# 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

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**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.

Last updated date: 28th Sep 2023

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