For all complex numbers z of the form 1 + i$\alpha$, $\alpha$ $\epsilon$ R, if z = x + iy, then :
- $y^2 - 4x + 2 = 0$
- $y^2 + 4x - 4 = 0$
- $y^2 - 4x + 4 = 0$
- $y^2 + 4x + 2 = 0$
The Correct Option is B
Solution and Explanation
$1-\alpha^{2}+2 i \alpha= x + i y$
so $x=1-\alpha^{2}, y=2 \alpha$
putting $ \alpha= y / 2$
$x=1-\left(\frac{y}{2}\right)^{2} $
$\Rightarrow y^{2}+4 x-4=0$
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Concepts Used:
Complex Numbers and Quadratic Equations
Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.
Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.
