The equation \(\arg\left(\frac{z - 1}{z + 1}\right) = \frac{\pi}{4}\) represents a circle with :
Show Hint
For \(\arg\left(\frac{z - z_1}{z - z_2}\right) = \alpha\), the points \(z_1\) and \(z_2\) are the ends of the chord. The centre of the circle lies on the perpendicular bisector of the segment joining \(z_1\) and \(z_2\).
Step 1: Understanding the Concept:
The locus of a complex number \(z\) such that \(\arg\left(\frac{z - z_1}{z - z_2}\right) = \alpha\) is an arc of a circle. If \(\alpha = \pi/2\), it is a semicircle; otherwise, it's a major or minor arc. Step 2: Key Formula or Approach:
Let \(z = x + iy\). Then evaluate the expression inside the argument:
\[ \frac{z - 1}{z + 1} = \frac{(x-1) + iy}{(x+1) + iy} \times \frac{(x+1) - iy}{(x+1) - iy} \]
Use \(\arg(w) = \tan^{-1}\left(\frac{\text{Im}(w)}{\text{Re}(w)}\right)\). Step 3: Detailed Explanation:
Expand the rationalized expression:
\[ \frac{(x^2 + x - x - 1 + y^2) + i(y(x+1) - y(x-1))}{(x+1)^2 + y^2} = \frac{(x^2 + y^2 - 1) + i(2y)}{(x+1)^2 + y^2} \]
The argument is \(\pi/4\), so:
\[ \tan\left(\frac{\pi}{4}\right) = \frac{\text{Im}}{\text{Re}} = \frac{2y}{x^2 + y^2 - 1} \]
Since \(\tan(\pi/4) = 1\):
\[ 1 = \frac{2y}{x^2 + y^2 - 1} \implies x^2 + y^2 - 1 = 2y \implies x^2 + y^2 - 2y - 1 = 0 \]
Completing the square for \(y\):
\[ x^2 + (y - 1)^2 - 1 - 1 = 0 \implies x^2 + (y - 1)^2 = 2 \]
This is a circle with centre \((0, 1)\) and radius \(\sqrt{2}\). Step 4: Final Answer:
The locus represents a circle with centre at (0, 1) and radius \(\sqrt{2}\).
