Question:
If $ (3+i)\,(z+\bar{z})-(2+i)(z-\bar{z})+14i\,=0, $ then $ z\bar{z} $ is equal to
If $ (3+i)\,(z+\bar{z})-(2+i)(z-\bar{z})+14i\,=0, $ then $ z\bar{z} $ is equal to
Updated On: Jun 23, 2024
- $ 5 $
- $ 8 $
- $ 10 $
- $ 40 $
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The Correct Option is C
Solution and Explanation
Let $ z=x+iy, $ then $ \bar{z}=x-iy $
$ \therefore $ $ z+\bar{z}=2x $ and $ z-\bar{z}=2iy $
Given, $ (3+i)(z+\bar{z})-(2+i)(z-\bar{z})+14i=0 $
$ \Rightarrow $ $ (3+i)2x-(2+i)2iy+14i=0 $
$ \Rightarrow $ $ 6x+2ix-4yi+2y+14i=0+0i $
On comparing real and imaginary part, we get
$ 6x+2y=0 $ and $ 2x-4y+14=0 $
On solving, we get
$ x=-1,\,y=3 $
$ \therefore $ $ z\bar{z}=|z{{|}^{2}}={{(\sqrt{{{(-1)}^{2}}+{{(3)}^{2}}})}^{2}}=10 $
$ \therefore $ $ z+\bar{z}=2x $ and $ z-\bar{z}=2iy $
Given, $ (3+i)(z+\bar{z})-(2+i)(z-\bar{z})+14i=0 $
$ \Rightarrow $ $ (3+i)2x-(2+i)2iy+14i=0 $
$ \Rightarrow $ $ 6x+2ix-4yi+2y+14i=0+0i $
On comparing real and imaginary part, we get
$ 6x+2y=0 $ and $ 2x-4y+14=0 $
On solving, we get
$ x=-1,\,y=3 $
$ \therefore $ $ z\bar{z}=|z{{|}^{2}}={{(\sqrt{{{(-1)}^{2}}+{{(3)}^{2}}})}^{2}}=10 $
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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.
