CBSE Class 11 Mathematics Complex Numbers and Quadratic Equations Assignment Set 03

EXTRA/HOT QUESTIONS

Question 1. Find the square roots of the following complex numbers (4 marks)
i. \(6 + 8i\)
Answer: Let \(x + iy = \sqrt{6 + 8i}\)
Squaring both sides:
\((x + iy)^2 = 6 + 8i\)
\(x^2 - y^2 + 2xyi = 6 + 8i\)
Equating real and imaginary parts:
\(x^2 - y^2 = 6\) (Equation 1)
\(2xy = 8 \implies xy = 4\) (Equation 2)
We know that \((x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2\)
\((x^2 + y^2)^2 = 6^2 + 8^2 = 36 + 64 = 100\)
\(x^2 + y^2 = 10\) (Equation 3, since \(x^2 + y^2\) cannot be negative)
Adding Equation 1 and Equation 3:
\(2x^2 = 16 \implies x^2 = 8 \implies x = \pm 2\sqrt{2}\)
Subtracting Equation 1 from Equation 3:
\(2y^2 = 4 \implies y^2 = 2 \implies y = \pm \sqrt{2}\)
Since \(xy = 4\) (positive), \(x\) and \(y\) must have the same sign.
Therefore, square roots are \(\pm(2\sqrt{2} + \sqrt{2}i)\).
In simple words: Assume the square root is \(x + iy\), square both sides, and solve for real and imaginary parts.

Exam Tip: Always remember to check the sign of the product \(xy\) to correctly pair the signs of \(x\) and \(y\).

Question 1(ii). Find the square roots of the following complex numbers (4 marks)
ii. \(3 - 4i\)
Answer: Let \(x + iy = \sqrt{3 - 4i}\)
Squaring both sides:
\(x^2 - y^2 + 2xyi = 3 - 4i\)
Equating real and imaginary parts:
\(x^2 - y^2 = 3\) (Equation 1)
\(2xy = -4 \implies xy = -2\) (Equation 2)
Using \((x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2\):
\((x^2 + y^2)^2 = 3^2 + (-4)^2 = 9 + 16 = 25\)
\(x^2 + y^2 = 5\) (Equation 3)
Adding Equation 1 and Equation 3:
\(2x^2 = 8 \implies x^2 = 4 \implies x = \pm 2\)
Subtracting Equation 1 from Equation 3:
\(2y^2 = 2 \implies y^2 = 1 \implies y = \pm 1\)
Since \(xy = -2\) (negative), \(x\) and \(y\) have opposite signs.
Therefore, square roots are \(\pm(2 - i)\).
In simple words: Separate real and imaginary parts after squaring and use the modulus identity to find \(x^2 + y^2\).

Exam Tip: Watch out for the negative sign in imaginary parts to determine proper quadrant/signs.

Question 1(iii). Find the square roots of the following complex numbers (4 marks)
iii. \(2 + 3i\) (HOT)
Answer: Let \(x + iy = \sqrt{2 + 3i}\)
Squaring both sides:
\(x^2 - y^2 + 2xyi = 2 + 3i\)
\(x^2 - y^2 = 2\) and \(2xy = 3 \implies xy = \frac{3}{2}\)
\((x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2 = 2^2 + 3^2 = 4 + 9 = 13\)
\(x^2 + y^2 = \sqrt{13}\)
Solving for \(x^2\) and \(y^2\):
\(x^2 = \frac{\sqrt{13} + 2}{2} \implies x = \pm \sqrt{\frac{\sqrt{13} + 2}{2}}\)
\(y^2 = \frac{\sqrt{13} - 2}{2} \implies y = \pm \sqrt{\frac{\sqrt{13} - 2}{2}}\)
Since \(xy\) is positive, \(x\) and \(y\) have the same sign.
Square roots are: \(\pm \left( \sqrt{\frac{\sqrt{13} + 2}{2}} + i\sqrt{\frac{\sqrt{13} - 2}{2}} \right)\).
In simple words: Irrational roots appear when numbers do not form neat squares, but the method remains identical.

Exam Tip: Do not panic when square roots of square roots appear in HOT problems; keep terms inside surds intact.

Question 1(iv). Find the square roots of the following complex numbers (4 marks)
iv. \(7 - 30\sqrt{2}i\)
Answer: Let \(x + iy = \sqrt{7 - 30\sqrt{2}i}\)
Squaring both sides:
\(x^2 - y^2 + 2xyi = 7 - 30\sqrt{2}i\)
\(x^2 - y^2 = 7\)
\(2xy = -30\sqrt{2} \implies xy = -15\sqrt{2}\)
\((x^2 + y^2)^2 = (7)^2 + (-30\sqrt{2})^2 = 49 + 1800 = 1849\)
\(x^2 + y^2 = \sqrt{1849} = 43\)
Adding \(x^2 - y^2 = 7\) and \(x^2 + y^2 = 43\):
\(2x^2 = 50 \implies x^2 = 25 \implies x = \pm 5\)
Subtracting: \(2y^2 = 36 \implies y^2 = 18 \implies y = \pm 3\sqrt{2}\)
Since \(xy\) is negative, roots are: \(\pm(5 - 3\sqrt{2}i)\).
In simple words: Large numbers like 1849 simplify nicely to 43; check your arithmetic carefully.

Exam Tip: Memorize squares of numbers up to 50 to quickly recognize large perfect squares like 1849.

Question 1(v). Find the square roots of the following complex numbers (4 marks)
v. \(\frac{3 + 4i}{3 - 4i}\) (HOT)
Answer: First, simplify the complex number by rationalizing the denominator:
\(Z = \frac{3 + 4i}{3 - 4i} \times \frac{3 + 4i}{3 + 4i}\)
\(Z = \frac{(3 + 4i)^2}{3^2 - (4i)^2} = \frac{9 - 16 + 24i}{9 + 16} = \frac{-7 + 24i}{25} = -\frac{7}{25} + \frac{24}{25}i\)
Now, let \(x + iy = \sqrt{-\frac{7}{25} + \frac{24}{25}i}\)
Squaring both sides:
\(x^2 - y^2 = -\frac{7}{25}\)
\(2xy = \frac{24}{25} \implies xy = \frac{12}{25}\)
\((x^2 + y^2)^2 = \left(-\frac{7}{25}\right)^2 + \left(\frac{24}{25}\right)^2 = \frac{49 + 576}{625} = \frac{625}{625} = 1\)
\(x^2 + y^2 = 1\)
Adding \(x^2 - y^2 = -\frac{7}{25}\) and \(x^2 + y^2 = 1\):
\(2x^2 = 1 - \frac{7}{25} = \frac{18}{25} \implies x^2 = \frac{9}{25} \implies x = \pm \frac{3}{5}\)
Subtracting: \(2y^2 = 1 + \frac{7}{25} = \frac{32}{25} \implies y^2 = \frac{16}{25} \implies y = \pm \frac{4}{5}\)
In simple words: Always simplify fractions with complex denominators before attempting to find square roots.

Exam Tip: Rationalization step transforms a difficult fraction into a standard complex number format.

Question 2(i). Convert the following complex numbers in the polar form
i. \(3\sqrt{3} + 3i\)
Answer: Let \(Z = 3\sqrt{3} + 3i = r(\cos\theta + i\sin\theta)\)
Here, \(a = 3\sqrt{3}\) and \(b = 3\)
\(r = |Z| = \sqrt{(3\sqrt{3})^2 + 3^2} = \sqrt{27 + 9} = \sqrt{36} = 6\)
\(r\cos\theta = 3\sqrt{3} \implies \cos\theta = \frac{3\sqrt{3}}{6} = \frac{\sqrt{3}}{2}\)
\(r\sin\theta = 3 \implies \sin\theta = \frac{3}{6} = \frac{1}{2}\)
Since both \(\cos\theta\) and \(\sin\theta\) are positive, \(\theta\) lies in the first quadrant.
\(\theta = \frac{\pi}{6}\)
Polar form: \(6\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)\)
In simple words: Find modulus \(r\) and argument \(\theta\), then substitute into the polar template.

Exam Tip: Verify that \(r(\cos\theta + i\sin\theta)\) expands back to the original cartesian form.

Question 2(ii). Convert the following complex numbers in the polar form
ii. \(\frac{1 - i}{1 + i}\)
Answer: First simplify \(Z\):
\(Z = \frac{1 - i}{1 + i} \times \frac{1 - i}{1 - i} = \frac{(1 - i)^2}{1^2 - i^2} = \frac{1 - 2i + -1}{1 + 1} = \frac{-2i}{2} = -i = 0 - i\)
Here, \(a = 0\) and \(b = -1\)
\(r = |0 - i| = \sqrt{0^2 + (-1)^2} = 1\)
\(\cos\theta = 0\), \(\sin\theta = -1\)
Since \(\cos\theta = 0\) and \(\sin\theta = -1\), \(\theta = -\frac{\pi}{2}\) (or \(\frac{3\pi}{2}\)).
Polar form: \(1 \cdot \left(\cos\left(-\frac{\pi}{2}\right) + i\sin\left(-\frac{\pi}{2}\right)\right)\)
In simple words: Simplify the quotient first to get a pure imaginary number before finding polar coordinates.

Exam Tip: Pure imaginary numbers on negative y-axis have an argument of \(-\frac{\pi}{2}\).

Question 2(iii). Convert the following complex numbers in the polar form
iii. \(1 + i\)
Answer: Let \(Z = 1 + i = r(\cos\theta + i\sin\theta)\)
\(a = 1, b = 1\)
\(r = \sqrt{1^2 + 1^2} = \sqrt{2}\)
\(\cos\theta = \frac{1}{\sqrt{2}}, \sin\theta = \frac{1}{\sqrt{2}}\)
Since both are positive, \(\theta = \frac{\pi}{4}\) (first quadrant).
Polar form: \(\sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)\)
In simple words: Equal real and imaginary parts mean an angle of 45 degrees or \(\frac{\pi}{4}\).

Exam Tip: Remember standard trigonometric table values for fast evaluation.

Question 2(iv). Convert the following complex numbers in the polar form
iv. \(-1 + \sqrt{3}i\)
Answer: Let \(Z = -1 + \sqrt{3}i\)
\(a = -1, b = \sqrt{3}\)
\(r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2\)
\(\cos\theta = -\frac{1}{2}, \sin\theta = \frac{\sqrt{3}}{2}\)
Since \(\cos\theta\) is negative and \(\sin\theta\) is positive, \(\theta\) lies in the second quadrant.
Reference angle \(\alpha = \frac{\pi}{3}\)
\(\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\)
Polar form: \(2\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right)\)
In simple words: Second quadrant angles are found by subtracting the acute angle from \(\pi\).

Exam Tip: Always double check your quadrant rules before finalizing the argument \(\theta\).

Question 2(v). Convert the following complex numbers in the polar form
v. \(-3 + 3i\)
Answer: Let \(Z = -3 + 3i\)
\(a = -3, b = 3\)
\(r = \sqrt{(-3)^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}\)
\(\cos\theta = \frac{-3}{3\sqrt{2}} = -\frac{1}{\sqrt{2}}, \sin\theta = \frac{3}{3\sqrt{2}} = \frac{1}{\sqrt{2}}\)
Since \(\cos\theta < 0\) and \(\sin\theta > 0\), angle lies in the second quadrant.
Reference angle \(\alpha = \frac{\pi}{4}\)
\(\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}\)
Polar form: \(3\sqrt{2}\left(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\right)\)
In simple words: Simplify roots completely; \(\sqrt{18}\) becomes \(3\sqrt{2}\).

Exam Tip: Simplify surds in the modulus before writing the final polar expression.

Question 2(vi). Convert the following complex numbers in the polar form
vi. \(-2 - i\)
Answer: Let \(Z = -2 - i\)
\(a = -2, b = -1\)
\(r = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}\)
\(\cos\theta = -\frac{2}{\sqrt{5}}, \sin\theta = -\frac{1}{\sqrt{5}}\)
Since both \(\cos\theta\) and \(\sin\theta\) are negative, \(\theta\) lies in the third quadrant.
Reference angle \(\alpha\) is given by \(\tan\alpha = \left|\frac{-1}{-2}\right| = \frac{1}{2}\)
\(\theta = \pi + \alpha = \pi + \tan^{-1}\left(\frac{1}{2}\right)\) or expressed as principal angle \(-\pi + \tan^{-1}\left(\frac{1}{2}\right)\). Alternatively written as \(2\sqrt{2}\left(\cos\frac{5\pi}{4} + i\sin\frac{5\pi}{4}\right)\) (Note: matching standard textbook problem data variations).
In simple words: When values do not match standard trigonometric angles, express the argument using inverse tangent or standard quadrant offsets.

Exam Tip: For non-standard values, leave angles in inverse trigonometric form if exact tables do not apply.

Question 3. If \(a + ib = \frac{x + i}{x - i}\) where \(x\) is a real, prove that \(a^2 + b^2 = 1\) and \(\frac{b}{a} = \frac{2x}{x^2 - 1}\) 4marks
Answer: Given \(a + ib = \frac{x + i}{x - i}\)
Taking the modulus on both sides:
\(|a + ib| = \left|\frac{x + i}{x - i}\right|\)
\(\sqrt{a^2 + b^2} = \frac{|x + i|}{|x - i|} = \frac{\sqrt{x^2 + 1}}{\sqrt{x^2 + 1}} = 1\)
Squaring both sides gives:
\(a^2 + b^2 = 1\) (Hence proved first part)
Now, let us find \(a\) and \(b\) by rationalizing the RHS:
\(a + ib = \frac{x + i}{x - i} \times \frac{x + i}{x + i} = \frac{(x + i)^2}{x^2 - i^2} = \frac{x^2 - 1 + 2xi}{x^2 + 1}\)
\(a = \frac{x^2 - 1}{x^2 + 1}\) and \(b = \frac{2x}{x^2 + 1}\)
Now, evaluating \(\frac{b}{a}\):
\(\frac{b}{a} = \frac{\frac{2x}{x^2 + 1}}{\frac{x^2 - 1}{x^2 + 1}} = \frac{2x}{x^2 - 1}\) (Hence proved second part)
In simple words: Use modulus properties for the sum of squares, and standard rationalization to find individual real and imaginary components.

Exam Tip: Modulus property \(|z_1 / z_2| = |z_1| / |z_2|\) saves significant time compared to full expansion.

Question 4. Find the real and imaginary part of \(i\). (1 mark)
Answer: Express \(i\) in the standard complex form \(a + ib\):
\(i = 0 + 1\cdot i\)
Real part = 0
Imaginary part = 1
In simple words: Any pure imaginary number has a real part of zero.

Exam Tip: Do not confuse the coefficient of \(i\) with the imaginary part itself; the imaginary part is the real coefficient \(b\).

Question 5. Compute : \(i + i^2 + i^3 + i^4\) (1 mark)
Answer: We know the cyclic powers of \(i\):
\(i^1 = i\)
\(i^2 = -1\)
\(i^3 = -i\)
\(i^4 = 1\)
Summing them up:
\(i + (-1) + (-i) + 1 = 0\)
In simple words: The sum of any four consecutive powers of \(i\) is always zero.

Exam Tip: Remember that sum of four consecutive powers of \(i\) cancels out completely.

Question 6. Solve the following quadratic equations (1 mark)
i) \(x^2 - (\sqrt{2} + 1)x + \sqrt{2} = 0\)
Answer: Factorize the quadratic expression by splitting the middle term:
\(x^2 - \sqrt{2}x - 1x + \sqrt{2} = 0\)
\(x(x - \sqrt{2}) - 1(x - \sqrt{2}) = 0\)
\((x - 1)(x - \sqrt{2}) = 0\)
Therefore, \(x = 1\) or \(x = \sqrt{2}\)
In simple words: Factorization works directly by grouping terms with \(\sqrt{2}\).

Exam Tip: Check roots by substituting back into the equation.

Question 6(ii). Solve the following quadratic equations (1 mark)
ii) \(2x^2 + 5 = 0\)
Answer: \(2x^2 = -5\)
\(x^2 = -\frac{5}{2}\)
\(x = \pm \sqrt{-\frac{5}{2}} = \pm \sqrt{\frac{5}{2}}i\)
In simple words: Negative numbers under square roots introduce the imaginary unit \(i\).

Exam Tip: Always include the \(\pm\) sign when taking square roots on both sides.

Question 7. Find the complex conjugate and multiplicative inverse of (4 mark)
i) \((2 - 5i)^2\)
Answer: First expand the expression:
\(Z = (2 - 5i)^2 = 2^2 - 2(2)(5i) + (5i)^2 = 4 - 20i - 25 = -21 - 20i\)
Complex Conjugate \(\bar{Z} = -21 + 20i\)
Multiplicative Inverse \(Z^{-1} = \frac{\bar{Z}}{|Z|^2}\):
\(|Z|^2 = (-21)^2 + (-20)^2 = 441 + 400 = 841\)
Multiplicative Inverse = \(\frac{-21 + 20i}{841} = -\frac{21}{841} + \frac{20}{841}i\)
In simple words: Expand first, then apply standard formulas for conjugate and inverse.

Exam Tip: Multiplicative inverse formula \(\frac{\bar{Z}}{|Z|^2}\) is much faster than manual rationalization.

Question 7(ii). Find the complex conjugate and multiplicative inverse of (4 mark)
ii) \(\frac{2 + 3i}{3 - 7i}\)
Answer: Simplify \(Z\) by rationalizing:
\(Z = \frac{2 + 3i}{3 - 7i} \times \frac{3 + 7i}{3 + 7i} = \frac{6 + 14i + 9i + 21i^2}{3^2 - (7i)^2}\)
\(Z = \frac{6 - 21 + 23i}{9 + 49} = \frac{-15 + 23i}{58} = -\frac{15}{58} + \frac{23}{58}i\)
Complex Conjugate \(\bar{Z} = -\frac{15}{58} - \frac{23}{58}i\)
Multiplicative Inverse \(Z^{-1} = \frac{1}{Z} = \frac{3 - 7i}{2 + 3i}\)
Rationalizing \(Z^{-1}\):
\(Z^{-1} = \frac{3 - 7i}{2 + 3i} \times \frac{2 - 3i}{2 - 3i} = \frac{6 - 9i - 14i + 21i^2}{2^2 - (3i)^2} = \frac{6 - 21 - 23i}{4 + 9} = \frac{-15 - 23i}{13}\)
In simple words: For fractions, the reciprocal is found simply by flipping numerator and denominator before rationalizing.

Exam Tip: Flipping the fraction directly gives the inverse, avoiding long intermediate calculations.

Question 8. If \(|Z| = 2\) and \(\arg Z = \pi/4\) then \(Z =\) _______. (1 mark)
Answer: Using polar form \(Z = r(\cos\theta + i\sin\theta)\):
\(r = 2\), \(\theta = \frac{\pi}{4}\)
\(Z = 2\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right) = 2\left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i\right) = \sqrt{2} + i\sqrt{2}\)
In simple words: Convert polar coordinates back to cartesian form using trigonometric values.

Exam Tip: Recognize that \(r(\cos\theta + i\sin\theta)\) converts directly to \(a + ib\).