CBSE Class 11 Mathematics Complex Numbers And Quadratic Equation Worksheet Set 06

CBSE Class 11 Mathematics Worksheet - Complex Numbers and Quadratic Equation (5). 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. Where does \( z \) lies, if \( \left|\frac{z-5i}{z+5i}\right| = 1 \)?
Answer: Let \( z = x + iy \)

We have, \( \left|\frac{z-5i}{z+5i}\right| = 1 \)
\( \Rightarrow \frac{|z-5i|}{|z+5i|} = 1 \)
\( \Rightarrow |z - 5i| = |z + 5i| \)
\( \Rightarrow |x + iy - 5i| = |x + iy + 5i| \)
\( \Rightarrow |x + i(y - 5)| = |x + i(y + 5)| \)
\( \Rightarrow \sqrt{x^2 + (y - 5)^2} = \sqrt{x^2 + (y + 5)^2} \)

Squaring both sides:
\( \Rightarrow x^2 + (y - 5)^2 = x^2 + (y + 5)^2 \)
\( \Rightarrow x^2 + y^2 + 25 - 10y = x^2 + y^2 + 25 + 10y \)
\( \Rightarrow 20y = 0 \)
\( \Rightarrow y = 0 \)
\( \Rightarrow z = x \), which lies on the \( x \)-axis.

Question. Evaluate: \( 1 + i^2 + i^4 + i^6 + \dots + i^{2n} \)
Answer: \( = 1 - 1 + 1 - 1 + \dots + (-1)^n \)

This cannot be evaluated unless the value of \( n \) is known.

Question. If \( |z_1| = |z_2| \), is it necessary that \( z_1 = z_2 \)?
Answer: No. Example \( z_1 = 3 + 4i \) & \( z_2 = 4 + 3i \)
\( \Rightarrow |z_1| = \sqrt{9 + 16} = 5 \)
\( \Rightarrow |z_2| = \sqrt{16 + 9} = 5 \)

Clearly, \( |z_1| = |z_2| \) but \( z_1 \neq z_2 \).

Question. What is the polar form of complex number \( = (i^{25})^3 \)?
Answer: Let \( z = (i^{25})^3 \)
\( \Rightarrow z = (i^1)^3 \)   \( [ \because i^{25} = (i^4)^6 \cdot i = i ] \)
\( \Rightarrow z = -i \)
\( \Rightarrow z = 0 - 1i \)

Here, \( a = 0, b = -1 \)
\( \Rightarrow r = \sqrt{0^2 + (-1)^2} = 1 \)
\( \Rightarrow \tan \alpha = \left|\frac{-1}{0}\right| = \infty \Rightarrow \alpha = \frac{\pi}{2} \)

Since \( z \) lies on the negative imaginary axis, it corresponds to the \( 4^{\text{th}} \) quadrant boundary:
\( \Rightarrow \theta = -\alpha \Rightarrow \theta = \frac{-\pi}{2} \)

Polar form:
\( \Rightarrow z = r(\cos \theta + i \sin \theta) \)
\( \Rightarrow z = 1 \left(\cos \left(\frac{-\pi}{2}\right) + i \sin \left(\frac{-pi}{2}\right)\right) \)
\( \Rightarrow z = \cos \left(\frac{\pi}{2}\right) - i \sin \left(\frac{\pi}{2}\right) \) ans.

Question. Find the complex number satisfying the equation \( z + \sqrt{2}|z + 1| + i = 0 \)
Answer: We have, \( z + \sqrt{2}|z + 1| + i = 0 \)
Let \( z = x + i\gamma \)
\( \Rightarrow x + iy + \sqrt{2}|x + iy + 1| + i = 0 \)
\( \Rightarrow x + iy + \sqrt{2}|(x + 1) + iy| + i = 0 \)
\( \Rightarrow x + iy + \sqrt{2}\sqrt{(x + 1)^2 + y^2} + i = 0 \)
\( \Rightarrow x + \sqrt{2}\sqrt{(x + 1)^2 + y^2} + i(y + 1) = 0 + 0i \)

Equating real & imaginary parts, we get:
\( y + 1 = 0 \Rightarrow y = -1 \)
\( \Rightarrow x + \sqrt{2}\sqrt{(x + 1)^2 + y^2} = 0 \)

Substituting \( y = -1 \):
\( \Rightarrow x + \sqrt{2}\sqrt{x^2 + 1 + 2x + 1} = 0 \)
\( \Rightarrow x + \sqrt{2}\sqrt{x^2 + 2x + 2} = 0 \)
\( \Rightarrow \sqrt{2}\sqrt{x^2 + 2x + 2} = -x \)

Squaring both sides:
\( 2(x^2 + 2x + 2) = x^2 \)
\( \Rightarrow 2x^2 + 4x + 4 = x^2 \)
\( \Rightarrow x^2 + 4x + 4 = 0 \)
\( \Rightarrow (x + 2)^2 = 0 \)
\( \Rightarrow x = -2 \)
\( \therefore z = x + iy = -2 - i \) ans.

Question. If \( \arg(z - 1) = \arg(z + 3i) \) then find \( (x - 1) : y \) if \( z = x + iy \).
Answer: We have, \( z = x + iy \)
\( \Rightarrow z - 1 = (x - 1) + iy \)
And \( z + 3i = x + iy + 3i = x + i(y + 3) \)
\( \Rightarrow \arg(z - 1) = \tan \alpha = \frac{y}{x - 1} \)
\( \Rightarrow \arg(z + 3i) = \tan \alpha = \frac{y + 3}{x} \)

Given \( \arg(z - 1) = \arg(z + 3i) \):
\( \Rightarrow \frac{y}{x - 1} = \frac{y + 3}{x} \)
\( \Rightarrow xy = (x - 1)(y + 3) \)
\( \Rightarrow xy = xy + 3x - y - 3 \)
\( \Rightarrow y = 3(x - 1) \)
\( \Rightarrow \frac{x - 1}{y} = \frac{1}{3} \)
\( \Rightarrow (x - 1) : y = 1 : 3 \) ans.

Question. If \( a = \cos \theta + i \sin \theta \), find the value of \( \frac{1 + a}{1 - a} \).
Answer: We have, \( \frac{1 + a}{1 - a} = \frac{(1 + \cos \theta) + i \sin \theta}{(1 - \cos \theta) - i \sin \theta} \)
Using trigonometric identity \( 1 + \cos \theta = 2 \cos^2 \frac{\theta}{2} \), \( 1 - \cos \theta = 2 \sin^2 \frac{\theta}{2} \) and \( \sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \):
\( = \frac{2 \cos^2 \frac{\theta}{2} + i 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin^2 \frac{\theta}{2} - i 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}} \)
\( = \frac{2 \cos \left( \frac{\theta}{2} \right) \left[ \cos \frac{\theta}{2} + i \sin \frac{\theta}{2} \right]}{2 \sin \left( \frac{\theta}{2} \right) \left[ \sin \frac{\theta}{2} - i \cos \frac{\theta}{2} \right]} \)

Rationalizing the denominator:
\( = \frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}} \cdot \left[ \frac{\cos \frac{\theta}{2} + i \sin \frac{\theta}{2}}{-i \left[ \cos \frac{\theta}{2} + i \sin \frac{\theta}{2} \right]} \right] \)
\( = \cot \left( \frac{\theta}{2} \right) \cdot \frac{1}{-i} \)
\( = i \cot \left( \frac{\theta}{2} \right) \) ans.

Question. Evaluate: \( i^n + i^{n+1} + i^{n+2} + i^{n+3} \)
Answer: \( i^n + i^n \cdot i + i^n \cdot i^2 + i^n \cdot i^3 \)
\( = i^n(1 + i + i^2 + i^3) \)
\( = i^n(1 + i - 1 - i) \)
\( = i^n(0) \)
\( = 0 \) ans.