Understanding the Locus of a Complex Number

The question asks for the locus of any complex number \(z\) that satisfies the condition \(\arg \left(\frac{z-1}{z+1}\right)=\frac{\pi}{3}\). This condition relates the complex number \(z\) to two fixed points in the complex plane, represented by \(1\) and \(-1\).

The argument of a complex number \(w\), denoted as \(\arg(w)\), represents the angle that the vector corresponding to \(w\) makes with the positive real axis in the complex plane. The expression \(\arg \left(\frac{z-1}{z+1}\right)\) can be interpreted geometrically or algebraically.

Geometric Interpretation of the Condition

Let \(A\) be the point representing the complex number \(-1\) and \(B\) be the point representing the complex number \(1\) in the complex plane. Let \(P\) be the point representing the complex number \(z\). The complex number \(z-1\) corresponds to the vector \(\vec{BP}\) (from \(1\) to \(z\)), and the complex number \(z+1\) corresponds to the vector \(\vec{AP}\) (from \(-1\) to \(z\)).

The argument of a quotient of two complex numbers is the difference of their arguments:

\[ \arg \left(\frac{z-1}{z+1}\right) = \arg(z-1) - \arg(z+1) \]This difference represents the angle turned from the vector corresponding to \(z+1\) to the vector corresponding to \(z-1\). In terms of the points \(A, B, P\), this is the angle \(\angle APB\). The condition \(\arg \left(\frac{z-1}{z+1}\right)=\frac{\pi}{3}\) means the angle \(\angle APB = \frac{\pi}{3}\).

The locus of a point \(P\) such that the angle subtended by a fixed line segment \(AB\) at \(P\) is a constant value \(\theta\) is an arc of a circle passing through \(A\) and \(B\). Since the angle is a constant \(\frac{\pi}{3}\), the locus is part of a circle passing through the points \(-1\) and \(1\).

Algebraic Approach to Find the Locus

Let \(z = x + iy\), where \(x\) and \(y\) are real numbers. Then \(z-1 = (x-1) + iy\) and \(z+1 = (x+1) + iy\).

The complex number \(\frac{z-1}{z+1}\) is:

\[ \frac{z-1}{z+1} = \frac{(x-1)+iy}{(x+1)+iy} \] To find the argument, we first express this complex number in the form \(X + iY\):

\[ \frac{(x-1)+iy}{(x+1)+iy} \times \frac{(x+1)-iy}{(x+1)-iy} = \frac{((x-1)+iy)((x+1)-iy)}{(x+1)^2 + y^2} \] \[ = \frac{(x-1)(x+1) - i(x-1)y + i(x+1)y - i^2y^2}{(x+1)^2 + y^2} \] \[ = \frac{x^2 - 1 + y^2 + iy(-(x-1) + (x+1))}{(x+1)^2 + y^2} \] \[ = \frac{x^2 + y^2 - 1 + i(2y)}{(x+1)^2 + y^2} \] \[ = \frac{x^2 + y^2 - 1}{(x+1)^2 + y^2} + i \frac{2y}{(x+1)^2 + y^2} \] The condition \(\arg \left(\frac{z-1}{z+1}\right)=\frac{\pi}{3}\) means the argument of this complex number is \(\frac{\pi}{3}\). For a complex number \(X+iY\), \(\arg(X+iY) = \theta\) implies \(\tan(\theta) = \frac{Y}{X}\), provided \(X > 0\). If \(X=0\), it depends on the sign of \(Y\). If \(X<0\), it depends on the quadrant.

Since the argument is \(\frac{\pi}{3}\), which is in the first quadrant, both the real part \(X\) and the imaginary part \(Y\) must be positive. \[ X = \frac{x^2 + y^2 - 1}{(x+1)^2 + y^2} \] \[ Y = \frac{2y}{(x+1)^2 + y^2} \] The condition \(X > 0\) implies \(x^2 + y^2 - 1 > 0\) (since the denominator \((x+1)^2 + y^2 > 0\) for \(z \neq -1\)). This means \(|z| > 1\). The condition \(Y > 0\) implies \(2y > 0\), which means \(y > 0\). So, the locus lies in the upper half-plane and outside the unit circle.

Now, using \(\tan\left(\frac{\pi}{3}\right) = \frac{Y}{X}\):

\[ \sqrt{3} = \frac{\frac{2y}{(x+1)^2 + y^2}}{\frac{x^2 + y^2 - 1}{(x+1)^2 + y^2}} \] Assuming \((x+1)^2 + y^2 \neq 0\) (i.e., \(z \neq -1\)), we can cancel the denominator:

\[ \sqrt{3} = \frac{2y}{x^2 + y^2 - 1} \] Rearranging the terms:

\[ \sqrt{3}(x^2 + y^2 - 1) = 2y \] \[ \sqrt{3}x^2 + \sqrt{3}y^2 - \sqrt{3} = 2y \] \[ \sqrt{3}x^2 + \sqrt{3}y^2 - 2y - \sqrt{3} = 0 \] Dividing by \(\sqrt{3}\):

\[ x^2 + y^2 - \frac{2}{\sqrt{3}}y - 1 = 0 \] This equation is of the form \(x^2 + y^2 + 2gx + 2fy + c = 0\), which is the standard equation of a circle. Comparing the coefficients, we have \(2g = 0\), \(2f = -\frac{2}{\sqrt{3}}\), and \(c = -1\). The center of this circle is \((0, \frac{1}{\sqrt{3}})\) and the radius is \(\sqrt{0^2 + (\frac{1}{\sqrt{3}})^2 - (-1)} = \sqrt{\frac{1}{3} + 1} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}\).

The derived equation \(x^2 + y^2 - \frac{2}{\sqrt{3}}y - 1 = 0\) represents a circle. While the argument condition with a specific value \(\frac{\pi}{3}\) geometrically corresponds to an arc, the algebraic manipulation directly yields the equation of the full circle on which this arc lies. Given the options, 'circle' is the most appropriate description for the locus of a complex number satisfying this equation.

Conclusion on the Locus

Both the geometric interpretation and the algebraic derivation point to the locus being related to a circle. The algebraic equation \(x^2 + y^2 - \frac{2}{\sqrt{3}}y - 1 = 0\) is the equation of a circle.

Therefore, the locus of any complex number \(z\) satisfying the given condition is a circle.

The options provided are:

• straight line
• circle
• parabola
• hyperbola

Based on our derivation, the locus of the complex number is a circle.