
If \[|8 + z\left| + \right|z - 8| = 16\;\]where z is a complex number, then the point z will lie on [J & K\[2005\]]
A) A Circle
B) An Ellipse
C) A straight line
D) None of these
Answer
164.1k+ views
Hint: in this question we have to find where given complex number lies which satisfy given condition. First write the given complex number as a combination of real and imaginary number. Put z in form of real and imaginary number into the equation.
Formula Used: Equation of complex number is given by
\[z = x + iy\]
Where
z is a complex number
x represent 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: Equation in the form of complex number
Now we have complex number equation\[|8 + z\left| + \right|z - 8| = 16\;\]
We know that complex number is written as a combination of real and imaginary number.
\[z = x + iy\]
Where
z is a complex number
x represent real part of complex number
iy is a imaginary part of complex number
i is iota
Put this value in\[|8 + z\left| + \right|z - 8| = 16\;\]
\[|8 + (x + iy)\left| + \right|(x + iy) - 8| = 16\;\]
\[|(x + 8) + iy\left| + \right|(x - 8) + iy| = 16\]
We know that
\[\left| z \right| = \sqrt {{x^2} + {y^2}} \]
\[\sqrt {{{(x + 8)}^2} + {y^2}} + \sqrt {{{(x - 8)}^2} + {y^2}} = 16\]
\[\sqrt {{{(x + 8)}^2} + {y^2}} = 16 - \sqrt {{{(x - 8)}^2} + {y^2}} \]
Squaring both side
\[{(\sqrt {{{(x + 8)}^2} + {y^2}} )^2} = {(16 - \sqrt {{{(x - 8)}^2} + {y^2}} )^2}\]
\[{x^2} + 64 + 16x + {y^2} = 256 + {x^2} - 16x + 64 + {y^2} - 32\sqrt {{{(x - 8)}^2} + {y^2}} \]
\[ - 256 + 32x = - 32\sqrt {{{(x - 8)}^2} + {y^2}} \]
\[8 - x = \sqrt {{{(x - 8)}^2} + {y^2}} \]
Square both side
\[{(8 - x)^2} = {(\sqrt {{{(x - 8)}^2} + {y^2}} )^2}\]
\[{(8 - x)^2} = {(x - 8)^2} + {y^2}\]
\[{y^2} = 0\]
\[y = 0\]
This equation represents equation of line.
So z lies on straight line.
Option ‘C’ is correct
Note: Complex number is a number which is a combination of real and imaginary number. So in complex number questions, 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: Equation of complex number is given by
\[z = x + iy\]
Where
z is a complex number
x represent 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: Equation in the form of complex number
Now we have complex number equation\[|8 + z\left| + \right|z - 8| = 16\;\]
We know that complex number is written as a combination of real and imaginary number.
\[z = x + iy\]
Where
z is a complex number
x represent real part of complex number
iy is a imaginary part of complex number
i is iota
Put this value in\[|8 + z\left| + \right|z - 8| = 16\;\]
\[|8 + (x + iy)\left| + \right|(x + iy) - 8| = 16\;\]
\[|(x + 8) + iy\left| + \right|(x - 8) + iy| = 16\]
We know that
\[\left| z \right| = \sqrt {{x^2} + {y^2}} \]
\[\sqrt {{{(x + 8)}^2} + {y^2}} + \sqrt {{{(x - 8)}^2} + {y^2}} = 16\]
\[\sqrt {{{(x + 8)}^2} + {y^2}} = 16 - \sqrt {{{(x - 8)}^2} + {y^2}} \]
Squaring both side
\[{(\sqrt {{{(x + 8)}^2} + {y^2}} )^2} = {(16 - \sqrt {{{(x - 8)}^2} + {y^2}} )^2}\]
\[{x^2} + 64 + 16x + {y^2} = 256 + {x^2} - 16x + 64 + {y^2} - 32\sqrt {{{(x - 8)}^2} + {y^2}} \]
\[ - 256 + 32x = - 32\sqrt {{{(x - 8)}^2} + {y^2}} \]
\[8 - x = \sqrt {{{(x - 8)}^2} + {y^2}} \]
Square both side
\[{(8 - x)^2} = {(\sqrt {{{(x - 8)}^2} + {y^2}} )^2}\]
\[{(8 - x)^2} = {(x - 8)^2} + {y^2}\]
\[{y^2} = 0\]
\[y = 0\]
This equation represents equation of line.
So z lies on straight line.
Option ‘C’ is correct
Note: Complex number is a number which is a combination of real and imaginary number. So in complex number questions, 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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