
If\[\arg (z - a) = \dfrac{\pi }{4}\], where a∈R, then the locus of z∈C is a [MP PET\[1997\]]
A) Hyperbola
B) Parabola
C) Ellipse
D) Straight line
Answer
161.4k+ views
Hint: in this question we have to find locus of the point \[z\] which satisfy the given condition. First write the given complex number as a combination of real and imaginary numbers. Then apply the formula for argument.
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
\[\arg (z) = {\tan ^{ - 1}}(\dfrac{y}{x})\]
Complete step by step solution:Given: Argument of complex number is given
Now we have argument which is equal to\[\arg (z - a) = \dfrac{\pi }{4}\]
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
Put this value in\[\arg (z - a) = \dfrac{\pi }{4}\]
\[\arg ((x + iy) - a) = \dfrac{\pi }{4}\]
\[\arg ((x - a) + iy) = \dfrac{\pi }{4}\]
We know that \[\arg (z) = {\tan ^{ - 1}}(\dfrac{y}{x})\]
\[{\tan ^{ - 1}}(\dfrac{y}{{x - a}}) = \dfrac{\pi }{4}\]
\[\dfrac{y}{{x - a}} = \tan (\dfrac{\pi }{4})\]
We know that \[\tan (\dfrac{\pi }{4}) = 1\]
\[\dfrac{y}{{x - a}} = 1\]
\[x - a = y\]this is equation of straight line
Here \[x - a = y\] represent the equation of straight line therefore locus of point represent straight line.
Option ‘D’ 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: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
\[\arg (z) = {\tan ^{ - 1}}(\dfrac{y}{x})\]
Complete step by step solution:Given: Argument of complex number is given
Now we have argument which is equal to\[\arg (z - a) = \dfrac{\pi }{4}\]
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
Put this value in\[\arg (z - a) = \dfrac{\pi }{4}\]
\[\arg ((x + iy) - a) = \dfrac{\pi }{4}\]
\[\arg ((x - a) + iy) = \dfrac{\pi }{4}\]
We know that \[\arg (z) = {\tan ^{ - 1}}(\dfrac{y}{x})\]
\[{\tan ^{ - 1}}(\dfrac{y}{{x - a}}) = \dfrac{\pi }{4}\]
\[\dfrac{y}{{x - a}} = \tan (\dfrac{\pi }{4})\]
We know that \[\tan (\dfrac{\pi }{4}) = 1\]
\[\dfrac{y}{{x - a}} = 1\]
\[x - a = y\]this is equation of straight line
Here \[x - a = y\] represent the equation of straight line therefore locus of point represent straight line.
Option ‘D’ 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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