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 \( z_2 \).
∫ √(2x2 - 5x + 2) dx = ∫ (41/60) dx,
and
-1/2 > α > 0, then α = ?
In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ then $ \beta + \gamma - \alpha $ is equal to: