HOTS problems from NCERT book on complex numbers

Eleven Standard >> HOTS problems from NCERT book on complex numbers

Complex Numbers – Special Conditions and Properties

i) If \( z \) is a complex number such that \( |z| = 1 \), and \( \frac{z - 1}{z + 1} \) is purely imaginary, what would be the result if \( z = 1 \)?

Substitute \( z = 1 \) into the expression:
\[ \frac{z - 1}{z + 1} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0 \] Since 0 is a real number, not imaginary, the expression is not purely imaginary.

Conclusion: If \( z = 1 \), then \( \frac{z - 1}{z + 1} = 0 \), which is real. So the condition fails for \( z = 1 \).

ii) Find the least positive value of \( n \), if \( \left( \frac{1 + i}{1 - i} \right)^n = 1 \)

First, simplify the base expression:

\[ \frac{1 + i}{1 - i} = \frac{(1 + i)^2}{(1 - i)(1 + i)} = \frac{1 + 2i + i^2}{1^2 - i^2} = \frac{1 + 2i - 1}{1 + 1} = \frac{2i}{2} = i \]

So, the expression becomes: \[ i^n = 1 \]

Recall the powers of \( i \):

\( i^1 = i \)
\( i^2 = -1 \)
\( i^3 = -i \)
\( i^4 = 1 \)

Conclusion: The least positive value of \( n \) such that \( i^n = 1 \) is 4.

iii) If \( z = x + iy \) and \( \omega = \frac{1 - iz}{z - i} \), show that \( |\omega| = 1 \) implies that \( z \) is purely real.

Given:

\[ z = x + iy, \quad \omega = \frac{1 - iz}{z - i} \] Substitute \( z \) into \( \omega \):

\[ \omega = \frac{1 - i(x + iy)}{(x + iy) - i} = \frac{1 - ix - i^2y}{x + i(y - 1)} = \frac{1 - ix + y}{x + i(y - 1)} \]

Now compute the modulus:

\[ |\omega| = \frac{Modulus\, of\, numerator}{Modulus\, of\, denominator} = 1 \] That gives: \[ \frac{\sqrt{(1 + y)^2 + x^2}}{\sqrt{x^2 + (y - 1)^2}} = 1 \]

Squaring both sides:

\[ (1 + y)^2 + x^2 = x^2 + (y - 1)^2 \]

Cancel \( x^2 \):

\[ (1 + y)^2 = (y - 1)^2 \] \[ 1 + 2y + y^2 = y^2 - 2y + 1 \] \[ 4y = 0 \Rightarrow y = 0 \]

Conclusion: If \( |\omega| = 1 \), then \( y = 0 \Rightarrow z = x \), so \( z \) is purely real.