Question:
If $Z$ is a complex number with $I Z I = 1$ and $Z + \frac{1}{Z} = x + iy$, then $xy = $
If $Z$ is a complex number with $I Z I = 1$ and $Z + \frac{1}{Z} = x + iy$, then $xy = $
Updated On: May 11, 2024
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The Correct Option is A
Solution and Explanation
We have, $|Z| = 1$ and $Z + \frac{1}{Z} = x + iy$
Let $Z = P + iQ$
$\therefore \:\:\:\: \sqrt{P^2 + Q^2} = 1\:\:\: \Rightarrow \:\: P^2 + Q^2 = 1 $
Now $Z + \frac{1}{Z} = x + iy$
$ \therefore\:\:\:\: P +iQ + \frac{1}{P +iQ} = x + iy$
$ \Rightarrow P +iQ + \frac{P - iQ}{P^{2} +Q^{2}} =x +iy $
$\Rightarrow P+iQ + P - iQ = x + iy \left(\because\:\:\: P^{2} +Q^{2} = 1\right) $
$\Rightarrow 2P = x + iy $
Equating coefficients of real and imaginary parts, we get
$x = 2P$ and $ y = 0$
$\therefore \:\: xy = 0$
Let $Z = P + iQ$
$\therefore \:\:\:\: \sqrt{P^2 + Q^2} = 1\:\:\: \Rightarrow \:\: P^2 + Q^2 = 1 $
Now $Z + \frac{1}{Z} = x + iy$
$ \therefore\:\:\:\: P +iQ + \frac{1}{P +iQ} = x + iy$
$ \Rightarrow P +iQ + \frac{P - iQ}{P^{2} +Q^{2}} =x +iy $
$\Rightarrow P+iQ + P - iQ = x + iy \left(\because\:\:\: P^{2} +Q^{2} = 1\right) $
$\Rightarrow 2P = x + iy $
Equating coefficients of real and imaginary parts, we get
$x = 2P$ and $ y = 0$
$\therefore \:\: xy = 0$
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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.
