CBSE Class 11 Mathematics Complex Numbers And Quadratic Equation Worksheet Set 01

Question 1. Find the real values of x and y, if
(i) \((x + y) (2 - 3i) = 4 + i\)
(ii) \(\frac{(1 + i)x - 2i}{3 + i} + \frac{(2 - 3i)y + i}{3 - i} = i\)
(iii) \((x - iy) (3 + 5i)\) is the conjugate of \(6 + 24i\).
Answer:
(i) Given equation is \((x + y) (2 - 3i) = 4 + i\)
Multiplying and equating real and imaginary parts:
\(2(x + y) - 3(x + y)i = 4 + 1i\)
\(2(x + y) = 4 \implies x + y = 2\) ... (1)
\(-3(x + y) = 1 \implies x + y = -\frac{1}{3}\) ... (2)
Since \(x + y\) cannot take two different values, there is no real solution for this part.
(ii) Simplifying the given expression by taking LCM and rationalizing denominators:
Multiplying cross terms and simplifying gives two simultaneous equations in \(x\) and \(y\). Equating real and imaginary parts yields \(x = 2\) and \(y = -1\).
(iii) Given \((x - iy)(3 + 5i)\) is the conjugate of \(6 + 24i\).
Conjugate of \(6 + 24i\) is \(6 - 24i\).
So, \((x - iy)(3 + 5i) = 6 - 24i\)
\(x - iy = \frac{6 - 24i}{3 + 5i}\)
Rationalizing the denominator:
\(x - iy = \frac{(6 - 24i)(3 - 5i)}{(3 + 5i)(3 - 5i)} = \frac{18 - 30i - 72i + 120i^2}{9 - 25i^2}\)
\(x - iy = \frac{18 - 120 - 102i}{9 + 25} = \frac{-102 - 102i}{34} = -3 - 3i\)
Equating real and imaginary parts: \(x = -3\), \(y = 3\).
In simple words: Expand the complex expressions and compare the real and imaginary parts on both sides to find x and y.

Exam Tip: Always write complex numbers in standard form \(a + ib\) before equating real and imaginary parts.

Question 2. Find the modulus and arguments of each of the following and represent it in the argand plane (i) \(z = -1 - i\sqrt{3}\) (ii) \(z = -\sqrt{3} + i\) (iii) \(z = 1 - i\) (iv) \(z = \frac{1 + 2i}{1 - 2i}\)
Answer:
(i) \(z = -1 - i\sqrt{3}\)
Modulus \(|z| = \sqrt{(-1)^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2\)
Here, \(x = -1\) and \(y = -\sqrt{3}\) (third quadrant).
Acute angle \(\alpha = \tan^{-1}\left|\frac{y}{x}\right| = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}\)
Argument \(\theta = -\left(\pi - \frac{\pi}{3}\right) = -\frac{2\pi}{3}\)
(ii) \(z = -\sqrt{3} + i\)
Modulus \(|z| = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2\)
Here, \(x = -\sqrt{3}\) and \(y = 1\) (second quadrant).
Argument \(\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\)
(iii) \(z = 1 - i\)
Modulus \(|z| = \sqrt{1^2 + (-1)^2} = \sqrt{2}\)
Argument \(\theta = -\frac{\pi}{4}\)
(iv) \(z = \frac{1 + 2i}{1 - 2i} = \frac{(1 + 2i)(1 + 2i)}{(1 - 2i)(1 + 2i)} = \frac{1 - 4 + 4i}{1 + 4} = \frac{-3 + 4i}{5} = -\frac{3}{5} + \frac{4}{5}i\)
Modulus \(|z| = \sqrt{\left(-\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2} = 1\)
Argument \(\theta = \pi - \tan^{-1}\left(\frac{4}{3}\right)\)
In simple words: Find modulus using \(\sqrt{x^2 + y^2}\) and argument by locating the correct quadrant.

Exam Tip: Be very careful with quadrant signs when finding arguments.

Question 3. Express the following complex numbers in polar form:
(i) \(\sin 50^{\circ} + i \cos 50^{\circ}\)
(ii) \(\cos 70^{\circ} + i \cos 20^{\circ}\)
Answer:
(i) Given \(z = \sin 50^{\circ} + i \cos 50^{\circ}\)
Rewrite using trigonometric identities: \(\sin 50^{\circ} = \cos(90^{\circ} - 50^{\circ}) = \cos 40^{\circ}\) and \(\cos 50^{\circ} = \sin 40^{\circ}\).
So, \(z = \cos 40^{\circ} + i \sin 40^{\circ}\), which is the polar form with modulus 1 and argument \(40^{\circ}\).
(ii) Given \(z = \cos 70^{\circ} + i \cos 20^{\circ} = \cos 70^{\circ} + i \sin(90^{\circ} - 20^{\circ}) = \cos 70^{\circ} + i \sin 70^{\circ}\).
Modulus is 1 and argument is \(70^{\circ}\).
In simple words: Convert sine to cosine and cosine to sine using complementary angle relations to get the standard \(\cos\theta + i\sin\theta\) form.

Exam Tip: Polar form strictly requires \(\cos\theta + i\sin\theta\) with the same angle \(\theta\) for both terms.

Question 4. Solve each of the following equations:
(i) \(\sqrt{3}x^2 - \sqrt{2}x + 2\sqrt{3} = 0\)
(ii) \(x^2 + \frac{x}{\sqrt{2}} + 1 = 0\)
(iii) \(3x^2 - 4x + \frac{20}{3} = 0\)
(iv) \(x^2 - 2xi + 12 = 0\)
(v) \(3x^2 - 7xi + 6 = 0\)
Answer:
(i) For \(\sqrt{3}x^2 - \sqrt{2}x + 2\sqrt{3} = 0\), using quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
Discriminant \(D = (-\sqrt{2})^2 - 4(\sqrt{3})(2\sqrt{3}) = 2 - 24 = -22\)
\(x = \frac{\sqrt{2} \pm \sqrt{-22}}{2\sqrt{3}} = \frac{\sqrt{2} \pm i\sqrt{22}}{2\sqrt{3}}\)
(ii) For \(x^2 + \frac{x}{\sqrt{2}} + 1 = 0\), multiply by \(\sqrt{2}\) or apply formula directly:
\(D = \left(\frac{1}{\sqrt{2}}\right)^2 - 4(1)(1) = \frac{1}{2} - 4 = -\frac{7}{2}\)
\(x = \frac{-\frac{1}{\sqrt{2}} \pm i\sqrt{\frac{7}{2}}}{2} = \frac{-1 \pm i\sqrt{7}}{2\sqrt{2}}\)
(iii) For \(3x^2 - 4x + \frac{20}{3} = 0\), multiply equation by 3: \(9x^2 - 12x + 20 = 0\)
\(D = (-12)^2 - 4(9)(20) = 144 - 720 = -576\)
\(x = \frac{12 \pm \sqrt{-576}}{18} = \frac{12 \pm 24i}{18} = \frac{2 \pm 4i}{3}\)
(iv) For \(x^2 - 2xi + 12 = 0\):
\(x^2 - 2xi - 12i^2 = 0 \implies (x - 3i)(x + 4i) = 0 \implies x = 3i, -4i\)
(v) For \(3x^2 - 7xi + 6 = 0\):
\(3x^2 - 7xi - 6i^2 = 0 \implies 3x^2 - 9xi + 2xi - 6i^2 = 0 \implies 3x(x - 3i) + 2i(x - 3i) = 0\)
\((3x + 2i)(x - 3i) = 0 \implies x = -\frac{2}{3}i, 3i\)
In simple words: Use the quadratic formula and introduce i when the discriminant is negative.

Exam Tip: For negative discriminants, write \(\sqrt{-D}\) as \(i\sqrt{D}\).

Question 5. Find the conjugate of the following:
(i) \(\frac{1}{3 + 4i}\)
(ii) \(7 + 5i\)
(iii) \(\frac{1}{(2 - 5i)^2}\)
(iv) \(\frac{(3 - 2i)(2 + 3i)}{(1 + 2i)(2 - i)}\)
Answer:
(i) Let \(z = \frac{1}{3 + 4i} = \frac{3 - 4i}{3^2 + 4^2} = \frac{3 - 4i}{25} = \frac{3}{25} - \frac{4}{25}i\)
Conjugate \(\bar{z} = \frac{3}{25} + \frac{4}{25}i\)
(ii) Let \(z = 7 + 5i\)
Conjugate \(\bar{z} = 7 - 5i\)
(iii) Let \(z = \frac{1}{(2 - 5i)^2} = \frac{1}{4 - 25 - 20i} = \frac{1}{-21 - 20i} = \frac{-21 + 20i}{(-21)^2 + (-20)^2} = \frac{-21 + 20i}{441 + 400} = \frac{-21 + 20i}{841}\)
Conjugate \(\bar{z} = \frac{-21 - 20i}{841}\)
(iv) Let \(z = \frac{(3 - 2i)(2 + 3i)}{(1 + 2i)(2 - i)} = \frac{6 + 9i - 4i - 6i^2}{2 - i + 4i - 2i^2} = \frac{12 + 5i}{4 + 3i}\)
Rationalizing: \(z = \frac{(12 + 5i)(4 - 3i)}{4^2 + 3^2} = \frac{48 - 36i + 20i - 15i^2}{25} = \frac{63 - 16i}{25}\)
Conjugate \(\bar{z} = \frac{63 + 16i}{25}\)
In simple words: Simplify the complex number into standard \(a + ib\) form first, then change the sign of the imaginary part.

Exam Tip: It is much faster to simplify the expression completely before finding the conjugate.

Question 6. Find the multiplicative inverse of the following complex number:
(i) \((2 + 3i)^2\)
(ii) \(\frac{5 + \sqrt{2}i}{5 - \sqrt{2}i}\)
Answer:
(i) Let \(z = (2 + 3i)^2 = 4 + 9i^2 + 12i = 4 - 9 + 12i = -5 + 12i\)
Multiplicative inverse \(z^{-1} = \frac{1}{-5 + 12i} = \frac{-5 - 12i}{(-5)^2 + 12^2} = \frac{-5 - 12i}{25 + 144} = \frac{-5 - 12i}{169}\)
(ii) Let \(z = \frac{5 + \sqrt{2}i}{5 - \sqrt{2}i}\)
Rationalizing \(z\): \(z = \frac{(5 + \sqrt{2}i)(5 + \sqrt{2}i)}{5^2 - (\sqrt{2}i)^2} = \frac{25 - 2 + 10\sqrt{2}i}{25 + 2} = \frac{23 + 10\sqrt{2}i}{27}\)
Multiplicative inverse \(z^{-1} = \frac{27}{23 + 10\sqrt{2}i} = \frac{27(23 - 10\sqrt{2}i)}{23^2 - (10\sqrt{2})^2} = \frac{27(23 - 10\sqrt{2}i)}{529 - 200} = \frac{27(23 - 10\sqrt{2}i)}{329}\)
In simple words: The multiplicative inverse of z is \(\frac{\bar{z}}{|z|^2}\).

Exam Tip: Remember that the multiplicative inverse is simply \(\frac{1}{z}\).

Question 7. Express the following complex number in form a + ib
(i) \(\left(\frac{1}{3} + 3i\right)^3\)
(ii) \((1 - i)^4\)
(iii) \(\frac{6 + 3i}{2 - i}\)
(iv) \(\left(\frac{1}{3} + \frac{7}{3}i\right) + \left(4 + \frac{1}{3}i\right) - \left(-\frac{4}{3} + i\right)\)
(v) \(i^{26} + i^5\)
Answer:
(i) \(\left(\frac{1}{3} + 3i\right)^3 = \left(\frac{1}{3}\right)^3 + (3i)^3 + 3\left(\frac{1}{3}\right)(3i)\left(\frac{1}{3} + 3i\right)\)
\(= \frac{1}{27} - 27i + 3i\left(\frac{1}{3} + 3i\right) = \frac{1}{27} - 27i + i + 9i^2 = \left(\frac{1}{27} - 9\right) - 26i = -\frac{242}{27} - 26i\)
(ii) \((1 - i)^4 = \left((1 - i)^2\right)^2 = (1 - 1 - 2i)^2 = (-2i)^2 = 4i^2 = -4 + 0i\)
(iii) \(\frac{6 + 3i}{2 - i} = \frac{(6 + 3i)(2 + i)}{2^2 + 1^2} = \frac{12 + 6i + 6i - 3}{5} = \frac{9 + 12i}{5} = \frac{9}{5} + \frac{12}{5}i\)
(iv) \(\left(\frac{1}{3} + 4 + \frac{4}{3}\right) + i\left(\frac{7}{3} + \frac{1}{3} - 1\right) = \left(\frac{5}{3} + 4\right) + i\left(\frac{8}{3} - 1\right) = \frac{17}{3} + \frac{5}{3}i\)
(v) \(i^{26} + i^5 = (i^4)^6 \cdot i^2 + (i^4)^1 \cdot i = 1(-1) + 1(i) = -1 + i\)
In simple words: Expand powers and combine real and imaginary parts using standard algebraic rules and \(i^2 = -1\).

Exam Tip: Use powers of i (like \(i^4 = 1\)) to simplify large exponents quickly.

Question 8. If \(z = x + iy\) and \(w = \frac{1 - iz}{z - i}\), show that \(|w| = 1 \implies z\) purely real.
Answer:
Given \(|w| = 1 \implies \left|\frac{1 - iz}{z - i}\right| = 1 \implies |1 - iz| = |z - i|\)
Substitute \(z = x + iy\):
\(|1 - i(x + iy)| = |x + iy - i|\)
\(|1 - ix + y| = |x + i(y - 1)|\)
\(|(y + 1) - ix| = |x + i(y - 1)|\)
Squaring both sides:
\((y + 1)^2 + (-x)^2 = x^2 + (y - 1)^2\)
\(y^2 + 2y + 1 + x^2 = x^2 + y^2 - 2y + 1\)
\(2y = -2y \implies 4y = 0 \implies y = 0\)
Since the imaginary part \(y = 0\), \(z = x + 0i = x\), which means \(z\) is purely real.
In simple words: Substitute \(z = x + iy\) into the modulus condition and simplify to prove \(y = 0\).

Exam Tip: Remember that \(|a + ib| = \sqrt{a^2 + b^2}\), so squaring both sides removes square roots easily.

Question 9. Convert the complex number \(z = \frac{i - 1}{\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}}\) in the polar form.
Answer:
Numerator: \(-1 + i\), modulus \(|-1 + i| = \sqrt{2}\), argument \(\theta_1 = \pi - \frac{\pi}{4} = \frac{3\pi}{4}\)
Denominator: \(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\), modulus = 1, argument \(\theta_2 = \frac{\pi}{3}\)
In polar form, division of complex numbers subtracts their arguments and divides their moduli:
\(|z| = \frac{\sqrt{2}}{1} = \sqrt{2}\)
Argument \(\theta = \frac{3\pi}{4} - \frac{\pi}{3} = \frac{9\pi - 4\pi}{12} = \frac{5\pi}{12}\)
Thus, polar form is \(z = \sqrt{2}\left(\cos\frac{5\pi}{12} + i\sin\frac{5\pi}{12}\right)\).
In simple words: Convert numerator and denominator to polar form separately, then divide moduli and subtract angles.

Exam Tip: Division in polar form corresponds to subtraction of arguments.

Question 10. Show that the images of the complex numbers \(3 + 2i\), \(5i\), \(-3 + 2i\) and \(-i\) form a square.
Answer:
Let the points corresponding to the complex numbers be:
\(A(3, 2)\), \(B(0, 5)\), \(C(-3, 2)\), and \(D(0, -1)\).
Calculate the lengths of all four sides using the distance formula:
\(AB = \sqrt{(0 - 3)^2 + (5 - 2)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}\)
\(BC = \sqrt{(-3 - 0)^2 + (2 - 5)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}\)
\(CD = \sqrt{(0 - (-3))^2 + (-1 - 2)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}\)
\(DA = \sqrt{(3 - 0)^2 + (2 - (-1))^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}\)
All four sides are equal (\(AB = BC = CD = DA = 3\sqrt{2}\)).
Calculate the lengths of the diagonals:
\(AC = \sqrt{(-3 - 3)^2 + (2 - 2)^2} = \sqrt{36} = 6\)
\(BD = \sqrt{(0 - 0)^2 + (-1 - 5)^2} = \sqrt{36} = 6\)
Since all four sides are equal and both diagonals are equal (\(AC = BD = 6\)), the figure is a square.
In simple words: Plot the complex numbers as coordinates and prove all four sides and both diagonals are equal.

Exam Tip: To prove a quadrilateral is a square, always check both that all four sides are equal and that the diagonals are equal.

Question 11. For a complex number z, what is the value of \(\text{Arg } z + \text{Arg } \bar{z}\) (\(z \neq 0\))?
Answer:
Let \(z = r(\cos\theta + i\sin\theta)\), so \(\text{Arg } z = \theta\).
The conjugate is \(\bar{z} = r(\cos\theta - i\sin\theta) = r(\cos(-\theta) + i\sin(-\theta))\), so \(\text{Arg } \bar{z} = -\theta\).
Therefore, \(\text{Arg } z + \text{Arg } \bar{z} = \theta + (-\theta) = 0\).
In simple words: Conjugation reflects the complex number across the real axis, negating its argument.

Exam Tip: Remember the property \(\text{Arg}(\bar{z}) = -\text{Arg}(z)\).

Question 12. Show that \(\left|\frac{z - 3}{z + 3}\right| = 2\) represent a circle.
Answer:
Let \(z = x + iy\).
Given \(\left|\frac{x + iy - 3}{x + iy + 3}\right| = 2 \implies \frac{|(x - 3) + iy|}{|(x + 3) + iy|} = 2\)
Squaring both sides:
\(\frac{(x - 3)^2 + y^2}{(x + 3)^2 + y^2} = 4\)
\((x - 3)^2 + y^2 = 4\left[(x + 3)^2 + y^2\right]\)
\(x^2 - 6x + 9 + y^2 = 4(x^2 + 6x + 9 + y^2)\)
\(x^2 - 6x + 9 + y^2 = 4x^2 + 24x + 36 + 4y^2\)
\(3x^2 + 3y^2 + 30x + 27 = 0\)
Divide by 3:
\(x^2 + y^2 + 10x + 9 = 0\)
This is the equation of a circle with center \((-5, 0)\) and radius \(\sqrt{5^2 - 9} = \sqrt{16} = 4\).
In simple words: Substitute \(z = x + iy\), square both sides, and simplify to get the standard equation of a circle.

Answer:
(i) Let \(\sqrt{3 + 4i} = x + iy\)
Squaring both sides: \(3 + 4i = (x^2 - y^2) + i(2xy)\)
Equating real and imaginary parts:
\(x^2 - y^2 = 3\) ... (1)
\(2xy = 4 \implies xy = 2\) ... (2)
Using identity \((x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2 = 3^2 + 4^2 = 9 + 16 = 25\)
\(x^2 + y^2 = 5\) ... (3)
Adding (1) and (3): \(2x^2 = 8 \implies x^2 = 4 \implies x = \pm 2\)
Subtracting (1) from (3): \(2y^2 = 2 \implies y^2 = 1 \implies y = \pm 1\)
Since \(xy = 2\) (positive), x and y have the same sign.
Thus, square roots are \(\pm(2 + i)\).
(ii) Let \(\sqrt{12 - 5i} = x + iy\)
Squaring both sides: \(12 - 5i = (x^2 - y^2) + i(2xy)\)
\(x^2 - y^2 = 12\) and \(2xy = -5 \implies xy = -\frac{5}{2}\)
\(x^2 + y^2 = \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13\)
Adding \(x^2 - y^2 = 12\) and \(x^2 + y^2 = 13\):
\(2x^2 = 25 \implies x^2 = \frac{25}{2} \implies x = \pm\frac{5}{\sqrt{2}}\)
\(2y^2 = 1 \implies y^2 = \frac{1}{2} \implies y = \pm\frac{1}{\sqrt{2}}\)
Since \(xy\) is negative, \(x\) and \(y\) have opposite signs.
Thus, square roots are \(\pm\left(\frac{5}{\sqrt{2}} - \frac{1}{\sqrt{2}}i\right)\).
In simple words: Equate the square root to \(x + iy\), find \(x^2 + y^2\) using modulus, and solve simultaneous equations for x and y.

Exam Tip: Always check the sign of the imaginary part to assign correct signs to x and y.

Complex Numbers MCQ Questions with Answers Class 11 Mathematics