Question

If the amplitude of \[z - 2 - 3i\;\] is \[\dfrac{\pi }{4}\], then the locus of \[z = x + iy\] is [EAMCET\[2003\]]
A) \[x + y - 1 = 0\]
B) \[x - y - 1 = 0\]
C) \[x + y + 1 = 0\]
D) \[x - y + 1 = 0\]

Answer 1

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 number. Amplitude is same as argument. Then apply 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: Amplitude of complex number is given
Now we have amplitude which is equal to\[z - 2 - 3i\;\]
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\[z - 2 - 3i\;\]
\[x + iy - 2 - 3i\;\]
\[x + iy - 2 - 3i\; = (x - 2) + i(y - 3)\]
\[\arg (z) = {\tan ^{ - 1}}(\dfrac{y}{x})\]
It is given in the question that amplitude is equal to\[\dfrac{\pi }{4}\]
\[{\tan ^{ - 1}} = \dfrac{{y - 3}}{{x - 2}} = \dfrac{\pi }{4}\]
\[\tan \dfrac{\pi }{4} = \dfrac{{y - 3}}{{x - 2}}\]
We know that
\[\tan \dfrac{\pi }{4} = 1\]
\[\dfrac{{y - 3}}{{x - 2}} = 1\]
\[y - 3 = x - 2\]
Now locus is given by
\[x - y + 1 = 0\]



Option ‘A’ is correct



Note: Here we have to remember that amplitude is equal to argument. 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.