Find the Cartesian equation of the locus of $a r g (\frac{z - 4}{z + 4}) = \frac{π}{4}$ $arg((z-4)/(z+4))=pi/4$ ?

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Find the Cartesian equation of the locus of $a r g (\frac{z - 4}{z + 4}) = \frac{π}{4}$ $arg((z-4)/(z+4))=pi/4$ ?

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Answer 1 · 2017-11-29T13:01:23

This equation is circle with centre $(- 4, 0)$ $(-4, 0)$ and its radius $r = 4 \sqrt{2}$ $r=4sqrt2$

Explanation:

Set $z = x + y i$ $z=x+yi$ , so $a r g (\frac{z - 4}{z + 4}) = \frac{π}{4}$ $arg((z-4)/(z+4))=pi/4$

$arctan (\frac{z - 4}{z + 4}) = \frac{π}{4}$ $arctan((z-4)/(z+4))=pi/4$

$arctan (\frac{x + y i - 4}{x + y i + 4}) = \frac{π}{4}$ $arctan((x+yi-4)/(x+yi+4))=pi/4$

$arctan (\frac{y - 4}{x}) - arctan (\frac{y + 4}{x}) = \frac{π}{4}$ $arctan((y-4)/x)-arctan((y+4)/x)=pi/4$

After taking tangent both sides,

$\frac{\frac{y - 4}{x} - \frac{y + 4}{x}}{1 + \frac{y - 4}{x} \cdot \frac{y + 4}{x}} = tan (\frac{π}{4})$ $[(y-4)/x-(y+4)/x]/[1+(y-4)/x*(y+4)/x]=tan(pi/4)$

$\frac{- \frac{8}{x}}{\frac{x^{2} + y^{2} - 16}{x^{2}}} = 1$ $(-8/x)/[(x^2+y^2-16)/x^2]=1$

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

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

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

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

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

This equation describes a circle with its centre $M (- 4, 0)$ $M (-4,0)$ and its radius $r = 4 \sqrt{2}$ $r=4sqrt2$ .

Answer 2 · 2017-11-29T14:15:01

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

Explanation:

We know that

$z - 4 = r_{1} e_{1}^{i ϕ_{1}}$ $z-4 = r_1 e^(i phi_1)$ with $ϕ_{1} = arctan (\frac{y}{x - 4})$ $phi_1 = arctan(y/(x-4))$
$z + 4 = r_{2} e_{2}^{i ϕ_{2}}$ $z+4 = r_2 e^(i phi_2)$ with $ϕ_{2} = arctan (\frac{y}{x + 4})$ $phi_2 = arctan(y/(x+4))$

then

$arg (\frac{z - 4}{z + 4}) = arg (\frac{r_{1}}{r_{2}} e^{i (ϕ_{1} - ϕ_{2})}) = ϕ_{1} - ϕ_{2} = \frac{π}{4}$ $"arg"((z-4)/(z+4)) = "arg"(r_1/r_2e^(i(phi_1-phi_2))) = phi_1-phi_2 = pi/4$ or

$arctan (\frac{y}{x - 4}) - arctan (\frac{y}{x + 4}) = \frac{π}{4}$ $arctan(y/(x-4))-arctan(y/(x+4))=pi/4$

now

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

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

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

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

NOTE

From $tan (ϕ_{1} - ϕ_{2}) = \frac{{tan ϕ}_{1} - {tan ϕ}_{2}}{1 + {tan ϕ}_{1} {tan ϕ}_{2}}$ $tan(phi_1-phi_2) = (tanphi_1-tanphi_2)/(1+tanphi_1 tanphi_2)$

making

$ϕ_{1} = arctan (\frac{y}{x - 4})$ $phi_1 = arctan(y/(x-4))$ and $ϕ_{2} = arctan (\frac{y}{x + 4})$ $phi_2 = arctan(y/(x+4))$

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

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

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Find the Cartesian equation of the locus of $a r g (\frac{z - 4}{z + 4}) = \frac{π}{4}$ $arg((z-4)/(z+4))=pi/4$ ?

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Explanation:

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Find the Cartesian equation of the locus of arg((z-4)/(z+4))=pi/4arg(z−4z+4)=π4arg((z-4)/(z+4))=pi/4?

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Explanation:

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Find the Cartesian equation of the locus of $a r g (\frac{z - 4}{z + 4}) = \frac{π}{4}$ $arg((z-4)/(z+4))=pi/4$ ?