CBSE Class 11 Mathematics Complex Numbers And Quadratic Equation Worksheet Set 04

CBSE Class 11 Mathematics Worksheet - Complex Numbers and Quadratic Equation (3). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Question. Convert the given complex number into polar form \( z = \frac{1+7i}{(2-i)^2} \)
Answer: \( z = \frac{1 + 7i}{(2 - i)^2} \)

First, we convert it to standard form:
\( z = \frac{1 + 7i}{4 + i^2 - 4i} \)
\( z = \frac{1 + 7i}{4 - 1 - 4i} \qquad [\because i^2 = -1] \)
\( z = \frac{1 + 7i}{3 - 4i} \)

Now, rationalize the fraction:
\( z = \frac{1 + 7i}{3 - 4i} \times \frac{3 + 4i}{3 + 4i} \)
\( z = \frac{3 + 4i + 21i + 28i^2}{9 - 16i^2} \)
\( z = \frac{3 + 25i - 28}{9 + 16} \)
\( z = \frac{-25 + 25i}{25} \)
\( z = -1 + i \) (standard form)

Here, \( a = -1 \) and \( b = 1 \).
Now, \( r = |z| = \sqrt{a^2 + b^2} \)
\( r = \sqrt{(-1)^2 + 1^2} = \sqrt{2} \)
\( \tan \alpha = \left|\frac{b}{a}\right| = \left|\frac{1}{-1}\right| = |-1| = 1 \)
\( \Rightarrow \tan \alpha = 1 \)
\( \Rightarrow \alpha = \frac{\pi}{4} \)

Since \( z \) is in the \( 2^{\text{nd}} \) quadrant:
\( \therefore \theta = \pi - \alpha \)
\( \theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \)

Polar form: \( z = r(\cos \theta + i \sin \theta) \)
\( z = \sqrt{2} \left( \cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4} \right) \) ans.

Question. Convert the given complex number in to polar form \( z = \frac{-16}{1+i\sqrt{3}} \)
Answer: We have, \( z = \frac{-16}{1+i\sqrt{3}} \)

Rationalizing \( z \):
\( z = \frac{-16}{1+i\sqrt{3}} \times \frac{1-i\sqrt{3}}{1-i\sqrt{3}} \)
\( z = \frac{-16(1-i\sqrt{3})}{1 - 3i^2} \)
\( z = \frac{-16 + i16\sqrt{3}}{4} \)
\( z = -4 + i4\sqrt{3} \)

Here, \( a = -4 \) and \( b = 4\sqrt{3} \).
Now, \( r = \sqrt{a^2 + b^2} = \sqrt{(-4)^2 + (4\sqrt{3})^2} = \sqrt{16 + 48} = \sqrt{64} = 8 \)
\( \therefore r = 8 \)
\( \tan \alpha = \left|\frac{b}{a}\right| = \left|\frac{4\sqrt{3}}{-4}\right| = |-\sqrt{3}| = \sqrt{3} \)
\( \Rightarrow \tan \alpha = \sqrt{3} \)
\( \Rightarrow \alpha = \frac{\pi}{3} \)

Since \( z \) is in the \( 2^{\text{nd}} \) quadrant:
\( \therefore \theta = \pi - \alpha \)
\( \theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \)

Polar form: \( z = r(\cos \theta + i \sin \theta) \)
\( z = 8 \left(\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}\right) \) ans.

Question. Convert the given complex number in to polar form \( z = \frac{i-1}{\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}} \)
Answer: We have, \( z = \frac{i-1}{\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}} \)
\( \Rightarrow z = \frac{-1+i}{\frac{1}{2} + \frac{i\sqrt{3}}{2}} \)
\( \Rightarrow z = \frac{-2+2i}{1+i\sqrt{3}} \)

Rationalizing \( z \):
\( z = \frac{-2+2i}{1+i\sqrt{3}} \times \frac{1-i\sqrt{3}}{1-i\sqrt{3}} \)
\( z = \frac{-2 + 2\sqrt{3}i + 2i - 2\sqrt{3}i^2}{1 - 3i^2} \)
\( z = \frac{-2 + 2\sqrt{3}i + 2i + 2\sqrt{3}}{1 + 3} \)
\( z = \frac{(2\sqrt{3}-2) + i(2\sqrt{3}+2)}{4} \)
\( z = \left(\frac{\sqrt{3}-1}{2}\right) + i\left(\frac{\sqrt{3}+1}{2}\right) \)

Here, \( a = \frac{\sqrt{3}-1}{2} \) and \( b = \frac{\sqrt{3}+1}{2} \).
Now, \( r = \sqrt{a^2 + b^2} \):
\( r = \sqrt{\left(\frac{\sqrt{3}-1}{2}\right)^2 + \left(\frac{\sqrt{3}+1}{2}\right)^2} \)
\( \Rightarrow r = \sqrt{\frac{3+1-2\sqrt{3}}{4} + \frac{3+1+2\sqrt{3}}{4}} \)
\( \Rightarrow r = \sqrt{\frac{8}{4}} = \sqrt{2} \)
\( \Rightarrow r = \sqrt{2} \)

\( \tan \alpha = \left|\frac{b}{a}\right| = \left|\frac{\frac{\sqrt{3} + 1}{2}}{\frac{\sqrt{3} - 1}{2}}\right| \)
\( \Rightarrow \tan \alpha = \frac{\sqrt{3}+1}{\sqrt{3}-1} \)
Dividing the numerator and denominator by \( \sqrt{3} \):
\( \Rightarrow \tan \alpha = \frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}} \)
\( \Rightarrow \tan \alpha = \frac{\tan\left(\frac{\pi}{4}\right)+\tan\left(\frac{\pi}{6}\right)}{1-\tan\left(\frac{\pi}{4}\right)\tan\left(\frac{\pi}{6}\right)} \qquad [ \because \tan(A+B) = \frac{\tan A+\tan B}{1-\tan A\tan B} ] \)
\( \Rightarrow \tan \alpha = \tan\left(\frac{\pi}{4} + \frac{\pi}{6}\right) \)
\( \Rightarrow \tan \alpha = \tan\left(\frac{5\pi}{12}\right) \)
\( \Rightarrow \alpha = \frac{5\pi}{12} \pmb{\text{ }} \)

Since \( z \) is in the \( 1^{\text{st}} \) quadrant:
\( \therefore \theta = \alpha = \frac{5\pi}{12} \)
Polar form: \( z = r(\cos \theta + i \sin \theta) \)
\( z = \sqrt{2} \left(\cos \frac{5\pi}{12} + i \sin \frac{5\pi}{12}\right) \) ans.

Question. Finding the value of a POLYNOMIAL:- If \( x = -5 + 2\sqrt{-4} \). Find the value of \( x^4 + 9x^3 + 35x^2 - x + 4 \)
Answer: We have, \( x = -5 + 2\sqrt{-4} \)
\( \Rightarrow x = -5 + 4i \qquad [ \because \sqrt{-4} = 2i ] \)
\( \Rightarrow x + 5 = 4i \)
Squaring both sides:
\( \Rightarrow (x + 5)^2 = (4i)^2 \)
\( \Rightarrow x^2 + 25 + 10x = -16 \)
\( \Rightarrow x^2 + 10x + 41 = 0 \) ............ (i)

Dividing \( x^4 + 9x^3 + 35x^2 - x + 4 \) by \( x^2 + 10x + 41 \) using long division:
We obtain:
\( x^4 + 9x^3 + 35x^2 - x + 4 = (x^2 + 10x + 41)(x^2 - x + 4) - 160 \)
\( \Rightarrow x^4 + 9x^3 + 35x^2 - x + 4 = 0(x^2 - x + 4) - 160 \qquad [ \text{using eq. (i)} ] \)
\( = -160 \) ans.

Question. Find the value of \( x^3 + 7x^2 - x + 16 \) when \( x = 1 + 2i \)
Answer: We have, \( x = 1 + 2i \)
\( \Rightarrow x - 1 = 2i \)
Squaring both sides:
\( \Rightarrow x^2 + 1 - 2x = -4 \)
\( \Rightarrow x^2 - 2x + 5 = 0 \) ............ eq.(i)

Dividing \( x^3 + 7x^2 - x + 16 \) by \( x^2 - 2x + 5 \) using long division:
\( = (x^2 - 2x + 5)(x + 9) + 12x - 29 \)
\( \Rightarrow x^3 + 7x^2 - x + 16 = 0(x + 9) + 12x - 29 \qquad [ \text{using eq. (i)} ] \)
\( = 12x - 29 \)
\( = 12(1 + 2i) - 29 \)
\( = 12 + 24i - 29 \)
\( = -17 + 24i \) ans.

Question. Square Root Of A Complex Number:- Find \( \sqrt{-15 - 8i} \)
Answer: Let \( x + iy = \sqrt{-15 - 8i} \)
Squaring both sides:
\( x^2 - y^2 + 2ixy = -15 - 8i \)
Equating real & imaginary parts:
\( \Rightarrow x^2 - y^2 = -15 \) ............ (i)
\( 2xy = -8 \) ............ (ii)

Using the identity:
\( (x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2 \)
\( (x^2 + y^2)^2 = (-15)^2 + (-8)^2 \)
\( (x^2 + y^2)^2 = 225 + 64 = 289 \)
\( \Rightarrow x^2 + y^2 = 17 \) ............ (ii)

Adding eq. (i) & (ii):
\( 2x^2 = -15 + 17 = 2 \)
\( \Rightarrow x^2 = 1 \Rightarrow x = \pm 1 \)

Substituting \( x^2 = 1 \) into eq. (ii):
\( 1 + y^2 = 17 \Rightarrow y^2 = 16 \Rightarrow y = \pm 4 \)

Since \( 2xy = -8 \) (imaginary part is negative), \( x \) and \( y \) must be of opposite signs.
\( \therefore \sqrt{-15 - 8i} = \pm(1 - 4i) \) ans.

Question. Find the value of square root of \( 1 - i \)
Answer: Let \( x + iy = \sqrt{1 - i} \)
Squaring both sides:
\( x^2 - y^2 + 2ixy = 1 - i \)
Equating real & imaginary parts:
\( \Rightarrow x^2 - y^2 = 1 \) ............ (i)
\( 2xy = -1 \)

Using the identity:
\( (x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2 \)
\( (x^2 + y^2)^2 = 1 + 1 = 2 \)
\( \Rightarrow x^2 + y^2 = \sqrt{2} \) ............ (ii)

Adding eq. (i) & (ii):
\( 2x^2 = \sqrt{2} + 1 \)
\( \Rightarrow x^2 = \frac{\sqrt{2} + 1}{2} \)
\( \Rightarrow x = \pm \sqrt{\frac{\sqrt{2} + 1}{2}} \)

Substituting \( x^2 = \frac{\sqrt{2} + 1}{2} \) into eq. (ii):
\( \frac{\sqrt{2} + 1}{2} + y^2 = \sqrt{2} \)
\( \Rightarrow y^2 = \sqrt{2} - \frac{\sqrt{2} + 1}{2} = \frac{\sqrt{2} - 1}{2} \)
\( \Rightarrow y = \pm \sqrt{\frac{\sqrt{2} - 1}{2}} \)

Since \( 2xy = -1 \) (imaginary part is negative), \( x \) and \( y \) must have opposite signs:
\( \therefore \sqrt{1 - i} = \pm \left( \sqrt{\frac{\sqrt{2} + 1}{2}} - i \sqrt{\frac{\sqrt{2} - 1}{2}} \right) \) ans.

Question. On Equality:- Find real value of \( x + y \) for which the complex numbers \( -3 + ix^2y \) and \( x^2 + y + 4i \) are conjugate of each other.
Answer: Given that \( -3 + ix^2y \) and \( x^2 + y + 4i \) are conjugate of each other:
\( \Rightarrow -3 + ix^2y = \overline{(x^2 + y) + 4i} \)
\( \Rightarrow -3 + ix^2y = (x^2 + y) - 4i \)
Equating real & imaginary parts:
\( \Rightarrow x^2 + y = -3 \) and \( x^2y = -4 \)
Substituting \( x^2 = -3 - y \) into the second equation:
\( \Rightarrow (-3 - y)y = -4 \)
\( \Rightarrow y^2 + 3y - 4 = 0 \)
\( \Rightarrow (y + 4)(y - 1) = 0 \)
\( \Rightarrow y = -4 \) or \( y = 1 \)

Now, solving for \( x \):
- When \( y = -4 \):
\( x^2(-4) = -4 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1 \)
- When \( y = 1 \):
\( x^2(1) = -4 \Rightarrow x^2 = -4 \Rightarrow x = \pm 2i \) (rejected as \( x \) must be real)
\( \therefore x = \pm 1 \) and \( y = -4 \) ans.

Question. If \( (x + iy)^{\frac{1}{3}} = a + ib \), show that \( \frac{x}{a} + \frac{y}{b} = 4(a^2 - b^2) \)
Answer: We have, \( (x + iy)^{\frac{1}{3}} = a + ib \)
Cubing both sides:
\( x + iy = (a + ib)^3 \)
\( \Rightarrow x + iy = a^3 + (ib)^3 + 3a^2(ib) + 3a(ib)^2 \)
\( \Rightarrow x + iy = a^3 - ib^3 + i3a^2b - 3ab^2 \)
\( \Rightarrow x + iy = (a^3 - 3ab^2) + i(3a^2b - b^3) \)

Equating real & imaginary parts:
\( x = a^3 - 3ab^2 \Rightarrow \frac{x}{a} = a^2 - 3b^2 \)
\( y = 3a^2b - b^3 \Rightarrow \frac{y}{b} = 3a^2 - b^2 \)

Adding the two simplified equations:
\( \frac{x}{a} + \frac{y}{b} = (a^2 - 3b^2) + (3a^2 - b^2) \)
\( = 4a^2 - 4b^2 \)
\( = 4(a^2 - b^2) = \text{R.H.S.} \) (proved)

Question. If \( a + ib = \frac{c+i}{c-i} \) show that \( \frac{b}{a} = \frac{2c}{c^2-1} \) and hence show \( a^2 + b^2 = 1 \)
Answer: We have, \( a + ib = \frac{c+i}{c-i} \)
Rationalizing the denominator:
\( a + ib = \frac{c + i}{c - i} \times \frac{c + i}{c + i} \)
\( \Rightarrow a + ib = \frac{(c+i)^2}{c^2 - i^2} \)
\( \Rightarrow a + ib = \frac{c^2 + i^2 + 2ic}{c^2 + 1} \)
\( \Rightarrow a + ib = \frac{c^2 - 1}{c^2 + 1} + i \frac{2c}{c^2 + 1} \)

Equating real & imaginary parts:
\( a = \frac{c^2 - 1}{c^2 + 1} \) and \( b = \frac{2c}{c^2 + 1} \)

Now, calculating \( \frac{b}{a} \):
\( \frac{b}{a} = \frac{\frac{2c}{c^2 + 1}}{\frac{c^2 - 1}{c^2 + 1}} = \frac{2c}{c^2 - 1} \) (proved)

Now, evaluating \( a^2 + b^2 \):
\( a^2 + b^2 = \left( \frac{c^2 - 1}{c^2 + 1} \right)^2 + \left( \frac{2c}{c^2 + 1} \right)^2 \)
\( = \frac{(c^2 - 1)^2 + (2c)^2}{(c^2 + 1)^2} \)
\( = \frac{c^4 + 1 - 2c^2 + 4c^2}{(c^2 + 1)^2} \)
\( = \frac{c^4 + 2c^2 + 1}{(c^2 + 1)^2} \)
\( = \frac{(c^2 + 1)^2}{(c^2 + 1)^2} = 1 \) (proved)