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प्रश्न
Show that the complex number z, satisfying the condition arg`((z - 1)/(z + 1)) = pi/4` lies on a circle.
उत्तर
Let z = x + iy
Given that: arg`((z - 1)/(z + 1)) = pi/4`
⇒ arg(z – 1) – arg(z + 1) = `pi/4` ......`[because "arg"(z_1) - "arg" (z_2) = "arg"z_1/z_2]`
⇒ arg[x + iy – 1] – arg[x + iy + 1] = `pi/4`
⇒ arg[(x – 1) + iy] – arg[(x + 1) + iy] = `pi/4`
⇒ `tan^-1 y/(x - 1) - tan^-1 y/(x + 1) = pi/4` ......`[because "arg" (x + yi) = tan^-1 y/x]`
⇒ `tan^-1 ((y/(x - 1) - y/(x + 1))/(1 + y/(x - 1) xx y/(x + 1))) = pi/4`
⇒ `(xy + y - xy + y)/(x^2 - 1 + y^2) = tan pi/4`
⇒ `(2y)/(x^2 + y^2 - 1)` = 1
⇒ x2 + y2 – 1 = 2y
⇒ x2 + y2 – 2y – 1 = 0
Which is a circle.
Hence, z lies on a circle.
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