Question

A variable complex number z=x+iy is such that the principal argument of (z-1)/(z+1) is pie/4. Show that x^2+y^2-2y-1=0.

$\boxed{\red{\bf\: moderators \: or \: brainly \: stars \: please \: answer \: this}}$


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Answer 1

Answer:

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Answer 2

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:z = x + iy$

Now, Consider

$\rm :\longmapsto\:\dfrac{z - 1}{z + 1}$

$\rm \: = \:\dfrac{x + iy - 1}{x + iy + 1}$

$\rm \: = \:\dfrac{x - 1+ iy}{(x + 1) + iy}$

On rationalizing the denominator, we get

$\rm \: = \:\dfrac{x - 1+ iy}{(x + 1) + iy} \times \dfrac{(x + 1) - iy}{(x + 1) - iy}$

$\rm \: = \:\dfrac{( {x}^{2} - 1) + iy(x + 1) - iy(x - 1) - {i}^{2} {y}^{2} }{ {(x + 1)}^{2} - {i}^{2} {y}^{2} }$

$\red{\bigg \{ \because \: (x + y)(x - y) = {x}^{2} - {y}^{2} \bigg \}}$

$\rm \: = \:\dfrac{ {x}^{2} - 1 + iy(x + 1 - x + 1) + {y}^{2} }{ {(x + 1)}^{2} + {y}^{2} }$

$\red{\bigg \{ \because \: {i}^{2} = - \: 1 \bigg \}}$

$\rm \: = \:\dfrac{ {x}^{2} + {y}^{2} - 1 + 2iy}{ {(x + 1)}^{2} + {y}^{2} }$

$\rm \: = \:\dfrac{ {x}^{2} + {y}^{2} - 1}{ {(x + 1)}^{2} + {y}^{2} } \: + i \: \dfrac{2y}{ {(x + 1)}^{2} + {y}^{2} }$

Thus,

$\rm :\longmapsto\:\dfrac{z - 1}{z + 1} = \:\dfrac{ {x}^{2} + {y}^{2} - 1}{ {(x + 1)}^{2} + {y}^{2} } \: + i \: \dfrac{2y}{ {(x + 1)}^{2} + {y}^{2} }$

We know,

$\boxed{ \bf{ \: If \: z = x + iy, \: then \: arg(z) = {tan}^{ - 1}\dfrac{y}{x}}}$

So,

$\rm :\longmapsto\:arg\bigg[\dfrac{z - 1}{z + 1}\bigg] = {tan}^{ - 1}\bigg[\dfrac{2y}{ {x}^{2} + {y}^{2} - 1} \bigg]$

But it is given that,

$\rm :\longmapsto\:arg\bigg[\dfrac{z - 1}{z + 1}\bigg] = \dfrac{\pi}{4}$

$\rm :\longmapsto\: {tan}^{ - 1}\bigg[\dfrac{2y}{ {x}^{2} + {y}^{2} - 1} \bigg] = \dfrac{\pi}{4}$

$\rm :\longmapsto\: \dfrac{2y}{ {x}^{2} + {y}^{2} - 1} =tan \dfrac{\pi}{4}$

$\rm :\longmapsto\: \dfrac{2y}{ {x}^{2} + {y}^{2} - 1} = 1$

$\rm :\longmapsto\: {x}^{2} + {y}^{2} - 1 = 2y$

$\rm :\longmapsto\: {x}^{2} + {y}^{2} - 2y - 1 = 0$

Hence, Proved