# Find the Cartesian equation of the locus of arg((z-4)/(z+4))=pi/4?

## Thanks!

Nov 29, 2017

This equation is circle with centre $\left(- 4 , 0\right)$ and its radius $r = 4 \sqrt{2}$

#### Explanation:

Set $z = x + y i$, so $a r g \left(\frac{z - 4}{z + 4}\right) = \frac{\pi}{4}$

$\arctan \left(\frac{z - 4}{z + 4}\right) = \frac{\pi}{4}$

$\arctan \left(\frac{x + y i - 4}{x + y i + 4}\right) = \frac{\pi}{4}$

$\arctan \left(\frac{y - 4}{x}\right) - \arctan \left(\frac{y + 4}{x}\right) = \frac{\pi}{4}$

After taking tangent both sides,

$\frac{\frac{y - 4}{x} - \frac{y + 4}{x}}{1 + \frac{y - 4}{x} \cdot \frac{y + 4}{x}} = \tan \left(\frac{\pi}{4}\right)$

$\frac{- \frac{8}{x}}{\frac{{x}^{2} + {y}^{2} - 16}{x} ^ 2} = 1$

$- \frac{8}{x} = \frac{{x}^{2} + {y}^{2} - 16}{x} ^ 2$

${x}^{2} + {y}^{2} - 16 = - 8 x$

${x}^{2} + 8 x + {y}^{2} - 16 = 0$

${x}^{2} + 8 x + 16 + {y}^{2} = 32$

${\left(x + 4\right)}^{2} + {y}^{2} = 32$

This equation describes a circle with its centre $M \left(- 4 , 0\right)$ and its radius $r = 4 \sqrt{2}$.

Nov 29, 2017

${x}^{2} + {\left(y - 4\right)}^{2} = 32$

#### Explanation:

We know that

$z - 4 = {r}_{1} {e}^{i {\phi}_{1}}$ with ${\phi}_{1} = \arctan \left(\frac{y}{x - 4}\right)$
$z + 4 = {r}_{2} {e}^{i {\phi}_{2}}$ with ${\phi}_{2} = \arctan \left(\frac{y}{x + 4}\right)$

then

$\text{arg"((z-4)/(z+4)) = "arg} \left({r}_{1} / {r}_{2} {e}^{i \left({\phi}_{1} - {\phi}_{2}\right)}\right) = {\phi}_{1} - {\phi}_{2} = \frac{\pi}{4}$ or

$\arctan \left(\frac{y}{x - 4}\right) - \arctan \left(\frac{y}{x + 4}\right) = \frac{\pi}{4}$

now

$\tan \left(\arctan \left(\frac{y}{x - 4}\right) - \arctan \left(\frac{y}{x + 4}\right)\right) = \tan \left(\frac{\pi}{4}\right) = 1$ or

$\frac{8 y}{{x}^{2} + {y}^{2} - 16} = 1$ or

${x}^{2} + {y}^{2} - 8 y - 16 = 0$ or

${x}^{2} + {\left(y - 4\right)}^{2} = 32$

NOTE

From $\tan \left({\phi}_{1} - {\phi}_{2}\right) = \frac{\tan {\phi}_{1} - \tan {\phi}_{2}}{1 + \tan {\phi}_{1} \tan {\phi}_{2}}$

making

${\phi}_{1} = \arctan \left(\frac{y}{x - 4}\right)$ and ${\phi}_{2} = \arctan \left(\frac{y}{x + 4}\right)$

and knowing that tan(arctan xi))= xi we have

$\frac{\left(\frac{y}{x - 4}\right) - \left(\frac{y}{x + 4}\right)}{1 + \left(\frac{y}{x - 4}\right) \left(\frac{y}{x + 4}\right)} = \frac{8 y}{{x}^{2} + {y}^{2} - 16}$