(1+i)\dfrac{(2+i)}{(3+i)}\right|$.

Solution. The modulus of a product or quotient is the product or quotient of moduli, so there is no need to multiply out:

$$\frac{|1+i|\,|2+i|}{|3+i|} = \frac{\sqrt2\cdot\sqrt5}{\sqrt{10}} = 1.$$

Question 33

Find the principal argument of $(1 + i\sqrt3)^2$.

Solution. Argument is no longer in the current NCERT chapter. $(1 + i\sqrt3)^2 = 1 – 3 + 2\sqrt3\,i = -2 + 2\sqrt3\,i$, which lies in the second quadrant with $\tan\theta = \dfrac{2\sqrt3}{-2} = -\sqrt3$ and reference angle $\frac\pi3$. So $\theta = \pi – \frac\pi3 = \frac{2\pi}3$. (Equivalently, $\arg(1 + i\sqrt3) = \frac\pi3$ and squaring doubles the argument.)

Question 34

Where does $z$ lie, if $\left|\dfrac{z – 5i}{z + 5i}\right| = 1$?

Solution. $|z – 5i| = |z + 5i|$ says $z$ is equally far from $(0, 5)$ and $(0, -5)$, so it lies on the perpendicular bisector of that segment, which is the $x$-axis. Algebraically $x^2 + (y-5)^2 = x^2 + (y+5)^2$ gives $y = 0$.

Common mistakes

• Question 25 (ii): writing $\sqrt{-25}\sqrt{-9} = \sqrt{225} = 15$. The product rule for square roots needs a non-negative number under at least one root; with both negative you must write $5i \cdot 3i = -15$.
• Question 25 (iii): expanding $(1-i)^3$ correctly but forgetting that $i^3 = -i$ in the denominator, getting $1 – i$ instead of $1 + i$.
• Question 25 (viii): taking the least value of $|z + 1|$ to be $|{-4} + 1| – 3 = 0$ and being unsure it is allowed. It is: the point $-1$ lies on the boundary circle, so the disc actually contains it.
• Question 26 (ii): remembering “rotation by $90^\circ$” but not the direction. Multiplying by $i$ is anticlockwise ($1 \to i$); by $-i$ it is clockwise.
• Question 26 (vii): reading equality in the triangle inequality as a statement about $z_1 – z_2$. It tells you $z_1$ and $z_2$ have the same argument, not that their difference has argument $0$.
• Question 27 (c), (d): swapping the two lines. $|z+2| = |z-2|$ separates points on the real axis, so its bisector is vertical (the $y$-axis); $|z + 2i| = |z – 2i|$ is the horizontal $x$-axis. Draw the two points before matching.

Practise next

• The other pages of this unit: Short and Long Answer, Q1–24 and Objective Type, Q35–50.
• NCERT Class 11 Maths Exercise 4.1 — Complex Numbers: powers of $i$, conjugates and standard form, the base for Questions 25 to 32.
• NCERT Class 11 Maths Miscellaneous Exercise — Complex Numbers: modulus and conjugate identities.