CBSE Class 11 Mathematics Complex Numbers And Quadratic Equation Worksheet Set 05

CBSE Class 11 Mathematics Worksheet - Complex Numbers and Quadratic Equation (4). 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.

Quadratic Equation:-

Question. Find all non-zero complex numbers \( z \) satisfying \( \bar{z} = iz^2 \)
Answer: Let \( z = x + iy \)
We have, \( \bar{z} = iz^2 \)
\( \Rightarrow x - iy = i(x + iy)^2 \)
\( \Rightarrow x - iy = i(x^2 - y^2 + 2ixy) \)
\( \Rightarrow x - iy = ix^2 - iy^2 - 2xy \)
\( \Rightarrow x - iy = -2xy + i(x^2 - y^2) \)

Equating real & imaginary parts:
\( \Rightarrow x = -2xy \) and \( -y = x^2 - y^2 \)
\( \Rightarrow x + 2xy = 0 \) and \( x^2 - y^2 + y = 0 \)

Consider \( x + 2xy = 0 \):
\( \Rightarrow x(1 + 2y) = 0 \)
\( \Rightarrow x = 0 \) or \( y = -\frac{1}{2} \)

Case 1: When \( x = 0 \)
\( \Rightarrow x^2 - y^2 + y = 0 \)
\( \Rightarrow -y^2 + y = 0 \)
\( \Rightarrow y(-y + 1) = 0 \)
\( \Rightarrow y = 0 \) or \( y = 1 \)
\( \therefore x = 0, y = 0 \) and \( x = 0, y = 1 \)
\( \Rightarrow z = 0 + 0i \) and \( z = 0 + 1i \)
Since we need non-zero complex numbers, \( z = 0 + 0i \) is rejected.
\( \therefore z = i \)

Case 2: When \( y = -\frac{1}{2} \)
\( \Rightarrow x^2 - y^2 + y = 0 \)
\( \Rightarrow x^2 - \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) = 0 \)
\( \Rightarrow x^2 - \frac{1}{4} - \frac{1}{2} = 0 \)
\( \Rightarrow x^2 = \frac{3}{4} \)
\( \Rightarrow x = \pm \frac{\sqrt{3}}{2} \)
\( \therefore x = \frac{\sqrt{3}}{2}, y = -\frac{1}{2} \) and \( x = -\frac{\sqrt{3}}{2}, y = -\frac{1}{2} \)
\( \Rightarrow z = \frac{\sqrt{3}}{2} - \frac{1}{2}i \) and \( z = -\frac{\sqrt{3}}{2} - \frac{1}{2}i \)

\( \therefore \) Required non-zero complex numbers are \( i \), \( \frac{\sqrt{3}}{2} - \frac{1}{2}i \) and \( -\frac{\sqrt{3}}{2} - \frac{1}{2}i \) ans.

Question. Solve the equation \( z^2 + |z| = 0 \).
Answer: Let \( z = x + iy \)
We have, \( z^2 + |z| = 0 \)
\( \Rightarrow (x + iy)^2 + \sqrt{x^2 + y^2} = 0 \)
\( \Rightarrow x^2 - y^2 + 2ixy + \sqrt{x^2 + y^2} = 0 \)
\( \Rightarrow (x^2 - y^2 + \sqrt{x^2 + y^2}) + i(2xy) = 0 + 0i \)

Equating real & imaginary parts:
\( \Rightarrow x^2 - y^2 + \sqrt{x^2 + y^2} = 0 \) and \( 2xy = 0 \)

Consider \( 2xy = 0 \Rightarrow x = 0 \) or \( y = 0 \)

Case 1: When \( x = 0 \)
Then \( x^2 - y^2 + \sqrt{x^2 + y^2} = 0 \)
\( \Rightarrow -y^2 + \sqrt{y^2} = 0 \)
\( \Rightarrow -y^2 + |y| = 0 \)

If \( y > 0 \), then \( |y| = y \):
\( \Rightarrow -y^2 + y = 0 \)
\( \Rightarrow y(-y + 1) = 0 \)
\( \Rightarrow y = 0 \) or \( y = 1 \)

If \( y < 0 \), then \( |y| = -y \):
\( \Rightarrow -y^2 - y = 0 \)
\( \Rightarrow -y(y + 1) = 0 \)
\( \Rightarrow y = 0 \) or \( y = -1 \)
\( \therefore x = 0, y = 0 \) or \( x = 0, y = -1 \) or \( x = 0, y = 1 \) ............ (ii)

Case 2: When \( y = 0 \)
Then \( x^2 - y^2 + \sqrt{x^2 + y^2} = 0 \)
\( \Rightarrow x^2 + \sqrt{x^2} = 0 \)
\( \Rightarrow x^2 + |x| = 0 \)

If \( x > 0 \), then \( |x| = x \):
\( \Rightarrow x^2 + x = 0 \)
\( \Rightarrow x(x + 1) = 0 \)
\( \Rightarrow x = 0 \) or \( x = -1 \) (rejected since \( x > 0 \))
\( \therefore x = 0, y = 0 \) ............ (iii)

If \( x < 0 \), then \( |x| = -x \):
\( \Rightarrow x^2 - x = 0 \)
\( \Rightarrow x(x - 1) = 0 \)
\( \Rightarrow x = 0 \) or \( x = 1 \) (rejected since \( x < 0 \))
\( \therefore x = 0, y = 0 \) ............ (iv)

From (ii), (iii) & (iv), the solutions are:
\( z = 0 + 0i \), \( z = 0 + i \), and \( z = 0 - i \) ans.

Question. If \( |z_1| = |z_2| = |z_3| = \dots = |z_n| = 1 \), show that \( |z_1 + z_2 + z_3 + \dots + z_n| = \left|\frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} + \dots + \frac{1}{z_n}\right| \)
Answer: We have, \( |z_1 + z_2 + z_3 + \dots + z_n| \)
\( = \left| \frac{z_1 \bar{z}_1}{\bar{z}_1} + \frac{z_2 \bar{z}_2}{\bar{z}_2} + \frac{z_3 \bar{z}_3}{\bar{z}_3} + \dots + \frac{z_n \bar{z}_n}{\bar{z}_n} \right| \)   [multiplying and dividing each term by its conjugate]
\( = \left| \frac{|z_1|^2}{\bar{z}_1} + \frac{|z_2|^2}{\bar{z}_2} + \frac{|z_3|^2}{\bar{z}_3} + \dots + \frac{|z_n|^2}{\bar{z}_n} \right| \)   \( [ \because z\bar{z} = |z|^2 ] \)
\( = \left| \frac{1}{\bar{z}_1} + \frac{1}{\bar{z}_2} + \frac{1}{\bar{z}_3} + \dots + \frac{1}{\bar{z}_n} \right| \)   \( [ \because |z_1| = |z_2| = \dots = |z_n| = 1 ] \)
\( = \left| \overline{ \left( \frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} + \dots + \frac{1}{z_n} \right) } \right| \)   \( [ \because \bar{z}_1 + \bar{z}_2 = \overline{z_1 + z_2} ] \)
\( = \left| \frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} + \dots + \frac{1}{z_n} \right| \)   \( [ \because |\bar{z}| = |z| ] \)

Question. If \( |z^2 - 1| = |z|^2 + 1 \), show that \( z \) lies on imaginary axis.
Answer: Let \( z = x + iy \)
Then, \( |z^2 - 1| = |z|^2 + 1 \)
\( \Rightarrow |(x + iy)^2 - 1| = \left(\sqrt{x^2 + y^2}\right)^2 + 1 \)
\( \Rightarrow |(x^2 - y^2 - 1) + 2ixy| = (x^2 + y^2) + 1 \)
\( \Rightarrow \sqrt{(x^2 - y^2 - 1)^2 + 4x^2y^2} = x^2 + y^2 + 1 \)

Squaring both sides:
\( \Rightarrow (x^2 - y^2 - 1)^2 + 4x^2y^2 = (x^2 + y^2 + 1)^2 \)
\( \Rightarrow x^4 + y^4 + 1 - 2x^2y^2 + 2y^2 - 2x^2 + 4x^2y^2 = x^4 + y^4 + 1 + 2x^2y^2 + 2y^2 + 2x^2 \)
\( \Rightarrow x^4 + y^4 + 1 + 2x^2y^2 + 2y^2 - 2x^2 = x^4 + y^4 + 1 + 2x^2y^2 + 2y^2 + 2x^2 \)
\( \Rightarrow -2x^2 = 2x^2 \)
\( \Rightarrow 4x^2 = 0 \)
\( \Rightarrow x^2 = 0 \Rightarrow x = 0 \)
\( \therefore z = 0 + iy \)

Clearly, \( z \) lies on the \( y \)-axis (imaginary axis).

Question. If the imaginary part of \( \frac{2z+1}{iz+1} \) is \( -2 \), then show that the locus of the point representing \( z \) in the argand plane is a straight line.
Answer: Let \( z = x + iy \)
Here, we have to prove that the equation containing \( x \) and \( y \) is linear (i.e., a straight line).
Now, \( \frac{2z+1}{iz+1} = \frac{2(x+iy)+1}{i(x+iy)+1} = \frac{(2x+1)+i2y}{(1-y)+ix} \)

Rationalizing the denominator:
\( = \frac{(2x + 1) + i2y}{(1 - y) + ix} \times \frac{(1 - y) - ix}{(1 - y) - ix} \)
\( = \frac{(2x + 1)(1 - y) - ix(2x + 1) + i2y(1 - y) - i^2 2xy}{(1 - y)^2 - i^2 x^2} \)
\( = \frac{2x - 2xy + 1 - y - i(2x^2 + x) + i(2y - 2y^2) + 2xy}{1 + y^2 - 2y + x^2} \)
\( = \frac{(2x + 1 - y) + i(2y - 2y^2 - 2x^2 - x)}{x^2 + y^2 - 2y + 1} \)
\( = \frac{2x + 1 - y}{x^2 + y^2 - 2y + 1} + i \frac{2y - 2y^2 - 2x^2 - x}{x^2 + y^2 - 2y + 1} \br />
Here, \( \operatorname{Im}\left(\frac{2z+1}{iz+1}\right) = \frac{2y - 2y^2 - 2x^2 - x}{x^2 + y^2 - 2y + 1} \)

Given, \( \operatorname{Im}\left(\frac{2z+1}{iz+1}\right) = -2 \):
\( \Rightarrow \frac{2y - 2y^2 - 2x^2 - x}{x^2 + y^2 - 2y + 1} = -2 \)
\( \Rightarrow 2y - 2y^2 - 2x^2 - x = -2x^2 - 2y^2 + 4y - 2 \)
\( \Rightarrow 2y - x = 4y - 2 \)
\( \Rightarrow x + 2y - 2 = 0 \)

Clearly, this represents the equation of a straight line.

Question. Let \( z_1 \) & \( z_2 \) be two complex numbers such that \( |z_1 + z_2| = |z_1| + |z_2| \), then show that \( \arg(z_1) - \arg(z_2) = 0 \).
Answer: Let \( z_1 = r_1(\cos \theta_1 + i \sin \theta_1) \) and \( z_2 = r_2(\cos \theta_2 + i \sin \theta_2) \).
Here, \( |z_1| = r_1 \), \( \arg(z_1) = \theta_1 \) and \( |z_2| = r_2 \), \( \arg(z_2) = \theta_2 \).

We have, \( |z_1 + z_2| = |z_1| + |z_2| \)
\( \Rightarrow |r_1(\cos \theta_1 + i \sin \theta_1) + r_2(\cos \theta_2 + i \sin \theta_2)| = r_1 + r_2 \)
\( \Rightarrow |(r_1 \cos \theta_1 + r_2 \cos \theta_2) + i(r_1 \sin \theta_1 + r_2 \sin \theta_2)| = r_1 + r_2 \)
\( \Rightarrow \sqrt{(r_1 \cos \theta_1 + r_2 \cos \theta_2)^2 + (r_1 \sin \theta_1 + r_2 \sin \theta_2)^2} = r_1 + r_2 \)

Squaring both sides:
\( \Rightarrow r_1^2 \cos^2 \theta_1 + r_2^2 \cos^2 \theta_2 + 2 r_1 r_2 \cos \theta_1 \cos \theta_2 + r_1^2 \sin^2 \theta_1 + r_2^2 \sin^2 \theta_2 + 2 r_1 r_2 \sin \theta_1 \sin \theta_2 = (r_1 + r_2)^2 \)
\( \Rightarrow r_1^2(\cos^2 \theta_1 + \sin^2 \theta_1) + r_2^2(\cos^2 \theta_2 + \sin^2 \theta_2) + 2 r_1 r_2(\cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2) = r_1^2 + r_2^2 + 2 r_1 r_2 \)
\( \Rightarrow r_1^2 + r_2^2 + 2 r_1 r_2 \cos(\theta_1 - \theta_2) = r_1^2 + r_2^2 + 2 r_1 r_2 \)
\( \Rightarrow \cos(\theta_1 - \theta_2) = 1 \)
\( \Rightarrow \theta_1 - \theta_2 = 0 \)
\( \Rightarrow \arg(z_1) - \arg(z_2) = 0 \)

Question. What does this equation \( |z + 1 - i| = |z - 1 + i| \) represents.
Answer: Let \( z = x + iy \)
Then, \( |z + 1 - i| = |z - 1 + i| \)
\( \Rightarrow |x + iy + 1 - i| = |x + iy - 1 + i| \)
\( \Rightarrow |(x + 1) + i(y - 1)| = |(x - 1) + i(y + 1)| \)
\( \Rightarrow \sqrt{(x + 1)^2 + (y - 1)^2} = \sqrt{(x - 1)^2 + (y + 1)^2} \)
\( \Rightarrow (x + 1)^2 + (y - 1)^2 = (x - 1)^2 + (y + 1)^2 \)
\( \Rightarrow x^2 + 1 + 2x + y^2 + 1 - 2y = x^2 + 1 - 2x + y^2 + 1 + 2y \)

Squaring and simplifying:
\( \Rightarrow x^2 + y^2 + 2x - 2y + 2 = x^2 + y^2 + 2y - 2x + 2 \)
\( \Rightarrow 4x - 4y = 0 \)
\( \Rightarrow x - y = 0 \)

Clearly, this represents the equation of a straight line.

Question. Evaluate: \( 1 + i^2 + i^4 + i^6 + \dots + i^{20} \)
Answer: We have, \( 1 + i^2 + i^4 + i^6 + \dots + i^{20} \)
\( = 1 - 1 + 1 - 1 + \dots + 1 \)
\( = 1 \) ans.

Question. Find the value of \( \frac{i^{4n+1} - i^{4n-1}}{2} \).
Answer: \( \Rightarrow \frac{i^{4n} \cdot i - i^{4n} \cdot i^{-1}}{2} \)
\( = \frac{1(i) - 1\left(\frac{1}{i}\right)}{2} \qquad [ \because i^{4n} = (i^4)^n = (1)^n = 1 ] \)
\( = \frac{i - (-i)}{2} \qquad \left[ \because \frac{1}{i} = \frac{1}{i} \times \frac{i}{i} = -i \right] \)
\( = \frac{2i}{2} = i \) ans.

Question. What is the smallest positive integer \( n \) for which \( (1 + i)^{2n} = (1 - i)^{2n} \).
Answer: We have, \( (1 + i)^{2n} = (1 - i)^{2n} \)
\( \Rightarrow \frac{(1 + i)^{2n}}{(1 - i)^{2n}} = 1 \)
\( \Rightarrow \left(\frac{1 + i}{1 - i}\right)^{2n} = 1 \)
Rationalizing the base:
\( \frac{1+i}{1-i} \times \frac{1+i}{1+i} = \frac{(1+i)^2}{1 - i^2} = \frac{1 + i^2 + 2i}{2} = \frac{2i}{2} = i \)
Thus, the equation becomes:
\( \Rightarrow (i)^{2n} = 1 \)
The smallest positive integer \( m \) for which \( i^m = 1 \) is \( 4 \).
Therefore:
\( 2n = 4 \Rightarrow n = 2 \).