complex numbers

Question

if the complex number P(w) lies on the standard unit circle in an Argand plane and  

z=(aw+b)(w-c)^-1 then ,find the locus of z and interpret it.Given a,b,c are real.

Answer 1

Z = (aw + b)(w - c)^-1 

Let w = x + iy

|w| = 1

\(\Rightarrow\) x² + y² = 1

|z|² = \(\frac{|a\text{w} + b|^2}{|\text{w} - c|^2}\) = \(\frac{|ax + iay + b|^2}{|x + iy - c|^2}\)

\(=\frac{(ax + b)^2 + a^2 y^2}{(x - c)^2 + y^2}\)

\(=\frac{a^2x^2 + b^2 + 2abx + a^2y^2}{x^2 + c^2 - 2cx + y^2}\)

\(=\frac{a^2 + b^2 + 2abx}{1 + c^2 - 2cx}\)

\(z = \frac{a\text{w} + b}{\text{w} - c}\) = \(\frac{a(x + iy) + b}{x + iy - c}\)

\(=\frac{ax + b + iay}{(x - c) + iy}\) × \(\frac{(x - c) - iy}{(x - c) - iy}\)

\(=\frac{(ax + b)(x - c) + ay^2 + i(axy - acy - axy - by)}{(x - c)^2 + y^2}\)

\(=\frac{ax^2 + bx - acx - bc + ay^2 - i(acy + by)}{x^2 - 2cx + c^2 + y^2}\)

\(=\frac{a(x^2 + y^2) + bx - acx - bc + i(xy - cy - axy - by)}{1 + c^2 - 2cx}\)

∴ Re(z) \(=\frac{a + bx - acx - bc}{1 + c^2 - 2cx}\) (∵ x² + y² = 1)

Now, (1 - c²) |z|² - 2(a + bc) Re z = (1 - c²) \((1 - c^2)\left(\frac{a^2 + b^2 + 2abx}{1 + c^2 - 2cx}\right) \) \(- 2(a + bc) \left(\frac{a + bx - acx - bc}{1 + c^2 - 2cx}\right)\)

\(=\frac{a^2 + b^2 - a^2c^2 - b^2c^2 + 2abx - 2abc^2x - 2a^2 - 2abx + 2a^2cx + 2abc - 2abc - 2b^2cx + 2abc^2x + 2b^2c^2}{1 + c^2 - 2cx}\)

\(=\frac{-a^2 + b^2 - a^2c^2 + b^2c^2 + 2a^2cx - 2b^2cx}{1 + c^2 - 2cx}\)

\(=\frac{-a^2(1 + c^2 - 2cx) + b^2(1 + c^2 - 2cx)}{1 + c^2 - 2cx}\)

\(=\frac{(-a^2 + b^2)(1 + c^2 - 2cx)}{1 + c^2 - 2cx}\)

\(=b^2 - a^2\)

Hence, (1 - c²) |z|² - 2(a + bc) Re z \(=b^2 - a^2\)

∴ Locus of z is (1 - c²) |z|² - 2(a + bc) Re z + a² - b² = 0.