Find-the-locus-of-z-if-arg-z-2-z-3-pi-4-

Question Number 19623 by Tinkutara last updated on 13/Aug/17

Find the locus of z if \arg (\frac{z - 2}{z - 3}) = \frac{π}{4}

Answered by ajfour last updated on 13/Aug/17

let z = x + iy

\Rightarrow \arg [\frac{x - 2 + iy}{x - 3 + iy}] = \frac{π}{4}

\Rightarrow \arg [\frac{(x - 2 + iy) (x - 3 - iy)}{{(x - 3)}^{2} + y^{2}}] = \frac{π}{4}

\Rightarrow \arg [(x - 2) (x - 3) + y^{2} - iy] = \frac{π}{4}

or \frac{- y}{(x^{2} + y^{2} - 5 x + 6)} = \tan \frac{π}{4}

\Rightarrow x^{2} + y^{2} - 5 x + y + 6 = 0

\Rightarrow {(x - \frac{5}{2})}^{2} + {(y + \frac{1}{2})}^{2} + 6 - \frac{26}{4} = 0

\Rightarrow {(x - \frac{5}{2})}^{2} + {(y + \frac{1}{2})}^{2} = {(\frac{1}{\sqrt{2}})}^{2}

or ∣ z - \frac{5}{2} + \frac{i}{2} ∣= \frac{1}{\sqrt{2}}

Locus is a circle with radius r = \frac{1}{\sqrt{2}}

and centre z_{0} = \frac{5}{2} - \frac{i}{2} .

Commented by Tinkutara last updated on 13/Aug/17

But answer is x^{2} + y^{2} - 5 x \pm y + 6 = 0

Why \pm y ?

Commented by ajfour last updated on 13/Aug/17

we can take \arg (\frac{z - 2}{z - 3}) = \pm \frac{π}{4} .

Commented by Tinkutara last updated on 13/Aug/17

Thank you very much Sir!

Commented by ajfour last updated on 13/Aug/17

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