plz help plz help
if z=x+jy,show that the locus arg{(z-1)/(z)}=pi/6 is a circle.find its centre and radius
This should help you out:
http://math.stackexchange.com/questions/1154727/how-to-find-the-locus-of-points-in-the-argand-plane-represented-as-follows
To start, let's simplify the expression ${(z-1)/(z)}$ in terms of $x$ and $y$.
Given: $z = x + jy$
We have ${(z-1)/(z)} = \frac{(x + jy - 1)}{(x + jy)}$
Next, we multiply the numerator and denominator by the conjugate of the denominator to eliminate the complex numbers in the denominator. The conjugate of $x + jy$ is $x - jy$.
$\frac{(x + jy - 1)}{(x + jy)} \times \frac{(x - jy)}{(x - jy)}$
This simplifies to:
$\frac{(x^2 - jxy - x + jxy)}{(x^2 + y^2)}$
The imaginary terms cancel out:
$\frac{(x^2 - x)}{(x^2 + y^2)} = \frac{x(x - 1)}{(x^2 + y^2)}$
The numerator is $x(x - 1)$, and the denominator is $x^2 + y^2$.
Now, let's find the locus of $\text{arg}(z)$ such that $\text{arg}\left(\frac{(z-1)}{(z)}\right) = \frac{\pi}{6}$.
The argument of a complex number represents the angle it makes with the positive real axis. In other words, it is the angle between the complex number and the positive x-axis.
For our locus to be $\frac{\pi}{6}$, the angle between the complex number and the positive x-axis should be $\frac{\pi}{6}$.
Let's denote the angle $\theta = \text{arg}(z)$. We can write:
$\frac{x(x - 1)}{(x^2 + y^2)} = \tan(\theta)$
From here, we need to rewrite the equation to express it in terms of $x$ and $y$ only. To do this, we need to eliminate $\theta$.
Using the trigonometric identity:
$\tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$
We can rewrite the equation as:
$\frac{x(x - 1)}{(x^2 + y^2)} = \frac{1}{\sqrt{3}}$
This equation represents a locus in the $(x, y)$ plane.
To determine if it is a circle, let's simplify it further:
$x(x - 1) = \frac{(x^2 + y^2)}{\sqrt{3}}$
Simplifying further, we have:
$x^2 - x = \frac{(x^2 + y^2)}{\sqrt{3}}$
Rearranging terms:
$x^2 - x - \frac{x^2 + y^2}{\sqrt{3}} = 0$
Multiplying through by $\sqrt{3}$:
$\sqrt{3}x^2 - \sqrt{3}x - x^2 - y^2 = 0$
Combine like terms:
$(\sqrt{3} - 1)x^2 - y^2 - \sqrt{3}x = 0$
Now, we can see that the equation indeed represents a circle in the form:
$(x - h)^2 + (y - k)^2 = r^2$
Comparing it with our equation:
$(\sqrt{3} - 1)x^2 - y^2 - \sqrt{3}x = 0$
We can determine the center and radius of the circle:
The center of the circle is given by coordinates $(h, k)$. In this case, $h = -\frac{\sqrt{3}}{2}$ and $k = 0$. Therefore, the center of the circle is $(-\frac{\sqrt{3}}{2}, 0)$.
The radius of the circle, $r$, can be calculated using the equation of the circle:
$r^2 = \frac{(h^2 + k^2)}{(\sqrt{3} - 1)^2}$
In our case, $r^2 = \frac{0}{(\sqrt{3} - 1)^2} = 0$. Therefore, the radius of the circle is 0, which means it is actually a single point.
Thus, we have shown that the locus $\text{arg}\left(\frac{(z-1)}{(z)}\right) = \frac{\pi}{6}$ represents a degenerate circle with a radius of 0, centered at $(-\frac{\sqrt{3}}{2}, 0)$.