Let \( z_1 = x_1 + i y_1 \) and \( z_2 = x_2 + i y_2 \), where \( x_1, x_2 \) are the real parts and \( y_1, y_2 \) are the imaginary parts of \( z_1 \) and \( z_2 \), respectively.
We are given that:
For \( z_1 z_2 = (x_1 + i y_1)(x_2 + i y_2) \), the real part is:
\( \text{Re}(z_1 z_2) = x_1 x_2 - y_1 y_2. \)
Thus, we have:
\( x_1 x_2 - y_1 y_2 = 0 \Rightarrow x_1 x_2 = y_1 y_2. \, \cdots (1) \)
For \( z_1 + z_2 = (x_1 + i y_1) + (x_2 + i y_2) \), the real part is:
\( \text{Re}(z_1 + z_2) = x_1 + x_2. \)
Thus, we have:
\( x_1 + x_2 = 0 \Rightarrow x_1 = -x_2. \, \cdots (2) \)
From equation (2), we know that \( x_1 = -x_2 \), meaning the real parts of \( z_1 \) and \( z_2 \) are opposite in sign.
From equation (1), we have \( x_1 x_2 = y_1 y_2 \), which means that the product of the real parts is equal to the product of the imaginary parts. For this to hold, \( y_1 \) and \( y_2 \) must have opposite signs, because the real parts are of opposite signs.
Thus, we conclude that \( y_1 \) and \( y_2 \) are of opposite signs, which means that the imaginary parts of \( z_1 \) and \( z_2 \) must satisfy the conditions:
The correct options are B and C, as they correspond to the valid cases for the imaginary parts of \( z_1 \) and \( z_2 \).
Let \( \alpha = \dfrac{-1 + i\sqrt{3}}{2} \) and \( \beta = \dfrac{-1 - i\sqrt{3}}{2} \), where \( i = \sqrt{-1} \). If
\[ (7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}, \] then the value of \( m \) is ___________.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,