
If the imaginary part of \[\dfrac{{2z + 1}}{{iz + 1}}\]is \[ - 2\], then the locus of the point representing in the complex plane is [DCE \[2001\]]
A) A circle
B) A straight line
C) A parabola
D) None of these
Answer
162k+ views
Hint: in this question we have to find the what the locus of the point representing in the complex plane represent. First write the given complex number as a combination of real and imaginary number. Then equate the imaginary part of complex number with given value.
Formula Used:\[z = x + iy\]
Where
z is a complex number
x is a real part of complex number
iy is a imaginary part of complex number
i is iota
Square of iota is equal to the negative of one
Complete step by step solution:Given: A complex number and imaginary part of complex number is \[ - 2\]
Now we have complex number \[\dfrac{{2z + 1}}{{iz + 1}}\]
We know that
\[z = x + iy\]
Put this value in \[\dfrac{{2z + 1}}{{iz + 1}}\]
\[\dfrac{{2(x + iy) + 1}}{{i(x + iy) + 1}}\]
Now rearrange the above expression
\[\dfrac{{(2x + 1) + 2iy}}{{(1 - y) + ix}}\]
\[\dfrac{{(2x + 1) + 2iy}}{{(1 - y) + ix}} = \dfrac{{[(2x + 1)(1 - y) + 2xy] + i(2y(1 - y) - x(2x + 1)}}{{{{(1 - y)}^2} + {x^2}}}\]
Imaginary part of complex number is \[\dfrac{{i[2y(1 - y) - x(2x + 1)]}}{{{{(1 - y)}^2} + {x^2}}}\]
Imaginary part is \[\dfrac{{i[2y - 2{y^2} - 2{x^2} - x]}}{{1 + {y^2} - 2y + {x^2}}}\]=\[\dfrac{{i[2y - 2{y^2} - 2{x^2} - x]}}{{{x^2} + {y^2} - 2y + 1}}\]
In question value of imaginary part is given as \[ - 2\]
\[\dfrac{{[2y - 2{y^2} - 2{x^2} - x]}}{{{x^2} + {y^2} - 2y + 1}} = - 2\]
\[2y - 2{y^2} - 2{x^2} - x = - 2({x^2} + {y^2} - 2y + 1)\]
\[2y - 2{y^2} - 2{x^2} - x = - 2{x^2} - 2{y^2} + 4y - 2\]
\[x + 2y = 2\]
Here degree of variable is one and also it represent the equation of line therefore locus of point represent straight line.
Option ‘B’ is correct
Note: Complex number is a number which is a combination of real and imaginary number. So in combination number question we have to represent number as a combination of real and its imaginary part. Imaginary part is known as iota. Square of iota is equal to negative one.
Formula Used:\[z = x + iy\]
Where
z is a complex number
x is a real part of complex number
iy is a imaginary part of complex number
i is iota
Square of iota is equal to the negative of one
Complete step by step solution:Given: A complex number and imaginary part of complex number is \[ - 2\]
Now we have complex number \[\dfrac{{2z + 1}}{{iz + 1}}\]
We know that
\[z = x + iy\]
Put this value in \[\dfrac{{2z + 1}}{{iz + 1}}\]
\[\dfrac{{2(x + iy) + 1}}{{i(x + iy) + 1}}\]
Now rearrange the above expression
\[\dfrac{{(2x + 1) + 2iy}}{{(1 - y) + ix}}\]
\[\dfrac{{(2x + 1) + 2iy}}{{(1 - y) + ix}} = \dfrac{{[(2x + 1)(1 - y) + 2xy] + i(2y(1 - y) - x(2x + 1)}}{{{{(1 - y)}^2} + {x^2}}}\]
Imaginary part of complex number is \[\dfrac{{i[2y(1 - y) - x(2x + 1)]}}{{{{(1 - y)}^2} + {x^2}}}\]
Imaginary part is \[\dfrac{{i[2y - 2{y^2} - 2{x^2} - x]}}{{1 + {y^2} - 2y + {x^2}}}\]=\[\dfrac{{i[2y - 2{y^2} - 2{x^2} - x]}}{{{x^2} + {y^2} - 2y + 1}}\]
In question value of imaginary part is given as \[ - 2\]
\[\dfrac{{[2y - 2{y^2} - 2{x^2} - x]}}{{{x^2} + {y^2} - 2y + 1}} = - 2\]
\[2y - 2{y^2} - 2{x^2} - x = - 2({x^2} + {y^2} - 2y + 1)\]
\[2y - 2{y^2} - 2{x^2} - x = - 2{x^2} - 2{y^2} + 4y - 2\]
\[x + 2y = 2\]
Here degree of variable is one and also it represent the equation of line therefore locus of point represent straight line.
Option ‘B’ is correct
Note: Complex number is a number which is a combination of real and imaginary number. So in combination number question we have to represent number as a combination of real and its imaginary part. Imaginary part is known as iota. Square of iota is equal to negative one.
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