CBSE Class 11 Mathematics Complex Numbers And Quadratic Equation Worksheet Set 02

Question 1. Solve the equation \(2x^2 + x + 1 = 0\). (2 marks)
Answer: Comparing the given quadratic equation with \(ax^2 + bx + c = 0\), we get \(a = 2\), \(b = 1\), and \(c = 1\).
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\(x = \frac{-1 \pm \sqrt{1^2 - 4(2)(1)}}{2(2)}\)
\(x = \frac{-1 \pm \sqrt{1 - 8}}{4}\)
\(x = \frac{-1 \pm \sqrt{-7}}{4}\)
\(x = \frac{-1 \pm i\sqrt{7}}{4}\)
In simple words: We used the quadratic formula for equations with negative discriminants to find complex roots containing \(i\).

Exam Tip: Always remember to substitute \(\sqrt{-1} = i\) when dealing with a negative value inside the square root.

Question 2. Evaluate: \(\left[i^{18} + \left(\frac{1}{i}\right)^{25}\right]^3\)
Answer: First, simplify each term using powers of \(i\):
\(i^{18} = i^{4 \times 4 + 2} = (i^4)^4 \cdot i^2 = (1)^4 \cdot (-1) = -1\)
\(\left(\frac{1}{i}\right)^{25} = \left(\frac{1}{i} \cdot \frac{i}{i}\right)^{25} = (-i)^{25} = -(i^{24} \cdot i) = -((i^4)^6 \cdot i) = -(1 \cdot i) = -i\)
Substitute these values back into the expression:
\(\left[i^{18} + \left(\frac{1}{i}\right)^{25}\right]^3 = (-1 + (-i))^3 = [-(1 + i)]^3 = -(1 + i)^3\)
Using the identity \((a + b)^3 = a^3 + b^3 + 3ab(a + b)\):
\((1 + i)^3 = 1^3 + i^3 + 3(1)(i)(1 + i) = 1 - i + 3i + 3i^2 = 1 + 2i - 3 = -2 + 2i\)
Therefore, \(-(1 + i)^3 = -(-2 + 2i) = 2 - 2i\)
In simple words: Simplify the powers of \(i\) using \(i^4 = 1\) and expand the cube of the binomial.

Exam Tip: Remember periodic properties of \(i\): \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), and \(i^4 = 1\).

Question 3. Convert the given complex number in polar form: \(-3\). (3 marks)
Answer: Let \(z = -3 + 0i\).
The modulus \(r\) is given by:
\(r = |z| = \sqrt{(-3)^2 + 0^2} = \sqrt{9} = 3\)
The polar form is \(z = r(\cos \theta + i \sin \theta)\).
Here, \(\cos \theta = \frac{-3}{3} = -1\) and \(\sin \theta = \frac{0}{3} = 0\).
Since the point \((-3, 0)\) lies on the negative real axis, the principal argument \(\theta\) is \(\pi\).
Therefore, the polar form is \(3(\cos \pi + i \sin \pi\)).
In simple words: Find the magnitude and the angle made with the positive real axis to write the complex number in trigonometric form.

Exam Tip: Always check the quadrant of the complex number before determining its argument.

Question 4. Express the given complex number in the form \(a + ib\): \((1 - i) - (-1 + i6)\). (1 mark)
Answer: Open the parentheses and combine real and imaginary parts:
\((1 - i) - (-1 + 6i) = 1 - i + 1 - 6i\)
\(= (1 + 1) + (-1 - 6)i\)
\(= 2 - 7i\)
In simple words: Distribute the negative sign and add like terms.

Exam Tip: Ensure the final answer is explicitly written in the standard \(a + ib\) format.

Question 5. Express \((-\sqrt{3} + \sqrt{-2})(2\sqrt{3} - i)\) in the form \(a + ib\).
Answer: Rewrite \(\sqrt{-2}\) as \(i\sqrt{2}\):
\((-\sqrt{3} + i\sqrt{2})(2\sqrt{3} - i)\)
Expand the product using standard binomial multiplication:
\(= (-\sqrt{3})(2\sqrt{3}) + (-\sqrt{3})(-i) + (i\sqrt{2})(2\sqrt{3}) + (i\sqrt{2})(-i)\)
\(= -6 + i\sqrt{3} + 2\sqrt{6}i - i^2\sqrt{2}\)
Since \(i^2 = -1\):
\(= -6 + i\sqrt{3} + 2\sqrt{6}i + \sqrt{2}\)
Group the real and imaginary terms:
\(= (-6 + \sqrt{2}) + (\sqrt{3} + 2\sqrt{6})i\)
In simple words: Convert negative square roots into imaginary numbers and multiply like ordinary binomials.

Exam Tip: Substitute \(i^2 = -1\) carefully during multiplication.

Question 6. Evaluate: \((- \sqrt{-1})^{4n + 3}\). (1 mark)
Answer: We know that \(\sqrt{-1} = i\). Therefore:
\((- \sqrt{-1})^{4n + 3} = (-i)^{4n + 3}\)
\(= (-1)^{4n + 3} \cdot i^{4n + 3}\)
Since \(4n + 3\) is odd, \((-1)^{4n + 3} = -1\).
For the power of \(i\):
\(i^{4n + 3} = i^{4n} \cdot i^3 = (i^4)^n \cdot (-i) = (1)^n \cdot (-i) = -i\)
Multiplying both parts together:
\(-1 \cdot (-i) = i\)
In simple words: Break down the exponents using rules of indices and simplify powers of \(i\).

Exam Tip: Remember that \(i^{4n} = 1\) for any integer \(n\).

Question 7. Express the given complex number in the form \(a + ib\): \(\left(\frac{1}{3} + 3i\right)^3\). (3 marks)
Answer: Use the binomial cube formula \((a + b)^3 = a^3 + b^3 + 3ab(a + b)\):
\(\left(\frac{1}{3} + 3i\right)^3 = \left(\frac{1}{3}\right)^3 + (3i)^3 + 3\left(\frac{1}{3}\right)(3i)\left(\frac{1}{3} + 3i\right)\)
\(= \frac{1}{27} + 27i^3 + 3i\left(\frac{1}{3} + 3i\right)\)
Since \(i^3 = -i\):
\(= \frac{1}{27} - 27i + i + 9i^2\)
Substitute \(i^2 = -1\):
\(= \frac{1}{27} - 26i - 9\)
Combine real terms:
\(= \left(\frac{1}{27} - 9\right) - 26i\)
\(= \left(\frac{1 - 243}{27}\right) - 26i\)
\(= -\frac{242}{27} - 26i\)
In simple words: Apply the algebraic cube formula and replace \(i^2\) with \(-1\) and \(i^3\) with \(-i\).

Exam Tip: Expand carefully and simplify fractions step by step to avoid arithmetic errors.

Question 8. Find the multiplicative inverse of the complex number \(\sqrt{5} + 3i\). (2 marks)
Answer: Let \(z = \sqrt{5} + 3i\). The multiplicative inverse is given by \(z^{-1} = \frac{1}{z}\):
\(z^{-1} = \frac{1}{\sqrt{5} + 3i}\)
Rationalize the denominator by multiplying the numerator and denominator by the conjugate \(\sqrt{5} - 3i\):
\(z^{-1} = \frac{\sqrt{5} - 3i}{(\sqrt{5} + 3i)(\sqrt{5} - 3i)}\)
Using \((a + b)(a - b) = a^2 - b^2\):
\(z^{-1} = \frac{\sqrt{5} - 3i}{(\sqrt{5})^2 - (3i)^2}\)
\(z^{-1} = \frac{\sqrt{5} - 3i}{5 - 9i^2}\)
Since \(i^2 = -1\):
\(z^{-1} = \frac{\sqrt{5} - 3i}{5 - 9(-1)}\)
\(z^{-1} = \frac{\sqrt{5} - 3i}{5 + 9} = \frac{\sqrt{5} - 3i}{14}\)
\(= \frac{\sqrt{5}}{14} - \frac{3}{14}i\)
In simple words: Multiply numerator and denominator by the complex conjugate of the denominator.

Exam Tip: The denominator will always simplify to a real number equal to the sum of squares of the real and imaginary parts.

Question 9. Express the given complex number in the form \(a + ib\): \((1 - i)^4\). (2 marks)
Answer: Rewrite \((1 - i)^4\) as \(\left[(1 - i)^2\right]^2\):
Expand \((1 - i)^2\):
\((1 - i)^2 = 1^2 - 2(1)(i) + i^2 = 1 - 2i - 1 = -2i\)
Now, square the result:
\(\left[(1 - i)^2\right]^2 = (-2i)^2 = 4i^2\)
Since \(i^2 = -1\):
\(= 4(-1) = -4\)
Written in \(a + ib\) form: \(-4 + 0i\)
In simple words: Squaring twice makes the calculation much easier than expanding with the binomial theorem.

Exam Tip: Whenever a high even power is given, square the base first to simplify.

Question 10. Find the multiplicative inverse of the complex number \(-i\). (1 mark)
Answer: Let \(z = -i\). The multiplicative inverse is:
\(z^{-1} = \frac{1}{-i}\)
Multiply numerator and denominator by \(i\):
\(z^{-1} = \frac{i}{-i^2}\)
Since \(i^2 = -1\):
\(z^{-1} = \frac{i}{-(-1)} = \frac{i}{1} = i\)
In simple words: Take the reciprocal and simplify using \(i^2 = -1\).

Exam Tip: The multiplicative inverse of \(-i\) is simply \(i\), and that of \(i\) is \(-i\).

Question 11. If \(x - iy = \sqrt{\frac{a - ib}{c - id}}\), then prove that \((x^2 + y^2)^2 = \frac{a^2 + b^2}{c^2 + d^2}\). (5 marks)
Answer: Given:\(x - iy = \sqrt{\frac{a - ib}{c - id}}\)
Taking the complex conjugate on both sides:
\(x + iy = \sqrt{\frac{a + ib}{c + id}}\)
Multiplying the given equation and its conjugate:
\((x - iy)(x + iy) = \left(\sqrt{\frac{a - ib}{c - id}}\right) \left(\sqrt{\frac{a + ib}{c + id}}\right)\)
\(x^2 - (iy)^2 = \sqrt{\frac{(a - ib)(a + ib)}{(c - id)(c + id)}}\)
\(x^2 + y^2 = \sqrt{\frac{a^2 + b^2}{c^2 + d^2}}\)
Squaring both sides:
\((x^2 + y^2)^2 = \frac{a^2 + b^2}{c^2 + d^2}\)
Hence proved.
In simple words: Multiply the given equation by its complex conjugate to eliminate imaginary units, then square both sides.

Exam Tip: This is a standard property of moduli: \(|z_1 \cdot z_2| = |z_1| \cdot |z_2|\).

Question 12. Convert the complex number \(z = \frac{1 - i}{\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}}\) in the polar form.
Answer: First, write the denominator in exponential/polar form:\(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3} = e^{i\pi/3}\).
Numerator \(1 - i\) in polar form:
Magnitude \(r_1 = \sqrt{1^2 + (-1)^2} = \sqrt{2}\)
Argument \(\theta_1 = -\frac{\pi}{4}\) (since it lies in the fourth quadrant).
So, \(1 - i = \sqrt{2}\left(\cos\left(-\frac{\pi}{4}\right) + i \sin\left(-\frac{\pi}{4}\right)\right)\).
Now, substitute both into \(z\):
\(z = \frac{\sqrt{2}\left(\cos\left(-\frac{\pi}{4}\right) + i \sin\left(-\frac{\pi}{4}\right)\right)}{\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}}\)
Using division in polar form (subtracting arguments):
Magnitude \(r = \sqrt{2}\)
Argument \(\theta = -\frac{\pi}{4} - \frac{\pi}{3} = -\frac{7\pi}{12}\)
Therefore, the polar form is \(\sqrt{2}\left(\cos\left(-\frac{7\pi}{12}\right) + i \sin\left(-\frac{7\pi}{12}\right)\right)\).
In simple words: Convert the numerator and denominator separately into polar form, then divide their moduli and subtract their arguments.

Exam Tip: When dividing complex numbers in polar form, remember to subtract the angle of the denominator from the angle of the numerator.

Question 13. Solve: \(x^2 + 2 = 0\)
Answer: Given equation: \(x^2 + 2 = 0\)
\(x^2 = -2\)
Taking square roots on both sides:
\(x = \pm \sqrt{-2}\)
\(x = \pm i\sqrt{2}\)
In simple words: Isolate \(x^2\) and introduce \(i\) for the negative sign.

Exam Tip: Always include both positive and negative roots when taking a square root.

Question 14. Solve \(4x^2 - 25i^2 = 0\). (1 mark)
Answer: Given equation: \(4x^2 - 25i^2 = 0\)
Substitute \(i^2 = -1\):
\(4x^2 - 25(-1) = 0\)
\(4x^2 + 25 = 0\)
\(4x^2 = -25\)
\(x^2 = -\frac{25}{4}\)
\(x = \pm \sqrt{-\frac{25}{4}}\)
\(x = \pm \frac{5}{2}i\)
In simple words: Replace \(i^2\) with \(-1\) and solve for \(x\).

Exam Tip: Recognizing \(i^2 = -1\) immediately simplifies expressions containing \(i^2\).

Question 15. Find the argument of \(1 + i\sqrt{3}\). (1 mark)
Answer: Let \(z = 1 + i\sqrt{3}\). Here, real part \(x = 1\) and imaginary part \(y = \sqrt{3}\).
Since both \(x\) and \(y\) are positive, the complex number lies in the first quadrant.
The acute angle \(\alpha\) is given by:
\(\tan \alpha = \left|\frac{y}{x}\right| = \left|\frac{\sqrt{3}}{1}\right| = \sqrt{3}\)
\(\alpha = \frac{\pi}{3}\)
Since it is in the first quadrant, the argument \(\theta = \alpha = \frac{\pi}{3}\).
In simple words: Find the reference angle using the ratio of the imaginary part to the real part.

Exam Tip: For the first quadrant, the argument is simply \(\tan^{-1}\left(\frac{y}{x}\right)\).

Question 16. Express \(\left[\left(\frac{1}{3} + i\frac{7}{3}\right) + \left(4 + i\frac{1}{3}\right)\right]\) in the form \(a + ib\).
Answer: Group the real parts together and the imaginary parts together:
\(= \left(\frac{1}{3} + 4\right) + i\left(\frac{7}{3} + \frac{1}{3}\right)\)
\(= \left(\frac{1 + 12}{3}\right) + i\left(\frac{8}{3}\right)\)
\(= \frac{13}{3} + \frac{8}{3}i\)
In simple words: Combine all real numbers together and all terms with \(i\) together.

Exam Tip: Simple addition of complex numbers only requires combining like terms.

Question 17. Express \(i^9 + i^{10} + i^{11} + i^{12}\) in the form \(a + bi\).
Answer: Simplify each term by dividing the exponent by 4:
\(i^9 = i^{4 \times 2 + 1} = i\)
\(i^{10} = i^{4 \times 2 + 2} = -1\)
\(i^{11} = i^{4 \times 2 + 3} = -i\)
\(i^{12} = i^{4 \times 3} = 1\)
Substitute these values into the expression:
\(i - 1 - i + 1 = 0\)
Written in \(a + bi\) form: \(0 + 0i\)
In simple words: The sum of four consecutive powers of \(i\) is always zero.

Exam Tip: Remember the general rule that \(i^n + i^{n+1} + i^{n+2} + i^{n+3} = 0\) for any integer \(n\).

Question 18. Express: \(i^9 + i^{19}\) in the form \(a + bi\).
Answer: Simplify each power of \(i\):
\(i^9 = i^{4 \times 2 + 1} = i\)
\(i^{19} = i^{4 \times 4 + 3} = -i\)
Add the simplified terms:
\(i + (-i) = 0\)
Written in \(a + bi\) form: \(0 + 0i\)
In simple words: Reduce powers of \(i\) by dividing exponents by 4 and taking the remainder.

Exam Tip: \(i^4 = 1\) is the key to simplifying any large power of \(i\).

Question 19. Solve the quadratic equation \(25x^2 - 30x + 11 = 0\). (2 marks)
Answer: Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 25\), \(b = -30\), and \(c = 11\):
\(x = \frac{-(-30) \pm \sqrt{(-30)^2 - 4(25)(11)}}{2(25)}\)
\(x = \frac{30 \pm \sqrt{900 - 1100}}{50}\)
\(x = \frac{30 \pm \sqrt{-200}}{50}\)
\(x = \frac{30 \pm 10i\sqrt{2}}{50}\)
Divide numerator and denominator by 10:
\(x = \frac{3 \pm i\sqrt{2}}{5} = \frac{3}{5} \pm \frac{\sqrt{2}}{5}i\)
In simple words: Apply the quadratic formula and simplify the resulting complex roots.

Exam Tip: Simplify radical terms fully by factoring out perfect squares.

Question 20. Write the conjugate of complex number \(-5 + 3i\). (1 mark)
Answer: The conjugate of a complex number \(z = a + ib\) is given by \(\bar{z} = a - ib\).
Therefore, the conjugate of \(-5 + 3i\) is \(-5 - 3i\).
In simple words: Change the sign of the imaginary part to find the conjugate.

Exam Tip: The real part remains unchanged when finding the complex conjugate.