Find the mirror image of the curve argz-iz-1-4- i-1 in the real line x-y-0

arg(z+iz+1)=π4 arg(z+1z i)=π4 arg(z iz+1)=π4 arg(z+iz 1)=π4 ∵ The image of z in the line #160; ...

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Position of a Point W.R.T Parabola

Find the mirr...

Question

Find the mirror image of the curve $a r g (\frac{z + i}{z - 1}) = \frac{π}{4}, i = \sqrt{- 1}$ in the real line $x - y = 0$

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a r g (\frac{z - 3}{z - i}) = \frac{π}{6}, i = \sqrt{- 1}

z

C o l u m n - I

a r g (z)

c o l u m n - I I

$C o l u m n - I$ (equation in z)	$C o l u m n - I I$ (principal value of arg(z))
$z^{2} - z + 1 = 0$	$- 2 π / 3$
$z^{2} + z + 1 = 0$	$- π / 3$
${2 z}^{2} + 1 + i \sqrt{3} = 0$	$π / 3$
${2 z}^{2} + 1 - i \sqrt{3} = 0$	$2 π / 3$

a r g (z + i) = \frac{π}{4}

| z + 4 - 3 i | + | z - 4 + 3 i |

z = \frac{π}{4} (1 + i)^{4} (\frac{1 - π i}{π + i} + \frac{π - i}{1 + π i})

(\frac{| z |}{a r g (z)})

a r g (z - i + 2) = \frac{π}{6}

a r g (z + 4 - 3 i) = - \frac{π}{4}

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