Question:
Let $ z=\frac{11-3i}{1+i}. $ If a is a real number such that $ z-i\alpha $ is real, then the value of $ \alpha $ is
Let $ z=\frac{11-3i}{1+i}. $ If a is a real number such that $ z-i\alpha $ is real, then the value of $ \alpha $ is
Updated On: May 22, 2024
- 4
- $ -\,4 $
- 7
- $ -\text{ }7 $
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The Correct Option is D
Solution and Explanation
$ \because $ $ z=\frac{11-3i}{1+i}\times \frac{1-i}{1-i} $
$=\frac{11-11i-3i-3}{1+1} $
$=\frac{8-14i}{2}=4-7i $
Also, $ \alpha $ is a real number such that $ z-i\alpha $ is real.
$ \therefore $ $ 4-7i-i\alpha $ is real, if $ \alpha =-7 $ .
$=\frac{11-11i-3i-3}{1+1} $
$=\frac{8-14i}{2}=4-7i $
Also, $ \alpha $ is a real number such that $ z-i\alpha $ is real.
$ \therefore $ $ 4-7i-i\alpha $ is real, if $ \alpha =-7 $ .
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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.
