First slide

Geometry of complex numbers

Question

$If | z - 2 - i | = | z | \sin (\frac{π}{4} - \arg z), then locus of z is$

Moderate

A

an ellipse
B

a circle
C

a parabola
D

a pair of straight lines

Solution

$\begin{array}{l} Let z = x + i y \\ Polar form of a complex number z = r (\cos θ + i \sin θ) \\ Where r = | z | and θ = \arg (z) \\ \Rightarrow x + \dot{t} y = r (\cos θ + i \sin θ) \\ \Rightarrow x = r \cos θ, y = r \sin θ \\ Given | z - 2 - i | = | z | \sin (\frac{π}{4} - \arg z) \\ \Rightarrow | x + i y - 2 - i | = r \sin (\frac{π}{4} - θ) \\ \Rightarrow | (x - 2) + i (y - 1) | = r (\frac{1}{\sqrt{2}} \cos θ - \frac{1}{\sqrt{2}} \sin θ) \\ \Rightarrow \sqrt{(x - 2)^{2} + (y - 1)^{2}} = \frac{1}{\sqrt{2}} r \cos θ - \frac{1}{\sqrt{2}} r \sin θ \\ \Rightarrow \sqrt{(x - 2)^{2} + (y - 1)^{2}} = \frac{1}{\sqrt{2}} (x - y) \end{array}$
$∴ Locus of z is a parabola with focus (2, 1) and directrix x - y = 0$
$(∵ Locus of a moving point ' P^{'} in a plane which such that the distance$
$From the point ' P^{'} to fixed point is equal to distance from the point ' P^{'} to fixed line is parabola)$

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