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 ___________.
Consider an A.P. $a_1,a_2,\ldots,a_n$; $a_1>0$. If $a_2-a_1=-\dfrac{3}{4}$, $a_n=\dfrac{1}{4}a_1$, and \[ \sum_{i=1}^{n} a_i=\frac{525}{2}, \] then $\sum_{i=1}^{17} a_i$ is equal to
